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A non-geodesic analogue of Reshetnyak’s majorization theorem

A non-geodesic analogue of Reshetnyak’s majorization theorem 1IntroductionTo find a characterization of those metric spaces that admit an isometric embedding into a CAT(κ){\rm{CAT}}\left(\kappa )space is a longstanding open problem posed by Gromov (see [3, Section 1.4], [9, Section 1.19+] and [10, §15]). On the other hand, various conditions on a general metric space are known to become equivalent to being CAT(κ){\rm{CAT}}\left(\kappa )under the assumption that the metric space is geodesic. The Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition defined by Gromov [10] is one of such conditions (see [10, §15]). In this article, we prove an analogue of Reshetnyak’s majorization theorem [14] for (possibly non-geodesic) metric spaces that satisfy the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition. Our result shows that metric spaces with the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition have more properties in common with CAT(κ){\rm{CAT}}\left(\kappa )spaces than expected. In particular, it follows from our result that every metric space with the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition satisfies the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions for all integers n≥5n\ge 5, although Gromov stated that this is apparently not true (see Subsection 1.1).For a real number κ\kappa , we denote by Mκ2{M}_{\kappa }^{2}the complete, simply connected, two-dimensional Riemannian manifold of constant Gaussian curvature κ\kappa , and by dκ{d}_{\kappa }the distance function on Mκ2{M}_{\kappa }^{2}. Let Dκ{D}_{\kappa }be the diameter of Mκ2{M}_{\kappa }^{2}. Thus, we have Dκ=πκifκ>0,∞ifκ≤0.{D}_{\kappa }=\left\{\begin{array}{ll}\frac{\pi }{\sqrt{\kappa }}\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \gt 0,\\ \infty \hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \le 0.\end{array}\right.For a positive integer nnand an integer mm, we denote by [m]n{\left[m]}_{n}the element of Z/nZ{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}represented by mm. We recall the definition of the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions introduced in [10].Definition 1.1Fix κ∈R\kappa \in {\mathbb{R}}and an integer n≥4n\ge 4. Let (X,dX)\left(X,{d}_{X})be a metric space. We say that XXis a Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )space or that XXsatisfies the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )condition if for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}\to Xwith (1.1)∑i∈Z/nZdX(f(i),f(i+[1]n))<2Dκ,\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },there exists a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}such that dκ(g(i),g(i+[1]n))≤dX(f(i),f(i+[1]n)),dκ(g(i),g(j))≥dX(f(i),f(j)){d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n}))\le {d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa }\left(g\left(i),g\left(j))\ge {d}_{X}(f\left(i),f\left(j))for any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with j≠i+[1]nj\ne i+{\left[1]}_{n}and i≠j+[1]ni\ne j+{\left[1]}_{n}.Remark 1.2The original definition of Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )spaces in [10, §7] requires the existence of a map g′:Z/nZ→Mκ′2g^{\prime} :{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa ^{\prime} }^{2}for some κ′≤κ\kappa ^{\prime} \le \kappa such that dκ′(g′(i),g′(i+[1]n))≤dX(f(i),f(i+[1]n)),dκ′(g′(i),g′(j))≥dX(f(i),f(j)){d}_{\kappa ^{\prime} }\left(g^{\prime} \left(i),g^{\prime} \left(i+{\left[1]}_{n}))\le {d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa ^{\prime} }\left(g^{\prime} \left(i),g^{\prime} \left(j))\ge {d}_{X}(f\left(i),f\left(j))for any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with j≠i+[1]nj\ne i+{\left[1]}_{n}and i≠j+[1]ni\ne j+{\left[1]}_{n}, instead of the existence of a map ggas in Definition 1.1. As mentioned in [10], this definition is equivalent to Definition 1.1. In fact, the existence of such a map g′g^{\prime} implies the existence of a map ggas in Definition 1.1 by Reshetnyak’s majorization theorem.Remark 1.3In [10, §7], assumption (1.1) is not stated explicitly. It is just remarked that we have to consider only maps f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xwith “small” images f(Z/nZ)f\left({\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}})when κ>0\kappa \gt 0.For the definition of CAT(κ){\rm{CAT}}\left(\kappa )spaces, see Definition 2.8 in this article. Concerning Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )spaces and CAT(κ){\rm{CAT}}\left(\kappa )spaces, Gromov [10] established the following fact.Theorem 1.4(Gromov [10]) Fix κ∈R\kappa \in {\mathbb{R}}. The following two assertions hold true. (1)A metric space (X,dX)\left(X,{d}_{X})is CAT(κ){\rm{CAT}}\left(\kappa )if and only if XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )and Dκ{D}_{\kappa }-geodesic. Here, we say X is Dκ{D}_{\kappa }-geodesic if any x,y∈Xx,y\in Xwith dX(x,y)<Dκ{d}_{X}\left(x,y)\lt {D}_{\kappa }can be joined by a geodesic segment in X.(2)Every CAT(κ){\rm{CAT}}\left(\kappa )space is Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )for all integers n≥4n\ge 4.On the other hand, the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition generally does not imply the isometric embeddability into a CAT(κ){\rm{CAT}}\left(\kappa )space without assuming that the metric space is Dκ{D}_{\kappa }-geodesic. In fact, Nina Lebedava constructed a 6-point Cycl4(0){{\rm{Cycl}}}_{4}\left(0)space that does not admit an isometric embedding into any CAT(0){\rm{CAT}}\left(0)space (see [1, §7.2]). Moreover, it follows from the result of Eskenazis, Mendel, and Naor [8] that there exists a Cycl4(0){{\rm{Cycl}}}_{4}\left(0)space that does not admit a coarse embedding into any CAT(0){\rm{CAT}}\left(0)space (see [17, p. 116]).The following theorem is our main result, which can be viewed as an analogue of Reshetnyak’s majorization theorem for Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )spaces.Theorem 1.5Let κ∈R\kappa \in {\mathbb{R}}. If X is a Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space, then for any integer n≥3n\ge 3, and for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xthat satisfies∑i∈Z/nZdX(f(i),f(i+[1]n))<2Dκ,f(j)≠f(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}f\left(j)\ne f\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, there exists a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}that satisfies the following two conditions: (1)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, we havedκ(g(i),g(i+[1]n))=dX(f(i),f(i+[1]n)),dκ(g(i),g(j))≥dX(f(i),f(j)).{d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa }\left(g\left(i),g\left(j))\ge {d}_{X}(f\left(i),f\left(j)).(2)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with i≠ji\ne j, we have [g(i),g(j)]∩[g(i−[1]n),g(i+[1]n)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n}),g\left(i+{\left[1]}_{n})]\ne \varnothing , where we denote by [a,b]\left[a,b]the line segment in Mκ2{M}_{\kappa }^{2}with endpoints a and b.Note that when the polygon with vertices g(i)g\left(i), i∈Z/nZi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}is non-degenerate, the condition (2) in Theorem 1.5 means that this polygon is convex.1.1Gromov’s remark about the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditionsTheorem 1.4 tells us that the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition implies the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions for all integers n≥5n\ge 5under the assumption that the metric space is Dκ{D}_{\kappa }-geodesic. In the study of upper curvature bound for general metric spaces, it is natural to ask whether this implication is true without assuming that the metric space is Dκ{D}_{\kappa }-geodesic. Concerning this question, Gromov [10, §15, Remarks (b)] stated “We shall see later on that Cycl4⇒Cyclk{{\rm{Cycl}}}_{4}\Rightarrow {{\rm{Cycl}}}_{k}for all k≥5k\ge 5in the geodesic case but this is apparently not so in general.” However, we prove as a direct consequence of Theorem 1.5 that this implication actually holds true without assuming that the metric space is Dκ{D}_{\kappa }-geodesic:Theorem 1.6Let κ∈R\kappa \in {\mathbb{R}}. Every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )for all integers n≥5n\ge 5.1.2Gromov’s question about the Wirtinger inequalitiesIn [10, §6], Gromov also introduced the following conditions on a metric space.Definition 1.7Fix an integer n≥4n\ge 4. We say that a metric space (X,dX)\left(X,{d}_{X})is a Wirn{{\rm{Wir}}}_{n}space if any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xsatisfies (1.2)0≤sin2jπn∑i∈Z/nZdX(f(i),f(i+[1]n))2−sin2πn∑i∈Z/nZdX(f(i),f(i+[j]n))20\le {\sin }^{2}\frac{j\pi }{n}\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}{(f\left(i),f\left(i+{\left[1]}_{n}))}^{2}-{\sin }^{2}\frac{\pi }{n}\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}{(f\left(i),f\left(i+{\left[j]}_{n}))}^{2}for every j∈Z∩[2,n−2]j\in {\mathbb{Z}}\cap \left[2,n-2].The family of inequalities (1.2) can be thought of as a discrete and nonlinear analogue of classical Wirtinger’s inequality for functions on S1{S}^{1}. Every Euclidean space is Wirn{{\rm{Wir}}}_{n}for all integers n≥4n\ge 4, which was first proved by Pech [13] before Gromov introduced the notion of Wirn{{\rm{Wir}}}_{n}spaces. Therefore, it follows from the definition of Cycln(0){{\rm{Cycl}}}_{n}\left(0)spaces that every Cycln(0){{\rm{Cycl}}}_{n}\left(0)space is Wirn{{\rm{Wir}}}_{n}for each integer n≥4n\ge 4. Thus, for general metric spaces, the following implications are true for each integer n≥4n\ge 4: CAT(0)⇒Cycln(0)⇒Wirn.{\rm{CAT}}\left(0)\Rightarrow {{\rm{Cycl}}}_{n}\left(0)\Rightarrow {{\rm{Wir}}}_{n}.In [10, p. 133, §25, Question], Gromov posed the question of whether the implication Cycl4(0)⇒Wirn{{\rm{Cycl}}}_{4}\left(0)\Rightarrow {{\rm{Wir}}}_{n}holds true for every integer n≥5n\ge 5without assuming that the metric space is geodesic. Kondo, Toyoda, and Uehara [11] answered this question affirmatively:Theorem 1.8[11] Every Cycl4(0){{\rm{Cycl}}}_{4}\left(0)space is Wirn{{\rm{Wir}}}_{n}for all integers n≥4n\ge 4.Since Theorem 1.6 implies the stronger implication Cycl4(0)⇒Cycln(0){{\rm{Cycl}}}_{4}\left(0)\Rightarrow {{\rm{Cycl}}}_{n}\left(0)for every integer n≥4n\ge 4, Theorem 1.6 strengthens Theorem 1.8 and gives another proof of it.1.3The ⊠\boxtimes -inequalitiesIt was remarked in [10, §7] that the Cycl4(0){{\rm{Cycl}}}_{4}\left(0)condition is equivalent to the validity of a certain family of inequalities, defined as follows.Definition 1.9We say that a metric space (X,dX)\left(X,{d}_{X})satisfies the ⊠\boxtimes -inequalities if for any t,s∈[0,1]t,s\in \left[0,1]and any x,y,z,w∈Xx,y,z,w\in X, we have0≤(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2.0\le \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}.Gromov [10] and Sturm [16] proved independently that every CAT(0){\rm{CAT}}\left(0)space satisfies the ⊠\boxtimes -inequalities. The name “⊠\boxtimes -inequalities” is based on a notation used by Gromov [10] and was used in [11] and [17]. Sturm [16] called these inequalities the weighted quadruple inequalities. Gromov stated the following fact in [10, §7].Theorem 1.10[10] A metric space is Cycl4(0){{\rm{Cycl}}}_{4}\left(0)if and only if it satisfies the ⊠\boxtimes -inequalities.For a proof of Theorem 1.10, see Section 6. The following two corollaries follow from Theorems 1.5 and 1.6 immediately.Corollary 1.11If a metric space X satisfies the ⊠\boxtimes -inequalities, then for any integer n≥3n\ge 3, and for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xthat satisfies f(j)≠f(j+[1]n)f\left(j)\ne f\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, there exists a map g:Z/nZ→R2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {{\mathbb{R}}}^{2}that satisfies the following two conditions: (1)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, we have‖g(i)−g(i+[1]n)‖=dX(f(i),f(i+[1]n)),‖g(i)−g(j)‖≥dX(f(i),f(j)).\Vert g\left(i)-g\left(i+{\left[1]}_{n})\Vert ={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}\Vert g\left(i)-g\left(j)\Vert \ge {d}_{X}(f\left(i),f\left(j)).(2)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with i≠ji\ne j, we have [g(i),g(j)]∩[g(i−[1]n),g(i+[1]n)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n}),g\left(i+{\left[1]}_{n})]\ne \varnothing , where we denote by [a,b]\left[a,b]the line segment in R2{{\mathbb{R}}}^{2}with endpoints a and b.Corollary 1.12If a metric space satisfies the ⊠\boxtimes -inequalities, then it is Cycln(0){{\rm{Cycl}}}_{n}\left(0)for every integer n≥4n\ge 4.1.4Graph comparisonWe can say that the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )condition is defined by comparing embeddings of the cycle graph with nnvertices into a given metric space with embeddings of the same graph into Mκ2{M}_{\kappa }^{2}(see Definition 1.4 in [17]). By replacing the cycle graph with another graph, and Mκ2{M}_{\kappa }^{2}with another space, we can define a large number of new conditions. In recent years, such graph comparison conditions have been used by Lebedeva, Petrunin, and Zolotov [12] and the present author [17], among others, and play an interesting role in the study of the geometry of metric spaces.1.5Outline of the proof of Theorem 1.5Our proof of Theorem 1.5 is based on the idea used by Ballmann in his lecture note [4] for proving Reshetnyak’s majorization theorem. We also recommend [2, Chapter 9, L] for the proof of Reshetnyak’s majorization theorem by using Ballmann’s idea. To use Ballmann’s idea in our setting, we need the following lemma, which we prove in Section 5.Lemma 1.13Let κ∈R\kappa \in {\mathbb{R}}, and let (X,dX)\left(X,{d}_{X})be a Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space. Suppose x,y,z,w∈Xx,y,z,w\in Xand x′,y′,z′,w′∈Mκ2x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x′,y′)+dκ(y′,z′)+dκ(z′,w′)+dκ(w′,x′)<2Dκ,dX(x,y)≤dκ(x′,y′),dX(y,z)≤dκ(y′,z′),dX(z,w)≤dκ(z′,w′),dX(w,x)≤dκ(w′,x′),dκ(x′,z′)≤dX(x,z).\begin{array}{l}{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )+{d}_{\kappa }\left(w^{\prime} ,x^{\prime} )\lt 2{D}_{\kappa },\\ {d}_{X}\left(x,y)\le {d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{X}(y,z)\le {d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{X}\left(z,w)\le {d}_{\kappa }\left(z^{\prime} ,w^{\prime} ),\\ {d}_{X}\left(w,x)\le {d}_{\kappa }\left(w^{\prime} ,x^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z).\end{array}Then we have dX(y,w)≤dκ(y′,p)+dκ(p,w′){d}_{X}(y,w)\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} )for every p∈[x′,z′]p\in \left[x^{\prime} ,z^{\prime} ].Once we have proved Lemma 1.13, we can prove Theorem 1.5 by just following the idea of Ballmann mentioned above. We show this in Section 3.Remark 1.14For the use of Lemma 1.13 to prove Theorem 1.5 by following Ballmann’s idea, it is no problem to replace the assumption dκ(x′,z′)≤dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z)in Lemma 1.13 with the equality dκ(x′,z′)=dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )={d}_{X}\left(x,z). However, proving the dκ(x′,z′)=dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )={d}_{X}\left(x,z)version of Lemma 1.13 still requires a similar argument as we do in this article for proving Lemma 1.13.For κ≤0\kappa \le 0, Lemma 1.13 can be proved by a straightforward computation (see Remark 2.7). To prove Lemma 1.13 for a general real number κ\kappa , we prove two lemmas in Section 4, which generalize Alexandrov’s lemma [6, p. 25]. By using those lemmas, we prove Lemma 1.13 in Section 5.Remark 1.15It was pointed out by an anonymous referee that the dκ(x′,z′)=dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )={d}_{X}\left(x,z)version of Lemma 1.13 can also be proved by using Zalgaller’s arm lemma [18] instead of the generalized Alexandrov’s lemma.1.6Organization of the paperThis article is organized as follows. In Section 2, we recall some definitions and results from metric geometry and the geometry of Mκ2{M}_{\kappa }^{2}. In Section 3, we prove Theorems 1.5 and 1.6 by using Lemma 1.13. In Section 4, we prove two lemmas, which generalize Alexandrov’s lemma [6, p. 25]. In Section 5, we prove Lemma 1.13 by using those lemmas proved in Section 4. In Section 6, we present a proof of Theorem 1.10 for completeness.2PreliminariesIn this section, we recall some definitions and results from metric geometry and the geometry of Mκ2{M}_{\kappa }^{2}.2.1GeodesicsLet (X,dX)\left(X,{d}_{X})be a metric space. A geodesic in XXis an isometric embedding of an interval of the real line into XX. For x,y∈Xx,y\in X, a geodesic segment with endpoints x and y is the image of a geodesic γ:[0,dX(x,y)]→X\gamma :\left[0,{d}_{X}\left(x,y)]\to Xwith γ(0)=x\gamma \left(0)=x, γ(dX(x,y))=y\gamma \left({d}_{X}\left(x,y))=y. If there exists a unique geodesic segment with endpoints xxand yy, we denote it by [x,y]\left[x,y]. We also denote the sets [x,y]⧹{x,y}\left[x,y]\setminus \left\{x,y\right\}, [x,y]⧹{x}\left[x,y]\setminus \left\{x\right\}and [x,y]⧹{y}\left[x,y]\setminus \{y\}by (x,y)\left(x,y), (x,y]\left(x,y]and [x,y)\left[x,y), respectively. A metric space XXis called geodesic if for any x,y∈Xx,y\in X, there exists a geodesic segment with endpoints xxand yy. Let D∈(0,∞)D\in \left(0,\infty ). We say that XXis D-geodesic if for any x,y∈Xx,y\in Xwith dX(x,y)<D{d}_{X}\left(x,y)\lt D, there exists a geodesic segment with endpoints xxand yy. A subset SSof XXis called convex if for any x,y∈Sx,y\in S, every geodesic segment in XXwith endpoints xxand yyis contained in SS. If this condition holds for any x,y∈Sx,y\in Swith dX(x,y)<D{d}_{X}\left(x,y)\lt D, then SSis called D-convex.2.2The geometry of Mκ2{M}_{\kappa }^{2}Let κ∈R\kappa \in {\mathbb{R}}. For any x,y∈Mκ2x,y\in {M}_{\kappa }^{2}with dκ(x,y)<Dκ{d}_{\kappa }\left(x,y)\lt {D}_{\kappa }, there exists a unique geodesic segment [x,y]\left[x,y]with endpoints xxand yy. We mean by a line in Mκ2{M}_{\kappa }^{2}the image of an isometric embedding of R{\mathbb{R}}into Mκ2{M}_{\kappa }^{2}if κ≤0\kappa \le 0, and a great circle in Mκ2{M}_{\kappa }^{2}if κ>0\kappa \gt 0. Then for any two distinct points x,y∈Mκ2x,y\in {M}_{\kappa }^{2}with dκ(x,y)<Dκ{d}_{\kappa }\left(x,y)\lt {D}_{\kappa }, there exits a unique line through xxand yy, which we denote by ℓ(x,y)\ell \left(x,y). For any line ℓ\ell in Mκ2{M}_{\kappa }^{2}, Mκ2⧹ℓ{M}_{\kappa }^{2}\setminus \ell consists of exactly two connected components. We call each connected component of Mκ2⧹ℓ{M}_{\kappa }^{2}\setminus \ell a side of ℓ\ell . One side of ℓ\ell is called the opposite side of the other. If x,y∈Mκ2x,y\in {M}_{\kappa }^{2}lie on the same side of a line, then dκ(x,y)<Dκ{d}_{\kappa }\left(x,y)\lt {D}_{\kappa }. Each side of a line is a convex subset of Mκ2{M}_{\kappa }^{2}. For a subset SSof a side of some line in Mκ2{M}_{\kappa }^{2}, the convex hull of S is the intersection of all convex subsets of Mκ2{M}_{\kappa }^{2}containing SS, or equivalently, the minimal convex subset of Mκ2{M}_{\kappa }^{2}containing SS. We denote the convex hull of SSby conv(S){\rm{conv}}\left(S). We recall the following well-known fact, which is trivial when κ≤0\kappa \le 0.Proposition 2.1Let κ∈R\kappa \in {\mathbb{R}}, and let n≥3n\ge 3be an integer. Suppose f:Z/nZ→Mκ2f:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}is a map such that ∑i∈Z/nZdκ(f(i),f(i+[1]n))<2Dκ{\sum }_{i\in {\mathbb{Z}}\text{/}n{\mathbb{Z}}}{d}_{\kappa }(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa }. Then there exists a line L in Mκ2{M}_{\kappa }^{2}such that all f(i)f\left(i), i∈Z/nZi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}lie on one side of L, and thus, conv(f(Z/nZ)){\rm{conv}}(f\left({\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}))is contained in one side of L.For x,y,z∈Mκ2x,y,z\in {M}_{\kappa }^{2}with 0<dκ(x,y)<Dκ0\lt {d}_{\kappa }\left(x,y)\lt {D}_{\kappa }and 0<dκ(y,z)<Dκ0\lt {d}_{\kappa }(y,z)\lt {D}_{\kappa }, we denote by ∠xyz∈[0,π]\angle xyz\in \left[0,\pi ]the interior angle measure at yyof the (possibly degenerate) triangle with vertices xx, yy, and zz. By the law of cosines for Mκ2{M}_{\kappa }^{2}(see [6, p. 24]), ∠xyz∈[0,π]\angle xyz\in \left[0,\pi ]satisfies the following formula: cos∠xyz=dκ(x,y)2+dκ(y,z)2−dκ(z,x)22dκ(x,y)dκ(y,z),ifκ=0,cosh(−κdκ(x,y))cosh(−κdκ(y,z))−cosh(−κdκ(z,x))sinh(−κdκ(x,y))sinh(−κdκ(y,z)),ifκ<0,cos(κdκ(z,x))−cos(κdκ(x,y))cos(κdκ(y,z))sin(κdκ(x,y))sin(κdκ(y,z)),ifκ>0.\cos \angle xyz=\left\{\begin{array}{ll}\frac{{{\rm{d}}}_{\kappa }{\left(x,y)}^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(z,x)}^{2}}{2{d}_{\kappa }\left(x,y){d}_{\kappa }(y,z)},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa =0,\\ \frac{\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))-\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(z,x))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\sinh (\sqrt{-\kappa }{d}_{\kappa }(y,z))},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \lt 0,\\ \frac{\cos (\sqrt{\kappa }{d}_{\kappa }\left(z,x))-\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\sin (\sqrt{\kappa }{d}_{\kappa }(y,z))},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \gt 0.\end{array}\right.The following proposition follows immediately from the law of cosines.Proposition 2.2Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,x′,y′,z′∈Mκ2x,y,z,x^{\prime} ,y^{\prime} ,z^{\prime} \in {M}_{\kappa }^{2}are points such that0<dκ(x,y)=dκ(x′,y′)<Dκ,0<dκ(y,z)=dκ(y′,z′)<Dκ.0\lt {d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )\lt {D}_{\kappa },\hspace{1.0em}0\lt {d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} )\lt {D}_{\kappa }.Then dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )if and only if ∠xyz≤∠x′y′z′\angle xyz\le \angle x^{\prime} y^{\prime} z^{\prime} . Moreover, dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} )if and only if ∠xyz=∠x′y′z′\angle xyz=\angle x^{\prime} y^{\prime} z^{\prime} .The following proposition also follows from the law of cosines.Proposition 2.3Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z∈Mκ2x,y,z\in {M}_{\kappa }^{2}are three distinct points such thatdκ(x,y)+dκ(y,z)<Dκ,dκ(x,y)≤dκ(y,z).{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }\left(x,y)\le {d}_{\kappa }(y,z).Suppose x′,y′,z′∈Mκ2x^{\prime} ,y^{\prime} ,z^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x,y)=dκ(x′,y′),dκ(y,z)=dκ(y′,z′),dκ(z,x)≤dκ(z′,x′).{d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(z,x)\le {d}_{\kappa }\left(z^{\prime} ,x^{\prime} ).Then ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxz.ProofWe consider three cases.CASE 1: κ=0\kappa =0. In this case, we have ∠yzx≤∠yxz\angle yzx\le \angle yxzand ∠y′z′x′≤∠y′x′z′\angle y^{\prime} z^{\prime} x^{\prime} \le \angle y^{\prime} x^{\prime} z^{\prime} by hypothesis, and therefore, ∠yzx≤π/2\angle yzx\le \pi \hspace{0.1em}\text{/}\hspace{0.1em}2and ∠y′z′x′≤π/2\angle y^{\prime} z^{\prime} x^{\prime} \le \pi \hspace{0.1em}\text{/}\hspace{0.1em}2. It follows that 0≤dκ(x,z)2+dκ(y,z)2−dκ(x,y)2,0≤dκ(x′,z′)2+dκ(y′,z′)2−dκ(x′,y′)2=dκ(x′,z′)2+dκ(y,z)2−dκ(x,y)2,\begin{array}{rcl}0& \le & {d}_{\kappa }{\left(x,z)}^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2},\\ 0& \le & {d}_{\kappa }{\left(x^{\prime} ,z^{\prime} )}^{2}+{d}_{\kappa }{(y^{\prime} ,z^{\prime} )}^{2}-{d}_{\kappa }{\left(x^{\prime} ,y^{\prime} )}^{2}={d}_{\kappa }{\left(x^{\prime} ,z^{\prime} )}^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2},\end{array}and therefore, we have (2.1)0≤α2+dκ(y,z)2−dκ(x,y)20\le {\alpha }^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2}for any α∈[dκ(x,z),dκ(x′,z′)]\alpha \in \left[{d}_{\kappa }\left(x,z),{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )]. Define a function f1:(0,∞)→R{f}_{1}:\left(0,\infty )\to {\mathbb{R}}by f1(α)=dκ(x,y)2+α2−dκ(y,z)22dκ(x,y)α.{f}_{1}\left(\alpha )=\frac{{{\rm{d}}}_{\kappa }{\left(x,y)}^{2}+{\alpha }^{2}-{d}_{\kappa }{(y,z)}^{2}}{2{d}_{\kappa }\left(x,y)\alpha }.Then we have ddαf1(α)=12dκ(x,y)α2(α2+dκ(y,z)2−dκ(x,y)2)≥0\frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{1}\left(\alpha )=\frac{1}{2{d}_{\kappa }\left(x,y){\alpha }^{2}}({\alpha }^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2})\ge 0for any α∈[dκ(x,z),dκ(x′,z′)]\alpha \in \left[{d}_{\kappa }\left(x,z),{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )]by (2.1), which implies that ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxzbecause f1(dκ(x,z))=cos∠yxz,f1(dκ(x′,z′))=cos∠y′x′z′{f}_{1}\left({d}_{\kappa }\left(x,z))=\cos \angle yxz,\hspace{1.0em}{f}_{1}\left({d}_{\kappa }\left(x^{\prime} ,z^{\prime} ))=\cos \angle y^{\prime} x^{\prime} z^{\prime} by the law of cosines.CASE 2: κ<0\kappa \lt 0. In this case, we define a function f2:(0,∞)→R{f}_{2}:\left(0,\infty )\to {\mathbb{R}}by f2(α)=cosh(−κdκ(x,y))cosh(−κα)−cosh(−κdκ(y,z))sinh(−κdκ(x,y))sinh(−κα).{f}_{2}\left(\alpha )=\frac{\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\cosh (\sqrt{-\kappa }\alpha )-\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\sinh (\sqrt{-\kappa }\alpha )}.Then we have ddαf2(α)=−κ(cosh(−κdκ(y,z))cosh(−κα)−cosh(−κdκ(x,y)))sinh(−κdκ(x,y))sinh2(−κα)≥−κ(cosh(−κdκ(y,z))−cosh(−κdκ(x,y)))sinh(−κdκ(x,y))sinh2(−κα)≥0\begin{array}{rcl}\frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{2}\left(\alpha )& =& \frac{\sqrt{-\kappa }(\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))\cosh (\sqrt{-\kappa }\alpha )-\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)){\sinh }^{2}(\sqrt{-\kappa }\alpha )}\\ & \ge & \frac{\sqrt{-\kappa }(\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))-\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)){\sinh }^{2}(\sqrt{-\kappa }\alpha )}\ge 0\end{array}for any α∈(0,∞)\alpha \in \left(0,\infty )by hypothesis. This implies that ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxzbecause f2(dκ(x,z))=cos∠yxz,f2(dκ(x′,z′))=cos∠y′x′z′{f}_{2}\left({d}_{\kappa }\left(x,z))=\cos \angle yxz,\hspace{1.0em}{f}_{2}\left({d}_{\kappa }\left(x^{\prime} ,z^{\prime} ))=\cos \angle y^{\prime} x^{\prime} z^{\prime} by the law of cosines.CASE 3: κ>0\kappa \gt 0. In this case, we define a function f3:(0,Dκ)→R{f}_{3}:\left(0,{D}_{\kappa })\to {\mathbb{R}}by f3(α)=cos(κdκ(y,z))−cos(κdκ(x,y))cos(κα)sin(κdκ(x,y))sin(κα).{f}_{3}\left(\alpha )=\frac{\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))-\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\cos (\sqrt{\kappa }\alpha )}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\sin (\sqrt{\kappa }\alpha )}.Then ddαf3(α)=κ(cos(κdκ(x,y))−cos(κdκ(y,z))cos(κα))sin(κdκ(x,y))sin2(κα).\frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{3}\left(\alpha )=\frac{\sqrt{\kappa }(\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\cos (\sqrt{\kappa }\alpha ))}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,y)){\sin }^{2}(\sqrt{\kappa }\alpha )}.If dκ(y,z)≤Dκ/2{d}_{\kappa }(y,z)\le {D}_{\kappa }\hspace{-0.08em}\text{/}\hspace{-0.02em}2, then 0<κdκ(x,y)≤κdκ(y,z)≤π20\lt \sqrt{\kappa }{d}_{\kappa }\left(x,y)\le \sqrt{\kappa }{d}_{\kappa }(y,z)\le \frac{\pi }{2}by hypothesis, and therefore, cos(κdκ(x,y))−cos(κdκ(y,z))cos(κα)≥cos(κdκ(x,y))−cos(κdκ(y,z))≥0\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\cos (\sqrt{\kappa }\alpha )\ge \cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\ge 0for any α∈(0,Dκ)\alpha \in \left(0,{D}_{\kappa }). If dκ(y,z)>Dκ/2{d}_{\kappa }(y,z)\gt {D}_{\kappa }\hspace{0.1em}\text{/}\hspace{0.1em}2, then 0<κdκ(x,y)<π2<κdκ(y,z)<π,κdκ(x,y)<π−κdκ(y,z)0\lt \sqrt{\kappa }{d}_{\kappa }\left(x,y)\lt \frac{\pi }{2}\lt \sqrt{\kappa }{d}_{\kappa }(y,z)\lt \pi ,\hspace{1.0em}\sqrt{\kappa }{d}_{\kappa }\left(x,y)\lt \pi -\sqrt{\kappa }{d}_{\kappa }(y,z)by hypothesis, and therefore, cos(κdκ(x,y))−cos(κdκ(y,z))cos(κα)≥cos(κdκ(x,y))+cos(κdκ(y,z))=cos(κdκ(x,y))−cos(π−κdκ(y,z))>cos(κdκ(x,y))−cos(κdκ(x,y))=0\begin{array}{l}\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\cos (\sqrt{\kappa }\alpha )\ge \cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))+\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\\ \hspace{1.0em}=\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\pi -\sqrt{\kappa }{d}_{\kappa }(y,z))\gt \cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))=0\end{array}for any α∈(0,Dκ)\alpha \in \left(0,{D}_{\kappa }). Thus, we always have 0≤ddαf3(α)0\le \frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{3}\left(\alpha )for any α∈(0,Dκ)\alpha \in \left(0,{D}_{\kappa }). This implies that ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxzbecause f3(dκ(x,z))=cos∠yxz,f3(dκ(x′,z′))=cos∠y′x′z′{f}_{3}\left({d}_{\kappa }\left(x,z))=\cos \angle yxz,\hspace{1.0em}{f}_{3}\left({d}_{\kappa }\left(x^{\prime} ,z^{\prime} ))=\cos \angle y^{\prime} x^{\prime} z^{\prime} by the law of cosines.The aforementioned three cases exhaust all possibilities.□The following formulas follow from straightforward computation.Proposition 2.4Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z∈Mκ2x,y,z\in {M}_{\kappa }^{2}are points such that x≠zx\ne zand dκ(a,b)<Dκ{d}_{\kappa }\left(a,b)\lt {D}_{\kappa }for any a,b∈{x,y,z}a,b\in \left\{x,y,z\right\}. Suppose γ:[0,dκ(x,z)]→Mκ2\gamma :\left[0,{d}_{\kappa }\left(x,z)]\to {M}_{\kappa }^{2}is the geodesic such that γ(0)=x\gamma \left(0)=xand γ(dκ(x,z))=z\gamma \left({d}_{\kappa }\left(x,z))=z. Let t∈[0,1]t\in \left[0,1], and let p=γ(tdκ(x,z))p=\gamma (t{d}_{\kappa }\left(x,z)). If κ=0\kappa =0, thendκ(y,p)2=(1−t)dκ(x,y)2+tdκ(y,z)2−t(1−t)dκ(x,z)2.{d}_{\kappa }{(y,p)}^{2}=\left(1-t){d}_{\kappa }{\left(x,y)}^{2}+t{d}_{\kappa }{(y,z)}^{2}-t\left(1-t){d}_{\kappa }{\left(x,z)}^{2}.If κ<0\kappa \lt 0, thencosh(−κdκ(y,p))=1sinh(−κdκ(x,z))(sinh(−κ(1−t)dκ(x,z))cosh(−κdκ(x,y))+sinh(−κtdκ(x,z))cosh(−κdκ(y,z))).\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,p))=\frac{1}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,z))}(\sinh (\sqrt{-\kappa }\left(1-t){d}_{\kappa }\left(x,z))\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))+\sinh (\sqrt{-\kappa }t{d}_{\kappa }\left(x,z))\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))).If κ>0\kappa \gt 0, thencos(κdκ(y,p))=sin(κ(1−t)dκ(x,z))cos(κdκ(x,y))+sin(κtdκ(x,z))cos(κdκ(y,z))sin(κdκ(x,z)).\cos (\sqrt{\kappa }{d}_{\kappa }(y,p))=\frac{\sin (\sqrt{\kappa }\left(1-t){d}_{\kappa }\left(x,z))\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))+\sin (\sqrt{\kappa }t{d}_{\kappa }\left(x,z))\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,z))}.The next corollary follows immediately from Proposition 2.4Corollary 2.5Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,y˜∈Mκ2x,y,z,\tilde{y}\in {M}_{\kappa }^{2}are points such thatdκ(x,y)≤dκ(x,y˜)<Dκ,dκ(y,z)≤dκ(y˜,z)<Dκ,0<dκ(x,z)<Dκ.{d}_{\kappa }\left(x,y)\le {d}_{\kappa }\left(x,\tilde{y})\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }(y,z)\le {d}_{\kappa }(\tilde{y},z)\lt {D}_{\kappa },\hspace{1.0em}0\lt {d}_{\kappa }\left(x,z)\lt {D}_{\kappa }.Then we have dκ(y,p)≤dκ(y˜,p){d}_{\kappa }(y,p)\le {d}_{\kappa }(\tilde{y},p)for any p∈[x,z]p\in \left[x,z].Although it is not necessary for our purpose, it is worth noting that for κ∈(−∞,0]\kappa \in \left(-\infty ,0], Proposition 2.4 also implies the following corollary.Corollary 2.6Let κ∈(−∞,0]\kappa \in \left(-\infty ,0]. Suppose x,y,z,x˜,y˜,z˜∈Mκ2x,y,z,\tilde{x},\tilde{y},\tilde{z}\in {M}_{\kappa }^{2}are points such thatdκ(x,y)≤dκ(x˜,y˜)<Dκ,dκ(y,z)≤dκ(y˜,z˜)<Dκ,0<dκ(x˜,z˜)≤dκ(x,z)<Dκ.{d}_{\kappa }\left(x,y)\le {d}_{\kappa }\left(\tilde{x},\tilde{y})\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }(y,z)\le {d}_{\kappa }(\tilde{y},\tilde{z})\lt {D}_{\kappa },\hspace{1.0em}0\lt {d}_{\kappa }\left(\tilde{x},\tilde{z})\le {d}_{\kappa }\left(x,z)\lt {D}_{\kappa }.Let γ:[0,dκ(x,z)]→Mκ2\gamma :\left[0,{d}_{\kappa }\left(x,z)]\to {M}_{\kappa }^{2}and γ˜:[0,dκ(x˜,z˜)]→Mκ2\tilde{\gamma }:\left[0,{d}_{\kappa }\left(\tilde{x},\tilde{z})]\to {M}_{\kappa }^{2}be the geodesics such thatγ(0)=x,γ(dκ(x,z))=z,γ˜(0)=x˜,γ˜(dκ(x˜,z˜))=z˜.\gamma \left(0)=x,\hspace{1.0em}\gamma \left({d}_{\kappa }\left(x,z))=z,\hspace{1.0em}\tilde{\gamma }\left(0)=\tilde{x},\hspace{1.0em}\tilde{\gamma }\left({d}_{\kappa }\left(\tilde{x},\tilde{z}))=\tilde{z}.Fix t∈[0,1]t\in \left[0,1], and set p=γ(tdκ(x,z))p=\gamma \left(t{d}_{\kappa }\left(x,z)), p˜=γ˜(tdκ(x˜,z˜))\tilde{p}=\tilde{\gamma }\left(t{d}_{\kappa }\left(\tilde{x},\tilde{z})). Then dκ(y,p)≤dκ(y˜,p˜){d}_{\kappa }(y,p)\le {d}_{\kappa }(\tilde{y},\tilde{p}).Remark 2.7For the case in which κ∈(−∞,0]\kappa \in \left(-\infty ,0], we can prove Lemma 1.13 easily by using Corollary 2.6. To prove Lemma 1.13 for general κ∈R\kappa \in {\mathbb{R}}, we will prove a generalization of Alexandrov’s lemma [6, p. 25] in Section 4.2.3CAT(κ){\rm{CAT}}\left(\kappa )spacesA geodesic triangle in a metric space XXis a triple △=(γ1,γ2,γ3)\bigtriangleup =\left({\gamma }_{1},{\gamma }_{2},{\gamma }_{3})of geodesics γi:[ai,bi]→X{\gamma }_{i}:\left[{a}_{i},{b}_{i}]\to Xsuch that γ1(b1)=γ2(a2){\gamma }_{1}\left({b}_{1})={\gamma }_{2}\left({a}_{2}), γ2(b2)=γ3(a3){\gamma }_{2}\left({b}_{2})={\gamma }_{3}\left({a}_{3})and γ3(b3)=γ1(a1){\gamma }_{3}\left({b}_{3})={\gamma }_{1}\left({a}_{1}). Let κ∈R\kappa \in {\mathbb{R}}. If the perimeter ∑i=13∣bi−ai∣{\sum }_{i=1}^{3}| {b}_{i}-{a}_{i}| of the geodesic triangle △\bigtriangleup is less than 2Dκ2{D}_{\kappa }, then there exists a geodesic triangle △κ=(γ1κ,γ2κ,γ3κ){\bigtriangleup }^{\kappa }=\left({\gamma }_{1}^{\kappa },{\gamma }_{2}^{\kappa },{\gamma }_{3}^{\kappa }), γiκ:[ai,bi]→Mκ2{\gamma }_{i}^{\kappa }:\left[{a}_{i},{b}_{i}]\to {M}_{\kappa }^{2}in Mκ2{M}_{\kappa }^{2}. Such a geodesic triangle △κ{\bigtriangleup }^{\kappa }is unique up to isometry of Mκ2{M}_{\kappa }^{2}. The geodesic triangle △\bigtriangleup is said to be κ\kappa -thin if dX(γi(s),γj(t))≤dκ(γiκ(s),γjκ(t)){d}_{X}\left({\gamma }_{i}\left(s),{\gamma }_{j}\left(t))\le {d}_{\kappa }\left({\gamma }_{i}^{\kappa }\left(s),{\gamma }_{j}^{\kappa }\left(t))for any i,j∈{1,2,3}i,j\in \left\{1,2,3\right\}, any s∈[ai,bi]s\in \left[{a}_{i},{b}_{i}], and any t∈[aj,bj]t\in \left[{a}_{j},{b}_{j}].Definition 2.8Let κ∈R\kappa \in {\mathbb{R}}. A metric space XXis called a CAT(κ){\rm{CAT}}\left(\kappa )space if XXis Dκ{D}_{\kappa }-geodesic, and any geodesic triangle in XXwith perimeter <2Dκ\lt 2{D}_{\kappa }is κ\kappa -thin.By definition, Mκ2{M}_{\kappa }^{2}is a CAT(κ){\rm{CAT}}\left(\kappa )space. It is easily observed that if (X,dX)\left(X,{d}_{X})is a CAT(κ){\rm{CAT}}\left(\kappa )space, then for any x,y∈Xx,y\in Xwith dX(x,y)<Dκ{d}_{X}\left(x,y)\lt {D}_{\kappa }, there exists the unique geodesic segment with endpoints xxand yy. Every Dκ{D}_{\kappa }-convex subset of a CAT(κ){\rm{CAT}}\left(\kappa )space equipped with the induced metric is a CAT(κ){\rm{CAT}}\left(\kappa )space. For a detailed exposition of CAT(κ){\rm{CAT}}\left(\kappa )spaces, see [2,6,7]. We note that in [2, p. 96, Definition 9.3], the term CAT(κ){\rm{CAT}}\left(\kappa )space is used with somewhat different meaning.3Ballmann’s argument for our settingIn this section, we prove Theorems 1.5 and 1.6 by using Lemma 1.13. As we mentioned in Section 1, our proof is based on the idea used by Ballmann in his lecture note [4] for proving Reshetnyak’s majorization theorem. We will prove Lemma 1.13 later in Section 5. First, we define the following conditions by slightly modifying the definition of the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions.Definition 3.1Fix κ∈R\kappa \in {\mathbb{R}}and a positive integer nn. We say that a metric space (X,dX)\left(X,{d}_{X})is a Cycln′(κ){{\rm{Cycl}}}_{n}^{^{\prime} }\left(\kappa )space if for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xthat satisfies ∑i∈Z/nZdX(f(i),f(i+[1]n))<2Dκ,f(j)≠f(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}f\left(j)\ne f\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, there exists a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}that satisfies the following two conditions: (1)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, dκ(g(i),g(i+[1]n))=dX(f(i),f(i+[1]n)),dκ(g(i),g(j))≥dX(f(i),f(j)).{d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa }\left(g\left(i),g\left(j))\ge {d}_{X}(f\left(i),f\left(j)).(2)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with i≠ji\ne j, [g(i),g(j)]∩[g(i−[1]n),g(i+[1]n)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n}),g\left(i+{\left[1]}_{n})]\ne \varnothing .We call such a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}that satisfies the aforementioned two conditions a comparison map for ff.Note that when the polygon with vertices g(i)g\left(i), i∈Z/nZi\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}is non-degenerate, the condition (2) in Definition 3.1 means that this polygon is convex. By definition, every metric space is Cycl1′(κ){{\rm{Cycl}}}_{1}^{^{\prime} }\left(\kappa ), Cycl2′(κ){{\rm{Cycl}}}_{2}^{^{\prime} }\left(\kappa ), and Cycl3′(κ){{\rm{Cycl}}}_{3}^{^{\prime} }\left(\kappa )for all κ∈R\kappa \in {\mathbb{R}}. It is also easily seen that for any κ∈R\kappa \in {\mathbb{R}}and any integer n≥4n\ge 4, if a metric space is Cyclm′(κ){{\rm{Cycl}}}_{m}^{^{\prime} }\left(\kappa )for all m∈Z∩[1,n]m\in {\mathbb{Z}}\cap \left[1,n], then it is Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa ). To prove Theorem 1.5, it clearly suffices to prove that every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycln′(κ){{\rm{Cycl}}}_{n}^{^{\prime} }\left(\kappa )for all positive integers nn.Proof of Theorem 1.5 by using Lemma 1.13Fix κ∈R\kappa \in {\mathbb{R}}. We will prove that every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycln′(κ){{\rm{Cycl}}}_{n}^{^{\prime} }\left(\kappa )for all positive integers nnby induction on nn. As we mentioned above, every metric space is Cycl1′(κ){{\rm{Cycl}}}_{1}^{^{\prime} }\left(\kappa ), Cycl2′(κ){{\rm{Cycl}}}_{2}^{^{\prime} }\left(\kappa ), and Cycl3′(κ){{\rm{Cycl}}}_{3}^{^{\prime} }\left(\kappa )trivially. Fix an integer n≥3n\ge 3, and assume that every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycll′(κ){{\rm{Cycl}}}_{l}^{^{\prime} }\left(\kappa )for all l∈Z∩[1,n]l\in {\mathbb{Z}}\cap \left[1,n]. Let (X,dX)\left(X,{d}_{X})be an arbitrary Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space, and let f:Z/(n+1)Z→Xf:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}\to Xbe an arbitrary map that satisfies (3.1)∑i∈Z/(n+1)ZdX(f(i),f(i+[1]n+1))<2Dκ,f(j)≠f(j+[1]n+1)\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n+1}))\lt 2{D}_{\kappa },\hspace{1.0em}f\left(j)\ne f\left(j+{\left[1]}_{n+1})for every j∈Z/(n+1)Zj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n+1){\mathbb{Z}}. Then what we need to prove is the existence of a comparison map for ff. Define a map f0:Z/nZ→X{f}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xby f0([m]n)=f([m]n+1),m∈Z∩[0,n−1].{f}_{0}\left({\left[m]}_{n})=f\left({\left[m]}_{n+1}),\hspace{1.0em}m\in {\mathbb{Z}}\cap \left[0,n-1].We consider two cases.CASE 1: The map f satisfies f([n−1]n+1)≠f([0]n+1)f\left({\left[n-1]}_{n+1})\ne f\left({\left[0]}_{n+1}). In this case, it follows from (3.1) that ∑i∈Z/nZdX(f0(i),f0(i+[1]n))<2Dκ,f0(j)≠f0(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}({f}_{0}\left(i),{f}_{0}\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}{f}_{0}\left(j)\ne {f}_{0}\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, and so there exists a comparison map g0:Z/nZ→Mκ2{g}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}for f0{f}_{0}by the inductive hypothesis. Since dκ(g0([n−1]n),g0([0]n))=dX(f([n−1]n+1),f([0]n+1)),{d}_{\kappa }\left({g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n}))={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[0]}_{n+1})),there exists p∈Mκ2p\in {M}_{\kappa }^{2}such that dκ(g0([n−1]n),p)=dX(f([n−1]n+1),f([n]n+1)),dκ(p,g0([0]n))=dX(f([n]n+1),f([0]n+1)).{d}_{\kappa }\left({g}_{0}\left({\left[n-1]}_{n}),p)={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1})),\hspace{1em}{d}_{\kappa }\left(p,{g}_{0}\left({\left[0]}_{n}))={d}_{X}(f\left({\left[n]}_{n+1}),f\left({\left[0]}_{n+1})).Because g0{g}_{0}is a comparison map for f0{f}_{0}, the condition (2) in Definition 3.1 guarantees that for any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, g0(i){g}_{0}\left(i)and g0(j){g}_{0}\left(j)do not lie on opposite sides of ℓ(g0([n−1]n),g0([0]n))\ell \left({g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n})). So we may assume that ppis not on the same side of ℓ(g0([n−1]n),g0([0]n))\ell \left({g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n}))as g0(i){g}_{0}\left(i)for every i∈Z/nZi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}. Define a map g:Z/(n+1)Z→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n+1){\mathbb{Z}}\to {M}_{\kappa }^{2}by g([m]n+1)=g0([m]n),ifm∈Z∩[0,n−1],p,ifm=n.g\left({\left[m]}_{n+1})=\left\{\phantom{\rule[-1.33em]{}{0ex}}\begin{array}{ll}{g}_{0}\left({\left[m]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-1],\\ p,\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n.\end{array}\right.Then it is straightforward to see that (3.2)dκ(g([l]n+1),g([m]n+1))≥dX(f([l]n+1),f([m]n+1)){d}_{\kappa }(g\left({\left[l]}_{n+1}),g\left({\left[m]}_{n+1}))\ge {d}_{X}(f\left({\left[l]}_{n+1}),f\left({\left[m]}_{n+1}))for any l,m∈Z∩[0,n−1]l,m\in {\mathbb{Z}}\cap \left[0,n-1], (3.3)dκ(g(i),g(i+[1]n+1))=dX(f(i),f(i+[1]n+1)){d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n+1}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n+1}))for any i∈Z/(n+1)Zi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n+1){\mathbb{Z}}, and (3.4)dκ(g([n−1]n+1),g([0]n+1))=dX(f([n−1]n+1),f([0]n+1)).{d}_{\kappa }\left(g\left({\left[n-1]}_{n+1}),g\left({\left[0]}_{n+1}))={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[0]}_{n+1})).We divide CASE 1 into the following two subcases.SUBCASE 1A: The map g satisfies∠g([n−2]n+1)g([n−1]n+1)g([0]n+1)+∠g([0]n+1)g([n−1]n+1)g([n]n+1)≤π,∠g([1]n+1)g([0]n+1)g([n−1]n+1)+∠g([n−1]n+1)g([0]n+1)g([n]n+1)≤π.\begin{array}{rcl}\angle g\left({\left[n-2]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})+\angle g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[n]}_{n+1})& \le & \pi ,\\ \angle g\left({\left[1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})+\angle g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n]}_{n+1})& \le & \pi .\end{array}In this subcase, since g0{g}_{0}is a comparison map for f0{f}_{0}, it follows from elementary geometry that we have [g(i),g(j)]∩[g(i−[1]n+1),g(i+[1]n+1)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n+1}),g\left(i+{\left[1]}_{n+1})]\ne \varnothing for any i,j∈Z/(n+1)Zi,j\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}with i≠ji\ne j. In particular, we have [g([n−1]n+1),g([0]n+1)]∩[g([m]n+1),g([n]n+1)]≠∅\left[g\left({\left[n-1]}_{n+1}),g\left({\left[0]}_{n+1})]\cap \left[g\left({\left[m]}_{n+1}),g\left({\left[n]}_{n+1})]\ne \varnothing for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]. Clearly, we also have dκ(g([0]n+1),g([m]n+1))+dκ(g([m]n+1),g([n−1]n+1))+dκ(g([n−1]n+1),g([n]n+1))+dκ(g([n]n+1),g([0]n+1))<2Dκ.{d}_{\kappa }\left(g\left({\left[0]}_{n+1}),g\left({\left[m]}_{n+1}))+{d}_{\kappa }\left(g\left({\left[m]}_{n+1}),g\left({\left[n-1]}_{n+1}))+{d}_{\kappa }\left(g\left({\left[n-1]}_{n+1}),g\left({\left[n]}_{n+1}))+{d}_{\kappa }\left(g\left({\left[n]}_{n+1}),g\left({\left[0]}_{n+1}))\lt 2{D}_{\kappa }.Therefore, Lemma 1.13 implies dκ(g([m]n+1),g([n]n+1))≥dX(f([m]n+1),f([n]n+1)){d}_{\kappa }\left(g\left({\left[m]}_{n+1}),g\left({\left[n]}_{n+1}))\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n]}_{n+1}))for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]because XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), and we have dX(f([0]n+1),f([m]n+1))≤dκ(g([0]n+1),g([m]n+1)),dX(f([m]n+1),f([n−1]n+1))≤dκ(g([m]n+1),g([n−1]n+1)),dX(f([n−1]n+1),f([n]n+1))=dκ(g([n−1]n+1),g([n]n+1)),dX(f([n]n+1),f([0]n+1))=dκ(g([n]n+1),g([0]n+1)),dX(f([0]n+1),f([n−1]n+1))=dκ(g([0]n+1),g([n−1]n+1))\begin{array}{rcl}{d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[m]}_{n+1}))& \le & {d}_{\kappa }\left(g\left({\left[0]}_{n+1}),g\left({\left[m]}_{n+1})),\\ {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n-1]}_{n+1}))& \le & {d}_{\kappa }\left(g\left({\left[m]}_{n+1}),g\left({\left[n-1]}_{n+1})),\\ {d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1}))& =& {d}_{\kappa }\left(g\left({\left[n-1]}_{n+1}),g\left({\left[n]}_{n+1})),\\ {d}_{X}(f\left({\left[n]}_{n+1}),f\left({\left[0]}_{n+1}))& =& {d}_{\kappa }\left(g\left({\left[n]}_{n+1}),g\left({\left[0]}_{n+1})),\\ {d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[n-1]}_{n+1}))& =& {d}_{\kappa }\left(g\left({\left[0]}_{n+1}),g\left({\left[n-1]}_{n+1}))\end{array}by (3.2), (3.3), and (3.4). Thus, ggis a comparison map for ff.SUBCASE 1B: The map g satisfies(3.5)π<∠g([n−2]n+1)g([n−1]n+1)g([0]n+1)+∠g([0]n+1)g([n−1]n+1)g([n]n+1)\pi \lt \angle g\left({\left[n-2]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})+\angle g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[n]}_{n+1})or π<∠g([1]n+1)g([0]n+1)g([n−1]n+1)+∠g([n−1]n+1)g([0]n+1)g([n]n+1).\pi \lt \angle g\left({\left[1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})+\angle g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n]}_{n+1}).In this subcase, we may assume without loss of generality that we have (3.5). Let S=conv(g0(Z/nZ)),T=conv({g0([n−1]n),p,g0([0]n)}).S={\rm{conv}}\left({g}_{0}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}})),\hspace{1.0em}T={\rm{conv}}\left(\left\{{g}_{0}\left({\left[n-1]}_{n}),p,{g}_{0}\left({\left[0]}_{n})\right\}).Equip the subsets SSand TTof Mκ2{M}_{\kappa }^{2}with the induced metrics, and regard them as disjoint metric spaces. Define (R,dR)\left(R,{d}_{R})to be the metric space obtained by gluing SSand TTby identifying [g0([n−1]n),g0([0]n)]⊆S\left[{g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n})]\subseteq Swith [g0([n−1]n),g0([0]n)]⊆T\left[{g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n})]\subseteq T. Then RRis a CAT(κ){\rm{CAT}}\left(\kappa )space by Reshetnyak’s gluing theorem (see [14] or [6, Chapter II.11, Theorem 11.1]). We denote by rm{r}_{m}the point in RRrepresented by g0([m]n)∈S{g}_{0}\left({\left[m]}_{n})\in Sfor each m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1], and by rn{r}_{n}the point in RRrepresented by p∈Tp\in T(Figure 1). Define a map f1:Z/nZ→R{f}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Rby f1([m]n)=rm,ifm∈Z∩[0,n−2],rn,ifm=n−1.{f}_{1}\left({\left[m]}_{n})=\left\{\phantom{\rule[-1.33em]{}{0ex}}\begin{array}{ll}{r}_{m},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ {r}_{n},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1.\end{array}\right.Then it follows from (3.1) and the definition of f1{f}_{1}that ∑i∈Z/nZdR(f1(i),f1(i+[1]n))<2Dκ,f1(j)≠f1(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}}{d}_{R}({f}_{1}\left(i),{f}_{1}\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}{f}_{1}\left(j)\ne {f}_{1}\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}. Because RRis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )by Theorem 1.4, there exists a comparison map g1:Z/nZ→Mκ2{g}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}for f1{f}_{1}by the inductive hypothesis. It follows from (3.5) and Alexandrov’s lemma [6, p. 25] that dκ(g1([n−2]n),g1([n−1]n))=dR(rn−2,rn)=dR(rn−2,rn−1)+dR(rn−1,rn).{d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-2},{r}_{n})={d}_{R}\left({r}_{n-2},{r}_{n-1})+{d}_{R}\left({r}_{n-1},{r}_{n}).Therefore, there exists a point q∈[g1([n−2]n),g1([n−1]n)]q\in \left[{g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n})]such that dκ(g1([n−2]n),q)=dR(rn−2,rn−1),dκ(q,g1([n−1]n))=dR(rn−1,rn).{d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),q)={d}_{R}\left({r}_{n-2},{r}_{n-1}),\hspace{1.0em}{d}_{\kappa }\left(q,{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-1},{r}_{n}).Define a map g2:Z/(n+1)Z→Mκ2{g}_{2}:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}\to {M}_{\kappa }^{2}by g2([m]n+1)=g1([m]n),ifm∈Z∩[0,n−2],q,ifm=n−1,g1([n−1]n),ifm=n.{g}_{2}\left({\left[m]}_{n+1})=\left\{\begin{array}{ll}{g}_{1}\left({\left[m]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ q,\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1,\\ {g}_{1}\left({\left[n-1]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n.\end{array}\right.Then, we clearly have [g2(i),g2(j)]∩[g2(i−[1]n+1),g2(i+[1]n+1)]≠∅\left[{g}_{2}\left(i),{g}_{2}\left(j)]\cap \left[{g}_{2}\left(i-{\left[1]}_{n+1}),{g}_{2}\left(i+{\left[1]}_{n+1})]\ne \varnothing for any i,j∈Z/(n+1)Zi,j\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}with i≠ji\ne j. It is straightforward to see that dκ(g2([l]n+1),g2([m]n+1))≥dX(f([l]n+1),f([m]n+1)){d}_{\kappa }\left({g}_{2}\left({\left[l]}_{n+1}),{g}_{2}\left({\left[m]}_{n+1}))\ge {d}_{X}(f\left({\left[l]}_{n+1}),f\left({\left[m]}_{n+1}))for any l,m∈Z∩[0,n−2]l,m\in {\mathbb{Z}}\cap \left[0,n-2], and dκ(g2(i),g2(i+[1]n+1))=dX(f(i),f(i+[1]n+1)){d}_{\kappa }\left({g}_{2}\left(i),{g}_{2}\left(i+{\left[1]}_{n+1}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n+1}))for any i∈Z/(n+1)Zi\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}. Lemma 1.13 implies (3.6)dκ(g1([m]n),q)≥dR(rm,rn−1){d}_{\kappa }\left({g}_{1}\left({\left[m]}_{n}),q)\ge {d}_{R}\left({r}_{m},{r}_{n-1})for every m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2]because RRis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), and we have dκ(g1([n−1]n),g1([m]n))≥dR(rn,rm),dκ(g1([m]n),g1([n−2]n))≥dR(rm,rn−2),dκ(g1([n−2]n),q)=dR(rn−2,rn−1),dκ(q,g1([n−1]n))=dR(rn−1,rn),dκ(g1([n−2]n),g1([n−1]n))=dR(rn−2,rn),q∈[g1([n−2]n),g1([n−1]n)]\begin{array}{l}{d}_{\kappa }\left({g}_{1}\left({\left[n-1]}_{n}),{g}_{1}\left({\left[m]}_{n}))\ge {d}_{R}\left({r}_{n},{r}_{m}),\hspace{1.0em}{d}_{\kappa }\left({g}_{1}\left({\left[m]}_{n}),{g}_{1}\left({\left[n-2]}_{n}))\ge {d}_{R}\left({r}_{m},{r}_{n-2}),\\ {d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),q)={d}_{R}\left({r}_{n-2},{r}_{n-1}),\hspace{1.0em}{d}_{\kappa }\left(q,{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-1},{r}_{n}),\\ {d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-2},{r}_{n}),\hspace{1.0em}q\in \left[{g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n})]\end{array}by definition of g1{g}_{1}. It follows from (3.6) and the definition of g2{g}_{2}that dκ(g2([m]n+1),g2([n−1]n+1))≥dX(f([m]n+1),f([n−1]n+1)){d}_{\kappa }\left({g}_{2}\left({\left[m]}_{n+1}),{g}_{2}\left({\left[n-1]}_{n+1}))\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n-1]}_{n+1}))for every m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2]. Together with the definition of RR, Lemma 1.13 also implies (3.7)dR(rm,rn)≥dX(f([m]n+1),f([n]n+1)){d}_{R}\left({r}_{m},{r}_{n})\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n]}_{n+1}))for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]because XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), and we have dR(r0,rm)≥dX(f([0]n+1),f([m]n+1)),dR(rm,rn−1)≥dX(f([m]n+1),f([n−1]n+1)),dR(rn−1,rn)=dX(f([n−1]n+1),f([n]n+1)),dR(rn,r0)=dX(f([n]n+1),f([0]n+1)),dR(r0,rn−1)=dX(f([0]n+1),f([n−1]n+1)).\begin{array}{l}{d}_{R}\left({r}_{0},{r}_{m})\ge {d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[m]}_{n+1})),\hspace{1.0em}{d}_{R}\left({r}_{m},{r}_{n-1})\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n-1]}_{n+1})),\\ {d}_{R}\left({r}_{n-1},{r}_{n})={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1})),\hspace{1.0em}{d}_{R}\left({r}_{n},{r}_{0})={d}_{X}(f\left({\left[n]}_{n+1}),f\left({\left[0]}_{n+1})),\\ {d}_{R}\left({r}_{0},{r}_{n-1})={d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[n-1]}_{n+1})).\end{array}It follows from (3.7) and the definition of g2{g}_{2}that dκ(g2([m]n+1),g2([n]n+1))≥dX(f([m]n+1),f([n]n+1)){d}_{\kappa }\left({g}_{2}\left({\left[m]}_{n+1}),{g}_{2}\left({\left[n]}_{n+1}))\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n]}_{n+1}))for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]. Thus, g2{g}_{2}is a comparison map for ff.CASE 2: The map f satisfies f([n−1]n+1)=f([0]n+1)f\left({\left[n-1]}_{n+1})=f\left({\left[0]}_{n+1}). In this case, we set d=dX(f([n−1]n+1),f([n]n+1))=dX(f([0]n+1),f([n]n+1)).d={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1}))={d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[n]}_{n+1})).Define a map f˜0:Z/(n−1)Z→X{\tilde{f}}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n-1){\mathbb{Z}}\to Xby f˜0([m]n−1)=f([m]n+1){\tilde{f}}_{0}\left({\left[m]}_{n-1})=f\left({\left[m]}_{n+1}), m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2]. Then there exists a comparison map g˜0:Z/(n−1)Z→Mκ2{\tilde{g}}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n-1){\mathbb{Z}}\to {M}_{\kappa }^{2}for f˜0{\tilde{f}}_{0}by the inductive hypothesis. Let S˜=conv(g˜0(Z/(n−1)Z)),T˜=[0,d].\tilde{S}={\rm{conv}}\left({\tilde{g}}_{0}\left({\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n-1){\mathbb{Z}})),\hspace{1.0em}\tilde{T}=\left[0,d].Equip S˜⊆Mκ2\tilde{S}\subseteq {M}_{\kappa }^{2}and T˜⊆R\tilde{T}\subseteq {\mathbb{R}}the induced metrics, and regard them as metric spaces in their own right. Define (R˜,dR˜)\left(\tilde{R},{d}_{\tilde{R}})to be the metric space obtained by gluing S˜\tilde{S}and T˜\tilde{T}by identifying {g˜0([0]n−1)}⊆S˜\left\{{\tilde{g}}_{0}\left({\left[0]}_{n-1})\right\}\subseteq \tilde{S}with {0}⊆T˜\left\{0\right\}\subseteq \tilde{T}. Then R˜\tilde{R}is a CAT(κ){\rm{CAT}}\left(\kappa )space by Reshetnyak’s gluing theorem. We denote by r˜m{\tilde{r}}_{m}the point in R˜\tilde{R}represented by g˜0([m]n−1)∈S˜{\tilde{g}}_{0}\left({\left[m]}_{n-1})\in \tilde{S}for each m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2], by r˜n−1{\tilde{r}}_{n-1}the point in R˜\tilde{R}represented by g˜0([0]n−1)∈S˜{\tilde{g}}_{0}\left({\left[0]}_{n-1})\in \tilde{S}, and by r˜n{\tilde{r}}_{n}the point in R˜\tilde{R}represented by d∈T˜d\in \tilde{T}. In particular, we have r˜0=r˜n−1{\tilde{r}}_{0}={\tilde{r}}_{n-1}. Define a map f˜1:Z/nZ→R˜{\tilde{f}}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to \tilde{R}by f˜1([m]n)=r˜m,ifm∈Z∩[0,n−2],r˜n,ifm=n−1.{\tilde{f}}_{1}\left({\left[m]}_{n})=\left\{\phantom{\rule[-1.33em]{}{0ex}}\begin{array}{ll}{\tilde{r}}_{m},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ {\tilde{r}}_{n},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1.\end{array}\right.Then since R˜\tilde{R}is Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )by Theorem 1.4, there exists a comparison map g˜1:Z/nZ→Mκ2{\tilde{g}}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}of f˜1{\tilde{f}}_{1}by the inductive hypothesis. By definition of the gluing of metric spaces, we have dκ(g˜1([n−2]n),g˜1([n−1]n))=dR˜(r˜n−2,r˜n)=dR˜(r˜n−2,r˜n−1)+dR˜(r˜n−1,r˜n),{d}_{\kappa }\left({\tilde{g}}_{1}\left({\left[n-2]}_{n}),{\tilde{g}}_{1}\left({\left[n-1]}_{n}))={d}_{\tilde{R}}\left({\tilde{r}}_{n-2},{\tilde{r}}_{n})={d}_{\tilde{R}}\left({\tilde{r}}_{n-2},{\tilde{r}}_{n-1})+{d}_{\tilde{R}}\left({\tilde{r}}_{n-1},{\tilde{r}}_{n}),and therefore, there exists a point q˜∈[g˜1([n−2]n),g˜1([n−1]n)]\tilde{q}\in \left[{\tilde{g}}_{1}\left({\left[n-2]}_{n}),{\tilde{g}}_{1}\left({\left[n-1]}_{n})]such that dκ(g˜1([n−2]n),q˜)=dR˜(r˜n−2,r˜n−1),dκ(q˜,g˜1([n−1]n))=dR˜(r˜n−1,r˜n).{d}_{\kappa }\left({\tilde{g}}_{1}\left({\left[n-2]}_{n}),\tilde{q})={d}_{\tilde{R}}\left({\tilde{r}}_{n-2},{\tilde{r}}_{n-1}),\hspace{1.0em}{d}_{\kappa }\left(\tilde{q},{\tilde{g}}_{1}\left({\left[n-1]}_{n}))={d}_{\tilde{R}}\left({\tilde{r}}_{n-1},{\tilde{r}}_{n}).Define a map g˜2:Z/(n+1)Z→Mκ2{\tilde{g}}_{2}:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}\to {M}_{\kappa }^{2}by g˜2([m]n+1)=g˜1([m]n),ifm∈Z∩[0,n−2],q˜,ifm=n−1,g˜1([n−1]n),ifm=n.{\tilde{g}}_{2}\left({\left[m]}_{n+1})=\left\{\begin{array}{ll}{\tilde{g}}_{1}\left({\left[m]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ \tilde{q},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1,\\ {\tilde{g}}_{1}\left({\left[n-1]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n.\end{array}\right.Then the same argument as in SUBCASE 1B shows that g˜2{\tilde{g}}_{2}is a comparison map for ff, which completes the proof.□Theorem 1.6 follows immediately from Theorem 1.5.Figure 1The CAT(κ){\rm{CAT}}\left(\kappa )space RR.Proof of Theorem 1.6Let XXbe a Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space. Then XXis Cyclm′(κ){{\rm{Cycl}}}_{m}^{^{\prime} }\left(\kappa )for every positive integer mmas shown in the proof of Theorem 1.5, which implies that XXis Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )for every integer n≥4n\ge 4.□4A generalization of Alexandrov’s lemmaIn this section, we prove two lemmas, Lemmas 4.6 and 4.10, which generalize Alexandrov’s lemma [6, p. 25]. We use these lemmas to prove Lemma 1.13 in Section 5.Before proving the first lemma, we recall some elementary facts about angle measure in Mκ2{M}_{\kappa }^{2}. We first recall the following three basic propositions about the sum of two adjacent angle measures in Mκ2{M}_{\kappa }^{2}.Proposition 4.1Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Then ∠xoz≤∠xoy+∠yoz\angle xoz\le \angle xoy+\angle yoz.Proposition 4.2Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Assume that ∠xoz=∠xoy+∠yoz\angle xoz=\angle xoy+\angle yoz. Then all of the following conditions are true: y and z do not lie on opposite sides of ℓ(o,x)\ell \left(o,x);x and y do not lie on opposite sides of ℓ(o,z)\ell \left(o,z);x and z do not lie on the same side of ℓ(o,y)\ell \left(o,y).Proposition 4.3Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Then the identity ∠xoz=∠xoy+∠yoz\angle xoz=\angle xoy+\angle yozholds if and only if yyand zzdo not lie on opposite sides of ℓ(o,x)\ell \left(o,x), and ∠xoy≤∠xoz\angle xoy\le \angle xoz.We recall two more elementary propositions.Proposition 4.4Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Then the following conditions are equivalent: (1)∠xoy+∠yoz+∠zox=2π\angle xoy+\angle yoz+\angle zox=2\pi ;(2)π≤∠yox+∠xoz\pi \le \angle yox+\angle xoz, and yyand zzdo not lie on the same side of ℓ(o,x)\ell \left(o,x);(3)π≤∠zoy+∠yox\pi \le \angle zoy+\angle yox, and zzand xxdo not lie on the same side of ℓ(o,y)\ell \left(o,y);(4)π≤∠xoz+∠zoy\pi \le \angle xoz+\angle zoy, and xxand yydo not lie on the same side of ℓ(o,z)\ell \left(o,z).When {o,x,y,z}\left\{o,x,y,z\right\}is contained in one side of some line, each condition in the statement of Proposition 4.4 means that oolies on the solid triangle with vertices xx, yy, and zz. Therefore, it is intuitively obvious that the following proposition holds true.Proposition 4.5Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are four distinct points such that {o,x,y,z}\left\{o,x,y,z\right\}is contained in one side of some line. Assume that one of (hence all of) the conditions (1), (2), (3), and (4) in the statement of Proposition 4.4holds true. Then all of the following conditions are true: o and x do not lie on opposite sides of ℓ(y,z)\ell (y,z);∠zxo+∠oxy=∠zxy\angle zxo+\angle oxy=\angle zxy;If o∈ℓ(x,y)∪ℓ(y,z)∪ℓ(z,x)o\in \ell \left(x,y)\cup \ell (y,z)\cup \ell \left(z,x), then o∈[x,y]∪[y,z]∪[z,x]o\in \left[x,y]\cup [y,z]\cup \left[z,x].We now start to prove the first lemma of this section, which was proved for κ=0\kappa =0in [17, Lemma 8.4].Lemma 4.6Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x,y)+dκ(y,z)+dκ(z,w)+dκ(w,x)<2Dκ,dκ(x,y)=dκ(x′,y′),dκ(y,z)=dκ(y′,z′),dκ(z,w)=dκ(z′,w′),dκ(w,x)=dκ(w′,x′),dκ(y,w)≤dκ(y′,w′).\begin{array}{l}{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)+{d}_{\kappa }\left(w,x)\lt 2{D}_{\kappa },\\ {d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(z,w)={d}_{\kappa }\left(z^{\prime} ,w^{\prime} ),\\ {d}_{\kappa }\left(w,x)={d}_{\kappa }\left(w^{\prime} ,x^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} ).\end{array}Whenever x, y, z, and w are distinct except the possibility that y=wy=w, we assume in addition that π≤∠yzx+∠xzw\pi \le \angle yzx+\angle xzw, and that y and w do not lie on the same side of ℓ(x,z)\ell \left(x,z). Thendκ(x,z)≤dκ(x′,z′).{d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).ProofIf one of the identities x=yx=y, x=zx=z, x=wx=w, y=zy=z, or z=wz=wholds, then it is straightforward to check the desired inequality holds true. So we assume that none of these identities hold. Suppose that y=wy=w. Then y∈ℓ(x,z)y\in \ell \left(x,z)because yyand wwdo not lie on the same side of ℓ(x,z)\ell \left(x,z)by hypothesis, and therefore, we have ∠xzy=∠xzw=0\angle xzy=\angle xzw=0or ∠xzy=∠xzw=π\angle xzy=\angle xzw=\pi . Because π≤∠yzx+∠xzw\pi \le \angle yzx+\angle xzwby hypothesis, it follows that (4.1)∠xzy=∠xzw=π.\angle xzy=\angle xzw=\pi .We also have (4.2)dκ(x,z)+dκ(z,y)<Dκ{d}_{\kappa }\left(x,z)+{d}_{\kappa }\left(z,y)\lt {D}_{\kappa }because otherwise κ\kappa would be greater than 0, and dκ(x,z)+dκ(z,y)+dκ(y,x){d}_{\kappa }\left(x,z)+{d}_{\kappa }\left(z,y)+{d}_{\kappa }(y,x)would be equal to 2Dκ2{D}_{\kappa }, which would imply that dκ(x,y)+dκ(y,z)+dκ(z,w)+dκ(w,x)≥dκ(x,y)+dκ(y,z)+dκ(z,x)=2Dκ,{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)+{d}_{\kappa }\left(w,x)\ge {d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,x)=2{D}_{\kappa },contradicting the hypothesis. It follows from (4.1) and (4.2) that z∈[x,y]z\in \left[x,y], and thus, dκ(x,z)=dκ(x,y)−dκ(y,z)=dκ(x′,y′)−dκ(y′,z′)≤dκ(x′,z′).{d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x,y)-{d}_{\kappa }(y,z)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).So henceforth we assume that xx, yy, zz, and wware distinct. We consider two cases.CASE 1: z∉ℓ(x,y)∪ℓ(y,w)∪ℓ(w,x)z\notin \ell \left(x,y)\cup \ell (y,w)\cup \ell \left(w,x). Let w˜∈Mκ2\tilde{w}\in {M}_{\kappa }^{2}be the point such that dκ(z,w˜)=dκ(z,w),∠yzw˜=∠y′z′w′,{d}_{\kappa }\left(z,\tilde{w})={d}_{\kappa }\left(z,w),\hspace{1.0em}\angle yz\tilde{w}=\angle y^{\prime} z^{\prime} w^{\prime} ,and w˜\tilde{w}is not on the opposite side of ℓ(y,z)\ell (y,z)from ww, as shown in Figure 2. Then Proposition 2.2 implies dκ(y,w˜)=dκ(y′,w′),{d}_{\kappa }(y,\tilde{w})={d}_{\kappa }(y^{\prime} ,w^{\prime} ),and therefore, by using Proposition 2.2 again, we obtain (4.3)∠w˜yz=∠w′y′z′.\angle \tilde{w}yz=\angle w^{\prime} y^{\prime} z^{\prime} .Since dκ(y,w)≤dκ(y′,w′){d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} )by hypothesis, Proposition 2.2 implies that ∠yzw≤∠y′z′w′=∠yzw˜,\angle yzw\le \angle y^{\prime} z^{\prime} w^{\prime} =\angle yz\tilde{w},and therefore, Proposition 4.3 implies that (4.4)∠yzw+∠wzw˜=∠yzw˜≤π.\angle yzw+\angle wz\tilde{w}=\angle yz\tilde{w}\le \pi .Because π≤∠yzx+∠xzw\pi \le \angle yzx+\angle xzw, and yyand wware not on the same side of ℓ(x,z)\ell \left(x,z), Proposition 4.4 implies that π≤∠yzw+∠wzx\pi \le \angle yzw+\angle wzx. Combining this with (4.4) yields ∠wzw˜≤∠wzx\angle wz\tilde{w}\le \angle wzx. Furthermore, w˜\tilde{w}and xxare not on opposite sides of ℓ(z,w)\ell \left(z,w)because (4.4) implies that w˜\tilde{w}is not on the same side of ℓ(z,w)\ell \left(z,w)as yyby Proposition 4.2, the hypothesis implies that xxis not on the same side of ℓ(z,w)\ell \left(z,w)as yyby Proposition 4.4, and y∉ℓ(z,w)y\notin \ell \left(z,w)by the assumption of CASE 1. Therefore, Proposition 4.3 implies that (4.5)∠wzx=∠wzw˜+∠w˜zx.\angle wzx=\angle wz\tilde{w}+\angle \tilde{w}zx.We have ∠w˜zx≤∠wzx\angle \tilde{w}zx\le \angle wzxby (4.5), and therefore, Proposition 2.2 implies that dκ(w˜,x)≤dκ(w,x)=dκ(w′,x′).{d}_{\kappa }\left(\tilde{w},x)\le {d}_{\kappa }\left(w,x)={d}_{\kappa }\left(w^{\prime} ,x^{\prime} ).Using Proposition 2.2 again, this implies that (4.6)∠w˜yx≤∠w′y′x′.\angle \tilde{w}yx\le \angle w^{\prime} y^{\prime} x^{\prime} .Because the hypothesis implies ∠yzw+∠wzx+∠xzy=2π\angle yzw+\angle wzx+\angle xzy=2\pi by Proposition 4.4, it follows from (4.4) and (4.5) that ∠w˜zx+∠xzy=2π−∠yzw˜≥π.\angle \tilde{w}zx+\angle xzy=2\pi -\angle yz\tilde{w}\ge \pi .Furthermore, yyand w˜\tilde{w}are not on the same side of ℓ(x,z)\ell \left(x,z)because yyis not on the same side of ℓ(x,z)\ell \left(x,z)as wwby hypothesis, (4.5) implies that w˜\tilde{w}is not on the opposite side of ℓ(x,z)\ell \left(x,z)from wwby Proposition 4.2, and w∉ℓ(x,z)w\notin \ell \left(x,z)by the assumption of CASE 1. Therefore, Proposition 4.5 implies that (4.7)∠w˜yx=∠w˜yz+∠zyx.\angle \tilde{w}yx=\angle \tilde{w}yz+\angle zyx.We also have (4.8)∠w′y′x′≤∠w′y′z′+∠z′y′x′\angle w^{\prime} y^{\prime} x^{\prime} \le \angle w^{\prime} y^{\prime} z^{\prime} +\angle z^{\prime} y^{\prime} x^{\prime} by Proposition 4.1. By combining (4.3), (4.6), (4.7), and (4.8), we obtain ∠zyx=∠w˜yx−∠w˜yz≤∠w′y′x′−∠w˜yz=∠w′y′x′−∠w′y′z′≤∠z′y′x′,\angle zyx=\angle \tilde{w}yx-\angle \tilde{w}yz\le \angle w^{\prime} y^{\prime} x^{\prime} -\angle \tilde{w}yz=\angle w^{\prime} y^{\prime} x^{\prime} -\angle w^{\prime} y^{\prime} z^{\prime} \le \angle z^{\prime} y^{\prime} x^{\prime} ,and therefore, Proposition 2.2 implies that dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).CASE 2: z∈ℓ(x,y)∪ℓ(y,w)∪ℓ(w,x)z\in \ell \left(x,y)\cup \ell (y,w)\cup \ell \left(w,x). In this case, z∈[x,y]∪[y,w]∪[w,x]z\in \left[x,y]\cup [y,w]\cup \left[w,x]by Proposition 4.5. If z∈[x,y]z\in \left[x,y], then dκ(x,z)=dκ(x,y)−dκ(y,z)=dκ(x′,y′)−dκ(y′,z′)≤dκ(x′,z′).{d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x,y)-{d}_{\kappa }(y,z)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).If z∈[w,x]z\in \left[w,x], then we obtain dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )similarly. So assume that z∈[y,w]z\in [y,w]. Then we have dκ(y,w)≤dκ(y′,w′)≤dκ(y′,z′)+dκ(z′,w′)=dκ(y,z)+dκ(z,w)=dκ(y,w),{d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} )\le {d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )={d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)={d}_{\kappa }(y,w),and thus, dκ(y,w)=dκ(y′,w′){d}_{\kappa }(y,w)={d}_{\kappa }(y^{\prime} ,w^{\prime} )and z′∈[y′,w′]z^{\prime} \in [y^{\prime} ,w^{\prime} ]. Hence, Proposition 2.2 implies ∠xyz=∠xyw=∠x′y′w′=∠x′y′z′,\angle xyz=\angle xyw=\angle x^{\prime} y^{\prime} w^{\prime} =\angle x^{\prime} y^{\prime} z^{\prime} ,and therefore, by using Proposition 2.2 again, we obtain dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).□Figure 2Proof of Lemma 4.6.Remark 4.7Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points that satisfy the hypothesis of Lemma 4.6. Alexandrov’s lemma [6, p. 25] states that if the identity dκ(y′,w′)=dκ(y′,z′)+dκ(z′,w′){d}_{\kappa }(y^{\prime} ,w^{\prime} )={d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )holds in addition, then we have dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).Before proving the second lemma of this section, we recall two additional elementary facts about the sum of two adjacent angle measures in Mκ2{M}_{\kappa }^{2}.Proposition 4.8Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}, and dκ(x,z)<Dκ{d}_{\kappa }\left(x,z)\lt {D}_{\kappa }. If [x,z]∩[o,y]≠∅\left[x,z]\cap \left[o,y]\ne \varnothing , then∠xoz=∠xoy+∠yoz.\angle xoz=\angle xoy+\angle yoz.Proposition 4.9Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z,w∈Mκ2o,x,y,z,w\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z,w}a\in \left\{x,y,z,w\right\}. If we have∠xoz=∠xoy+∠yoz,∠xow=∠xoz+∠zow,\angle xoz=\angle xoy+\angle yoz,\hspace{1.0em}\angle xow=\angle xoz+\angle zow,then we have∠yow=∠yoz+∠zow,∠xow=∠xoy+∠yow.\angle yow=\angle yoz+\angle zow,\hspace{1.0em}\angle xow=\angle xoy+\angle yow.We now prove the second lemma of this section.Lemma 4.10Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x′,y′)+dκ(y′,z′)+dκ(z′,w′)+dκ(w′,x′)<2Dκ,dκ(x,y)=dκ(x′,y′),dκ(y,z)=dκ(y′,z′),dκ(z,w)=dκ(z′,w′),dκ(w,x)=dκ(w′,x′),dκ(x′,z′)≤dκ(x,z),\begin{array}{l}{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )+{d}_{\kappa }\left(w^{\prime} ,x^{\prime} )\lt 2{D}_{\kappa },\\ {d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(z,w)={d}_{\kappa }\left(z^{\prime} ,w^{\prime} ),\\ {d}_{\kappa }\left(w,x)={d}_{\kappa }\left(w^{\prime} ,x^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z),\end{array}and [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing . Thendκ(y,w)≤dκ(y′,w′).{d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} ).ProofIf x′=z′x^{\prime} =z^{\prime} , then it follows from the hypothesis that x′∈[y′,w′]x^{\prime} \in [y^{\prime} ,w^{\prime} ], which implies the desired inequality by the triangle inequality. So we assume that x′≠z′x^{\prime} \ne z^{\prime} . We next consider the case in which y′∈ℓ(x′,z′)y^{\prime} \in \ell \left(x^{\prime} ,z^{\prime} ). If y′∈[x′,z′]y^{\prime} \in \left[x^{\prime} ,z^{\prime} ], then we have dκ(x,z)≤dκ(x,y)+dκ(y,z)=dκ(x′,y′)+dκ(y′,z′)=dκ(x′,z′)≤dκ(x,z),{d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )={d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z),and thus, dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} )and y∈[x,z]y\in \left[x,z]. Hence, Proposition 2.2 implies ∠yxw=∠zxw=∠z′x′w′=∠y′x′w′,\angle yxw=\angle zxw=\angle z^{\prime} x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} w^{\prime} ,and therefore, by using Proposition 2.2 again, we obtain dκ(y,w)=dκ(y′,w′).{d}_{\kappa }(y,w)={d}_{\kappa }(y^{\prime} ,w^{\prime} ).If y′∈ℓ(x′,z′)⧹[x′,z′]y^{\prime} \in \ell \left(x^{\prime} ,z^{\prime} )\setminus \left[x^{\prime} ,z^{\prime} ], then it follows from the hypothesis that x′x^{\prime} or z′z^{\prime} is on [y′,w′][y^{\prime} ,w^{\prime} ], which implies the desired inequality by the triangle inequality. Similarly, the desired inequality holds true in the case in which w′∈ℓ(x′,z′)w^{\prime} \in \ell \left(x^{\prime} ,z^{\prime} )as well. So henceforth we assume in addition that y′∉ℓ(x′,z′)y^{\prime} \notin \ell \left(x^{\prime} ,z^{\prime} )and w′∉ℓ(x′,z′)w^{\prime} \notin \ell \left(x^{\prime} ,z^{\prime} )In particular, this requires that x′x^{\prime} , y′y^{\prime} , z′z^{\prime} , and w′w^{\prime} are distinct.By hypothesis, one of the inequalities dκ(x′,y′)+dκ(y′,z′)<Dκ{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )\lt {D}_{\kappa }or dκ(z′,w′)+dκ(w′,x′)<Dκ{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )+{d}_{\kappa }\left(w^{\prime} ,x^{\prime} )\lt {D}_{\kappa }holds true. We may assume without loss of generality that dκ(x′,y′)+dκ(y′,z′)<Dκ,dκ(x′,y′)≤dκ(y′,z′).{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )\le {d}_{\kappa }(y^{\prime} ,z^{\prime} ).Then we have (4.9)∠yxz≤∠y′x′z′\angle yxz\le \angle y^{\prime} x^{\prime} z^{\prime} by Proposition 2.3. Let z˜∈Mκ2\tilde{z}\in {M}_{\kappa }^{2}be the point such that dκ(y′,z˜)=dκ(y′,z′),∠x′y′z˜=∠xyz,{d}_{\kappa }(y^{\prime} ,\tilde{z})={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}\angle x^{\prime} y^{\prime} \tilde{z}=\angle xyz,and z˜\tilde{z}is not on the opposite side of ℓ(x′,y′)\ell \left(x^{\prime} ,y^{\prime} )from z′z^{\prime} , as shown in Figure 3. Then Proposition 2.2 implies dκ(x′,z˜)=dκ(x,z),{d}_{\kappa }\left(x^{\prime} ,\tilde{z})={d}_{\kappa }\left(x,z),and therefore, by using Proposition 2.2 again, we obtain (4.10)∠y′x′z˜=∠yxz.\angle y^{\prime} x^{\prime} \tilde{z}=\angle yxz.Since dκ(x′,z′)≤dκ(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z), Proposition 2.2 implies (4.11)∠x′y′z′≤∠xyz=∠x′y′z˜.\angle x^{\prime} y^{\prime} z^{\prime} \le \angle xyz=\angle x^{\prime} y^{\prime} \tilde{z}.Because z˜\tilde{z}is not on the opposite side of ℓ(x′,y′)\ell \left(x^{\prime} ,y^{\prime} )from z′z^{\prime} , (4.11) implies (4.12)∠x′y′z˜=∠x′y′z′+∠z′y′z˜\angle x^{\prime} y^{\prime} \tilde{z}=\angle x^{\prime} y^{\prime} z^{\prime} +\angle z^{\prime} y^{\prime} \tilde{z}by Proposition 4.3. The hypothesis that [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing implies (4.13)∠x′y′z′=∠x′y′w′+∠w′y′z′\angle x^{\prime} y^{\prime} z^{\prime} =\angle x^{\prime} y^{\prime} w^{\prime} +\angle w^{\prime} y^{\prime} z^{\prime} by Proposition 4.8. By Proposition 4.9, (4.12) and (4.13) imply (4.14)∠x′y′z˜=∠x′y′w′+∠w′y′z˜.\angle x^{\prime} y^{\prime} \tilde{z}=\angle x^{\prime} y^{\prime} w^{\prime} +\angle w^{\prime} y^{\prime} \tilde{z}.By combining (4.11), (4.13), and (4.14), we obtain ∠w′y′z′≤∠w′y′z˜\angle w^{\prime} y^{\prime} z^{\prime} \le \angle w^{\prime} y^{\prime} \tilde{z}, and therefore, Proposition 2.2 implies dκ(w,z)=dκ(w′,z′)≤dκ(w′,z˜).{d}_{\kappa }\left(w,z)={d}_{\kappa }\left(w^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(w^{\prime} ,\tilde{z}).Using Proposition 2.2 again, this implies (4.15)∠zxw≤∠z˜x′w′.\angle zxw\le \angle \tilde{z}x^{\prime} w^{\prime} .The hypothesis that [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing implies (4.16)∠y′x′w′=∠y′x′z′+∠z′x′w′\angle y^{\prime} x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} w^{\prime} by Proposition 4.8. It follows from (4.9) and (4.10) that ∠y′x′z˜≤∠y′x′z′\angle y^{\prime} x^{\prime} \tilde{z}\le \angle y^{\prime} x^{\prime} z^{\prime} , which implies (4.17)∠y′x′z′=∠y′x′z˜+∠z˜x′z′\angle y^{\prime} x^{\prime} z^{\prime} =\angle y^{\prime} x^{\prime} \tilde{z}+\angle \tilde{z}x^{\prime} z^{\prime} by Proposition 4.3 because z˜\tilde{z}is not on the opposite side of ℓ(x′,y′)\ell \left(x^{\prime} ,y^{\prime} )from z′z^{\prime} . By Proposition 4.9, (4.16), and (4.17) imply (4.18)∠y′x′w′=∠y′x′z˜+∠z˜x′w′.\angle y^{\prime} x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} \tilde{z}+\angle \tilde{z}x^{\prime} w^{\prime} .We also have (4.19)∠yxw≤∠yxz+∠zxw\angle yxw\le \angle yxz+\angle zxwby Proposition 4.1. By combining (4.10), (4.15), (4.18), and (4.19), we obtain ∠yxw≤∠yxz+∠zxw≤∠y′x′z˜+∠z˜x′w′=∠y′x′w′,\angle yxw\le \angle yxz+\angle zxw\le \angle y^{\prime} x^{\prime} \tilde{z}+\angle \tilde{z}x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} w^{\prime} ,and therefore, Proposition 2.2 implies dκ(y,w)≤dκ(y′,w′){d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} ).□Figure 3Proof of Lemma 4.10.Recall that for any x,y,z,w∈Mκ2x,y,z,w\in {M}_{\kappa }^{2}such that dκ(x,y)+dκ(y,z)+dκ(z,w)+dκ(w,x)<2Dκ{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)+{d}_{\kappa }\left(w,x)\lt 2{D}_{\kappa }, and xx, yy, zz, and wware distinct except the possibility that y=wy=w, we have [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing if and only if ∠yxz+∠zxw≤π\angle yxz+\angle zxw\le \pi , ∠yzx+∠xzw≤π\angle yzx+\angle xzw\le \pi , and yyand wware not on the same side of ℓ(x,z)\ell \left(x,z).Lemmas 4.6 and 4.10 imply the following corollary.Corollary 4.11Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points that satisfy the hypothesis of Lemma 4.10. Assume in addition that x≠zx\ne zand that yyand wwdo not lie on the same side of ℓ(x,z)\ell \left(x,z). Then [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing .ProofWe have (4.20)dκ(y,w)≤dκ(y′,w′){d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} )by Lemma 4.10. If one of the identities x=yx=y, x=wx=w, y=zy=z, or z=wz=wholds, then we have [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing clearly. So we assume that xx, yy, zz, and wware distinct except the possibility that y=wy=w. Suppose for the sake of contradiction that [x,z]∩[y,w]=∅\left[x,z]\cap [y,w]=\varnothing . Since yyand wwdo not lie on the same side of ℓ(x,z)\ell \left(x,z), this requires that we have π<∠yxz+∠zxw\pi \lt \angle yxz+\angle zxwor π<∠yzx+∠xzw\pi \lt \angle yzx+\angle xzw. Therefore, (4.20) implies dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )by Lemma 4.6. Combining this with the hypothesis that dκ(x′,z′)≤dκ(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z)yields dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} ). Therefore, it follows from Proposition 2.2 that we have π<∠yxz+∠zxw=∠y′x′z′+∠z′x′w′\pi \lt \angle yxz+\angle zxw=\angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} w^{\prime} or π<∠yzx+∠xzw=∠y′z′x′+∠x′z′w′,\pi \lt \angle yzx+\angle xzw=\angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} w^{\prime} ,which implies [x′,z′]∩[y′,w′]=∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]=\varnothing , contradicting the hypothesis that [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing . Thus, we have [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing .□5Proof of Lemma 1.13In this section, we prove Lemma 1.13.Proof of Lemma 1.13Let κ\kappa , XX, xx, yy, zz, ww, x′x^{\prime} , y′y^{\prime} , z′z^{\prime} , and w′w^{\prime} be as in the hypothesis, and fix p∈[x′,z′]p\in \left[x^{\prime} ,z^{\prime} ]. If x′=z′x^{\prime} =z^{\prime} , then p=x′p=x^{\prime} , and therefore, dX(y,w)≤dX(y,x)+dX(x,w)≤dκ(y′,x′)+dκ(x′,w′)=dκ(y′,p)+dκ(p,w′).{d}_{X}(y,w)\le {d}_{X}(y,x)+{d}_{X}\left(x,w)\le {d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,w^{\prime} )={d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ).So henceforth we assume that x′≠z′x^{\prime} \ne z^{\prime} . Then we also have x≠zx\ne zsince dκ(x′,z′)≤dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z). Let w¯∈Mκ2\overline{w}\in {M}_{\kappa }^{2}be the point such that dκ(x′,w¯)=dκ(x′,w′),dκ(w¯,z′)=dκ(w′,z′),{d}_{\kappa }\left(x^{\prime} ,\overline{w})={d}_{\kappa }\left(x^{\prime} ,w^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(\overline{w},z^{\prime} )={d}_{\kappa }\left(w^{\prime} ,z^{\prime} ),and w¯\overline{w}is not on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} )as y′y^{\prime} . (If w′w^{\prime} is not on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} )as y′y^{\prime} , then w¯\overline{w}is w′w^{\prime} itself.) Then it is easily seen that dκ(p,w¯)=dκ(p,w′){d}_{\kappa }\left(p,\overline{w})={d}_{\kappa }\left(p,w^{\prime} ), and therefore, (5.1)dκ(y′,w¯)≤dκ(y′,p)+dκ(p,w¯)=dκ(y′,p)+dκ(p,w′).{d}_{\kappa }(y^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,\overline{w})={d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ).Because XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), there exist x0,y0,z0,w0∈Mκ2{x}_{0},{y}_{0},{z}_{0},{w}_{0}\in {M}_{\kappa }^{2}such that dκ(x0,y0)≤dX(x,y),dκ(y0,z0)≤dX(y,z),dκ(z0,w0)≤dX(z,w),dκ(w0,x0)≤dX(w,x),dX(x,z)≤dκ(x0,z0),dX(y,w)≤dκ(y0,w0).\begin{array}{l}{d}_{\kappa }\left({x}_{0},{y}_{0})\le {d}_{X}\left(x,y),\hspace{1.0em}{d}_{\kappa }({y}_{0},{z}_{0})\le {d}_{X}(y,z),\hspace{1.0em}{d}_{\kappa }\left({z}_{0},{w}_{0})\le {d}_{X}\left(z,w),\\ {d}_{\kappa }\left({w}_{0},{x}_{0})\le {d}_{X}\left(w,x),\hspace{1.0em}{d}_{X}\left(x,z)\le {d}_{\kappa }\left({x}_{0},{z}_{0}),\hspace{1.0em}{d}_{X}(y,w)\le {d}_{\kappa }({y}_{0},{w}_{0}).\end{array}These points satisfy ∣dκ(x′,y′)−dκ(y′,z′)∣≤dκ(x′,z′)≤dX(x,z)≤dκ(x0,z0)≤dκ(x0,y0)+dκ(y0,z0)≤dX(x,y)+dX(y,z)≤dκ(x′,y′)+dκ(y′,z′),| {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )| \le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z)\le {d}_{\kappa }\left({x}_{0},{z}_{0})\le {d}_{\kappa }\left({x}_{0},{y}_{0})+{d}_{\kappa }({y}_{0},{z}_{0})\le {d}_{X}\left(x,y)+{d}_{X}(y,z)\le {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} ),and thus, ∣dκ(x′,y′)−dκ(y′,z′)∣≤dκ(x0,z0)≤dκ(x′,y′)+dκ(y′,z′).| {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )| \le {d}_{\kappa }\left({x}_{0},{z}_{0})\le {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} ).This guarantees that there exists a point y˜∈Mκ2\tilde{y}\in {M}_{\kappa }^{2}such that dκ(x0,y˜)=dκ(x′,y′),dκ(y˜,z0)=dκ(y′,z′).{d}_{\kappa }\left({x}_{0},\tilde{y})={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(\tilde{y},{z}_{0})={d}_{\kappa }(y^{\prime} ,z^{\prime} ).Similarly, there also exists a point w˜∈Mκ2\tilde{w}\in {M}_{\kappa }^{2}such that dκ(x0,w˜)=dκ(x′,w′),dκ(w˜,z0)=dκ(w′,z′).{d}_{\kappa }\left({x}_{0},\tilde{w})={d}_{\kappa }\left(x^{\prime} ,w^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(\tilde{w},{z}_{0})={d}_{\kappa }\left(w^{\prime} ,z^{\prime} ).Clearly, we may assume that w˜\tilde{w}does not lie on the same side of ℓ(x0,z0)\ell \left({x}_{0},{z}_{0})as y˜\tilde{y}. We consider two cases.CASE 1: [x′,z′]∩[y′,w¯]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,\overline{w}]\ne \varnothing . In this case, it follows from Lemma 4.10 that (5.2)dκ(y˜,w˜)≤dκ(y′,w¯){d}_{\kappa }(\tilde{y},\tilde{w})\le {d}_{\kappa }(y^{\prime} ,\overline{w})because dκ(x0,y˜)=dκ(x′,y′),dκ(y˜,z0)=dκ(y′,z′),dκ(z0,w˜)=dκ(z′,w¯),dκ(w˜,x0)=dκ(w¯,x′),dκ(x′,z′)≤dX(x,z)≤dκ(x0,z0).\begin{array}{rcl}{d}_{\kappa }\left({x}_{0},\tilde{y})& =& {d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(\tilde{y},{z}_{0})={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left({z}_{0},\tilde{w})={d}_{\kappa }\left(z^{\prime} ,\overline{w}),\\ {d}_{\kappa }\left(\tilde{w},{x}_{0})& =& {d}_{\kappa }\left(\overline{w},x^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z)\le {d}_{\kappa }\left({x}_{0},{z}_{0}).\end{array}We also have [x0,z0]∩[y˜,w˜]≠∅\left[{x}_{0},{z}_{0}]\cap [\tilde{y},\tilde{w}]\ne \varnothing by Corollary 4.11. Choose p0∈[x0,z0]∩[y˜,w˜]{p}_{0}\in \left[{x}_{0},{z}_{0}]\cap [\tilde{y},\tilde{w}]. Then dκ(y0,p0)≤dκ(y˜,p0),dκ(p0,w0)≤dκ(p0,w˜){d}_{\kappa }({y}_{0},{p}_{0})\le {d}_{\kappa }(\tilde{y},{p}_{0}),\hspace{1.0em}{d}_{\kappa }\left({p}_{0},{w}_{0})\le {d}_{\kappa }\left({p}_{0},\tilde{w})by Corollary 2.5, and therefore, (5.3)dκ(y0,w0)≤dκ(y0,p0)+dκ(p0,w0)≤dκ(y˜,p0)+dκ(p0,w˜)=dκ(y˜,w˜).{d}_{\kappa }({y}_{0},{w}_{0})\le {d}_{\kappa }({y}_{0},{p}_{0})+{d}_{\kappa }\left({p}_{0},{w}_{0})\le {d}_{\kappa }(\tilde{y},{p}_{0})+{d}_{\kappa }\left({p}_{0},\tilde{w})={d}_{\kappa }(\tilde{y},\tilde{w}).It follows from (5.1), (5.2), and (5.3) that dX(y,w)≤dκ(y0,w0)≤dκ(y˜,w˜)≤dκ(y′,w¯)≤dκ(y′,p)+dκ(p,w′).{d}_{X}(y,w)\le {d}_{\kappa }({y}_{0},{w}_{0})\le {d}_{\kappa }(\tilde{y},\tilde{w})\le {d}_{\kappa }(y^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ).CASE 2: [x′,z′]∩[y′,w¯]=∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,\overline{w}]=\varnothing . In this case, we have x′≠y′x^{\prime} \ne y^{\prime} , x′≠w′x^{\prime} \ne w^{\prime} , y′≠z′y^{\prime} \ne z^{\prime} and z′≠w¯z^{\prime} \ne \overline{w}, and one of the inequalities π<∠y′x′z′+∠z′x′w¯\pi \lt \angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} \overline{w}or π<∠y′z′x′+∠x′z′w¯\pi \lt \angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} \overline{w}holds since y′y^{\prime} and w¯\overline{w}are not on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} ). We may assume without loss of generality that π<∠y′x′z′+∠z′x′w¯.\pi \lt \angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} \overline{w}.Then we have dκ(y′,x′)+dκ(x′,w¯)≤dκ(y′,p)+dκ(p,w¯){d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,\overline{w})by Alexandrov’s lemma [6, p. 25], and therefore, dX(y,w)≤dX(y,x)+dX(x,w)≤dκ(y′,x′)+dκ(x′,w′)=dκ(y′,x′)+dκ(x′,w¯)≤dκ(y′,p)+dκ(p,w¯)=dκ(y′,p)+dκ(p,w′),{d}_{X}(y,w)\le {d}_{X}(y,x)+{d}_{X}\left(x,w)\le {d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,w^{\prime} )={d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,\overline{w})={d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ),which completes the proof.□6Proof of Theorem 1.10In this section, we present a proof of Theorem 1.10 for completeness (see Remark 6.2). First, we recall the following fact, which was established by Sturm when he proved in [16, Theorem 4.9] that if a geodesic metric space satisfies the ⊠\boxtimes -inequalities, then it is CAT(0){\rm{CAT}}\left(0).Proposition 6.1Let (X,dX)\left(X,{d}_{X})be a metric space that satisfies the ⊠\boxtimes -inequalities. Suppose x,y,z∈Xx,y,z\in Xare points such that x≠zx\ne z, and(6.1)dX(x,z)=dX(x,y)+dX(y,z).{d}_{X}\left(x,z)={d}_{X}\left(x,y)+{d}_{X}(y,z).Set t=dX(x,y)/dX(x,z)t={d}_{X}\left(x,y)\hspace{0.1em}\text{/}\hspace{0.1em}{d}_{X}\left(x,z). Then we havedX(y,w)2≤(1−t)dX(x,w)2+tdX(z,w)2−t(1−t)dX(x,z)2.{d}_{X}{(y,w)}^{2}\le \left(1-t){d}_{X}{\left(x,w)}^{2}+t{d}_{X}{\left(z,w)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}.for any w∈Xw\in X.ProofBy the hypothesis (6.1), we compute (1−t)dX(x,y)2+tdX(y,z)2=dX(y,z)dX(x,z)dX(x,y)2+dX(x,y)dX(x,z)dX(y,z)2=dX(x,y)dX(y,z)dX(x,z)(dX(x,y)+dX(y,z))=dX(x,y)dX(y,z)=t(1−t)dX(x,z)2.\begin{array}{rcl}\left(1-t){d}_{X}{\left(x,y)}^{2}+t{d}_{X}{(y,z)}^{2}& =& \frac{{{\rm{d}}}_{X}(y,z)}{{d}_{X}\left(x,z)}{d}_{X}{\left(x,y)}^{2}+\frac{{{\rm{d}}}_{X}\left(x,y)}{{d}_{X}\left(x,z)}{d}_{X}{(y,z)}^{2}\\ & =& \frac{{{\rm{d}}}_{X}\left(x,y){d}_{X}(y,z)}{{d}_{X}\left(x,z)}({d}_{X}\left(x,y)+{d}_{X}(y,z))\\ & =& {d}_{X}\left(x,y){d}_{X}(y,z)\\ & =& t\left(1-t){d}_{X}{\left(x,z)}^{2}.\end{array}Combining this with the ⊠\boxtimes -inequality in XXyields 0≤(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+s(1−t)dX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2=(1−s)((1−t)dX(x,y)2+tdX(y,z)2)+tsdX(z,w)2+s(1−t)dX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2=(1−s)t(1−t)dX(x,z)2+tsdX(z,w)2+s(1−t)dX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2=tsdX(z,w)2+s(1−t)dX(w,x)2−st(1−t)dX(x,z)2−s(1−s)dX(y,w)2\begin{array}{rcl}0& \le & \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\\ & =& \left(1-s)(\left(1-t){d}_{X}{\left(x,y)}^{2}+t{d}_{X}{(y,z)}^{2})+ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\\ & =& \left(1-s)t\left(1-t){d}_{X}{\left(x,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\\ & =& ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-st\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\end{array}for every s∈[0,1]s\in \left[0,1]. For any s∈(0,1]s\in \left(0,1], dividing this by ss, we obtain (1−s)dX(y,w)2≤tdX(z,w)2+(1−t)dX(w,x)2−t(1−t)dX(x,z)2.\left(1-s){d}_{X}{(y,w)}^{2}\le t{d}_{X}{\left(z,w)}^{2}+\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}.Letting s→0s\to 0in this inequality yields the desired inequality.□We now prove Theorem 1.10.Proof of Theorem 1.10First, assume that a metric space (X,dX)\left(X,{d}_{X})is Cycl4(0){{\rm{Cycl}}}_{4}\left(0). Then for any x,y,z,w∈Xx,y,z,w\in X, there exist x0,y0,z0,w0∈R2{x}_{0},{y}_{0},{z}_{0},{w}_{0}\in {{\mathbb{R}}}^{2}such that ‖x0−y0‖≤dX(x,y),‖y0−z0‖≤dX(y,z),‖z0−w0‖≤dX(z,w),‖w0−x0‖≤dX(w,x),‖x0−z0‖≥dX(x,z),‖y0−w0‖≥dX(y,w).\begin{array}{l}\Vert {x}_{0}-{y}_{0}\Vert \le {d}_{X}\left(x,y),\hspace{1.0em}\Vert {y}_{0}-{z}_{0}\Vert \le {d}_{X}(y,z),\hspace{1.0em}\Vert {z}_{0}-{w}_{0}\Vert \le {d}_{X}\left(z,w),\\ \Vert {w}_{0}-{x}_{0}\Vert \le {d}_{X}\left(w,x),\hspace{1.0em}\Vert {x}_{0}-{z}_{0}\Vert \ge {d}_{X}\left(x,z),\hspace{1.0em}\Vert {y}_{0}-{w}_{0}\Vert \ge {d}_{X}(y,w).\end{array}Therefore, for any s,t∈[0,1]s,t\in \left[0,1], we have (1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)d(y,w)2≥(1−t)(1−s)‖x0−y0‖2+t(1−s)‖y0−z0‖2+ts‖z0−w0‖2+(1−t)s‖w0−x0‖2−t(1−t)‖x0−z0‖2−s(1−s)‖y0−w0‖2=‖((1−t)x′+tz′)−((1−s)y′+sw′)‖2≥0,\begin{array}{l}\left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s)d{(y,w)}^{2}\\ \hspace{1.0em}\ge \left(1-t)\left(1-s)\Vert {x}_{0}-{y}_{0}{\Vert }^{2}+t\left(1-s)\Vert {y}_{0}-{z}_{0}{\Vert }^{2}+ts\Vert {z}_{0}-{w}_{0}{\Vert }^{2}+\left(1-t)s\Vert {w}_{0}-{x}_{0}{\Vert }^{2}-t\left(1-t)\Vert {x}_{0}-{z}_{0}{\Vert }^{2}-s\left(1-s)\Vert {y}_{0}-{w}_{0}{\Vert }^{2}\\ \hspace{1.0em}=\Vert \left(\left(1-t)x^{\prime} +tz^{\prime} )-\left(\left(1-s)y^{\prime} +sw^{\prime} ){\Vert }^{2}\ge 0,\end{array}which means that XXsatisfies the ⊠\boxtimes -inequalities.For the converse, assume that (X,dX)\left(X,{d}_{X})satisfies the ⊠\boxtimes -inequalities. Fix x,y,z,w∈Xx,y,z,w\in X. If xx, yy, zz, and wware not distinct, then we can embed {x,y,z,w}\left\{x,y,z,w\right\}isometrically into R2{{\mathbb{R}}}^{2}. So we assume that xx, yy, zz, and wware distinct. Then there exist x′,y′,z′,w′∈R2x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {{\mathbb{R}}}^{2}such that ‖x′−y′‖=dX(x,y),‖y′−z′‖=dX(y,z),‖z′−w′‖=dX(z,x),‖x′−w′‖=dX(x,w),‖w′−z′‖=dX(w,z),\begin{array}{l}\Vert x^{\prime} -y^{\prime} \Vert ={d}_{X}\left(x,y),\hspace{1.0em}\Vert y^{\prime} -z^{\prime} \Vert ={d}_{X}(y,z),\hspace{1.0em}\Vert z^{\prime} -w^{\prime} \Vert ={d}_{X}\left(z,x),\\ \Vert x^{\prime} -w^{\prime} \Vert ={d}_{X}\left(x,w),\hspace{1.0em}\Vert w^{\prime} -z^{\prime} \Vert ={d}_{X}\left(w,z),\end{array}and y′y^{\prime} and w′w^{\prime} do not lie on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} ). We consider three cases.CASE 1: [x′,z′]∩(y′,w′)≠∅\left[x^{\prime} ,z^{\prime} ]\cap (y^{\prime} ,w^{\prime} )\ne \varnothing . In this case, there exist s∈(0,1)s\in \left(0,1)and t∈[0,1]t\in \left[0,1]such that (1−t)x′+tz′=(1−s)y′+sw′.\left(1-t)x^{\prime} +tz^{\prime} =\left(1-s)y^{\prime} +sw^{\prime} .It follows that 0=∥((1−t)x′+tz′)−((1−s)y′+sw′)∥2=(1−t)(1−s)‖x′−y′‖2+t(1−s)‖y′−z′‖2+ts‖z′−w′‖2+(1−t)s‖w′−x′‖2−t(1−t)‖x′−z′‖2−s(1−s)‖y′−w′‖2=(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)‖y′−w′‖2.\begin{array}{rcl}0& =& {\parallel (\left(1-t)x^{\prime} +tz^{\prime} )-(\left(1-s)y^{\prime} +sw^{\prime} )\parallel }^{2}\\ & =& \left(1-t)\left(1-s)\Vert x^{\prime} -y^{\prime} {\Vert }^{2}+t\left(1-s)\Vert y^{\prime} -z^{\prime} {\Vert }^{2}+ts\Vert z^{\prime} -w^{\prime} {\Vert }^{2}+\left(1-t)s\Vert w^{\prime} -x^{\prime} {\Vert }^{2}-t\left(1-t)\Vert x^{\prime} -z^{\prime} {\Vert }^{2}-s\left(1-s)\Vert y^{\prime} -w^{\prime} {\Vert }^{2}\\ & =& \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s)\Vert y^{\prime} -w^{\prime} {\Vert }^{2}.\end{array}On the other hand, we have 0≤(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)20\le \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}because XXsatisfies the ⊠\boxtimes -inequalities. Comparing these yields dX(y,w)≤‖y′−w′‖{d}_{X}(y,w)\le \Vert y^{\prime} -w^{\prime} \Vert .CASE 2: [x′,z′]∩{y′,w′}≠∅\left[x^{\prime} ,z^{\prime} ]\cap \{y^{\prime} ,w^{\prime} \}\ne \varnothing . In this case, we may assume without loss of generality that y′∈[x′,z′]y^{\prime} \in \left[x^{\prime} ,z^{\prime} ]. Then we can write y′=(1−c)x′+cz′,y^{\prime} =\left(1-c)x^{\prime} +cz^{\prime} ,where (6.2)c=‖x′−y′‖‖x′−z′‖=dX(x,y)dX(x,z)∈(0,1).c=\frac{\Vert x^{\prime} -y^{\prime} \Vert }{\Vert x^{\prime} -z^{\prime} \Vert }=\frac{{{\rm{d}}}_{X}\left(x,y)}{{d}_{X}\left(x,z)}\in \left(0,1).It follows that ‖y′−w′‖2=‖(1−c)x′+cz′−w′‖2=(1−c)‖x′−w′‖2+c‖z′−w′‖2−c(1−c)‖x′−z′‖2=(1−c)dX(x,w)2+cdX(z,w)2−c(1−c)dX(x,z)2.\begin{array}{rcl}\Vert y^{\prime} -w^{\prime} {\Vert }^{2}& =& \Vert \left(1-c)x^{\prime} +cz^{\prime} -w^{\prime} {\Vert }^{2}\\ & =& \left(1-c)\Vert x^{\prime} -w^{\prime} {\Vert }^{2}+c\Vert z^{\prime} -w^{\prime} {\Vert }^{2}-c\left(1-c)\Vert x^{\prime} -z^{\prime} {\Vert }^{2}\\ & =& \left(1-c){d}_{X}{\left(x,w)}^{2}+c{d}_{X}{\left(z,w)}^{2}-c\left(1-c){d}_{X}{\left(x,z)}^{2}.\end{array}On the other hand, it follows from (6.2) and Proposition 6.1 that dX(y,w)2≤(1−c)dX(x,w)2+cdX(z,w)2−c(1−c)dX(x,z)2{d}_{X}{(y,w)}^{2}\le \left(1-c){d}_{X}{\left(x,w)}^{2}+c{d}_{X}{\left(z,w)}^{2}-c\left(1-c){d}_{X}{\left(x,z)}^{2}because we have dX(x,z)=‖x′−z′‖=‖x′−y′‖+‖y′−z′‖=dX(x,y)+dX(y,z).{d}_{X}\left(x,z)=\Vert x^{\prime} -z^{\prime} \Vert =\Vert x^{\prime} -y^{\prime} \Vert +\Vert y^{\prime} -z^{\prime} \Vert ={d}_{X}\left(x,y)+{d}_{X}(y,z).Combining these yields dX(y,w)≤‖y′−w′‖{d}_{X}(y,w)\le \Vert y^{\prime} -w^{\prime} \Vert .CASE 3: [x′,z′]∩[y′,w′]=∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]=\varnothing . Since y′y^{\prime} and w′w^{\prime} do not lie on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} ), this requires that we have ∠y′z′x′+∠x′z′w′>π\angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} w^{\prime} \gt \pi or ∠y′x′z′+∠z′x′w′>π\angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} w^{\prime} \gt \pi . We may assume without loss of generality that ∠y′z′x′+∠x′z′w′>π\angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} w^{\prime} \gt \pi . Therefore, it follows from Alexandrov’s lemma [6, p. 25] that there exist x˜,y˜,w˜∈R2\tilde{x},\tilde{y},\tilde{w}\in {{\mathbb{R}}}^{2}and z˜∈[y˜,w˜]\tilde{z}\in [\tilde{y},\tilde{w}]such that ‖x˜−y˜‖=dX(x,y),‖x˜−w˜‖=dX(x,w),‖y˜−w˜‖=dX(y,z)+dX(z,w),‖y˜−z˜‖=dX(y,z),‖z˜−w˜‖=dX(z,w),\begin{array}{l}\Vert \tilde{x}-\tilde{y}\Vert ={d}_{X}\left(x,y),\hspace{1.0em}\Vert \tilde{x}-\tilde{w}\Vert ={d}_{X}\left(x,w),\hspace{1.0em}\Vert \tilde{y}-\tilde{w}\Vert ={d}_{X}(y,z)+{d}_{X}\left(z,w),\\ \Vert \tilde{y}-\tilde{z}\Vert ={d}_{X}(y,z),\hspace{1.0em}\Vert \tilde{z}-\tilde{w}\Vert ={d}_{X}\left(z,w),\end{array}and dX(x,z)=‖x′−z′‖≤‖x˜−z˜‖.{d}_{X}\left(x,z)=\Vert x^{\prime} -z^{\prime} \Vert \le \Vert \tilde{x}-\tilde{z}\Vert .The points x˜,y˜,z˜,w˜∈R2\tilde{x},\tilde{y},\tilde{z},\tilde{w}\in {{\mathbb{R}}}^{2}have the desired property.The aforementioned three cases exhaust all possibilities, and thus, XXis Cycl4(0){{\rm{Cycl}}}_{4}\left(0).□Remark 6.2A proof of Theorem 1.10 can also be found in [11, Lemma 2.6]. However, the case corresponding to CASE 2 in the aforementioned proof is omitted in the proof in [11]. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Analysis and Geometry in Metric Spaces de Gruyter

A non-geodesic analogue of Reshetnyak’s majorization theorem

Analysis and Geometry in Metric Spaces , Volume 11 (1): 1 – Jan 1, 2023

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de Gruyter
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© 2023 the author(s), published by De Gruyter
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2299-3274
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2299-3274
DOI
10.1515/agms-2022-0151
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Abstract

1IntroductionTo find a characterization of those metric spaces that admit an isometric embedding into a CAT(κ){\rm{CAT}}\left(\kappa )space is a longstanding open problem posed by Gromov (see [3, Section 1.4], [9, Section 1.19+] and [10, §15]). On the other hand, various conditions on a general metric space are known to become equivalent to being CAT(κ){\rm{CAT}}\left(\kappa )under the assumption that the metric space is geodesic. The Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition defined by Gromov [10] is one of such conditions (see [10, §15]). In this article, we prove an analogue of Reshetnyak’s majorization theorem [14] for (possibly non-geodesic) metric spaces that satisfy the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition. Our result shows that metric spaces with the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition have more properties in common with CAT(κ){\rm{CAT}}\left(\kappa )spaces than expected. In particular, it follows from our result that every metric space with the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition satisfies the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions for all integers n≥5n\ge 5, although Gromov stated that this is apparently not true (see Subsection 1.1).For a real number κ\kappa , we denote by Mκ2{M}_{\kappa }^{2}the complete, simply connected, two-dimensional Riemannian manifold of constant Gaussian curvature κ\kappa , and by dκ{d}_{\kappa }the distance function on Mκ2{M}_{\kappa }^{2}. Let Dκ{D}_{\kappa }be the diameter of Mκ2{M}_{\kappa }^{2}. Thus, we have Dκ=πκifκ>0,∞ifκ≤0.{D}_{\kappa }=\left\{\begin{array}{ll}\frac{\pi }{\sqrt{\kappa }}\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \gt 0,\\ \infty \hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \le 0.\end{array}\right.For a positive integer nnand an integer mm, we denote by [m]n{\left[m]}_{n}the element of Z/nZ{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}represented by mm. We recall the definition of the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions introduced in [10].Definition 1.1Fix κ∈R\kappa \in {\mathbb{R}}and an integer n≥4n\ge 4. Let (X,dX)\left(X,{d}_{X})be a metric space. We say that XXis a Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )space or that XXsatisfies the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )condition if for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}\to Xwith (1.1)∑i∈Z/nZdX(f(i),f(i+[1]n))<2Dκ,\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },there exists a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}such that dκ(g(i),g(i+[1]n))≤dX(f(i),f(i+[1]n)),dκ(g(i),g(j))≥dX(f(i),f(j)){d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n}))\le {d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa }\left(g\left(i),g\left(j))\ge {d}_{X}(f\left(i),f\left(j))for any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with j≠i+[1]nj\ne i+{\left[1]}_{n}and i≠j+[1]ni\ne j+{\left[1]}_{n}.Remark 1.2The original definition of Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )spaces in [10, §7] requires the existence of a map g′:Z/nZ→Mκ′2g^{\prime} :{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa ^{\prime} }^{2}for some κ′≤κ\kappa ^{\prime} \le \kappa such that dκ′(g′(i),g′(i+[1]n))≤dX(f(i),f(i+[1]n)),dκ′(g′(i),g′(j))≥dX(f(i),f(j)){d}_{\kappa ^{\prime} }\left(g^{\prime} \left(i),g^{\prime} \left(i+{\left[1]}_{n}))\le {d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa ^{\prime} }\left(g^{\prime} \left(i),g^{\prime} \left(j))\ge {d}_{X}(f\left(i),f\left(j))for any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with j≠i+[1]nj\ne i+{\left[1]}_{n}and i≠j+[1]ni\ne j+{\left[1]}_{n}, instead of the existence of a map ggas in Definition 1.1. As mentioned in [10], this definition is equivalent to Definition 1.1. In fact, the existence of such a map g′g^{\prime} implies the existence of a map ggas in Definition 1.1 by Reshetnyak’s majorization theorem.Remark 1.3In [10, §7], assumption (1.1) is not stated explicitly. It is just remarked that we have to consider only maps f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xwith “small” images f(Z/nZ)f\left({\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}})when κ>0\kappa \gt 0.For the definition of CAT(κ){\rm{CAT}}\left(\kappa )spaces, see Definition 2.8 in this article. Concerning Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )spaces and CAT(κ){\rm{CAT}}\left(\kappa )spaces, Gromov [10] established the following fact.Theorem 1.4(Gromov [10]) Fix κ∈R\kappa \in {\mathbb{R}}. The following two assertions hold true. (1)A metric space (X,dX)\left(X,{d}_{X})is CAT(κ){\rm{CAT}}\left(\kappa )if and only if XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )and Dκ{D}_{\kappa }-geodesic. Here, we say X is Dκ{D}_{\kappa }-geodesic if any x,y∈Xx,y\in Xwith dX(x,y)<Dκ{d}_{X}\left(x,y)\lt {D}_{\kappa }can be joined by a geodesic segment in X.(2)Every CAT(κ){\rm{CAT}}\left(\kappa )space is Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )for all integers n≥4n\ge 4.On the other hand, the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition generally does not imply the isometric embeddability into a CAT(κ){\rm{CAT}}\left(\kappa )space without assuming that the metric space is Dκ{D}_{\kappa }-geodesic. In fact, Nina Lebedava constructed a 6-point Cycl4(0){{\rm{Cycl}}}_{4}\left(0)space that does not admit an isometric embedding into any CAT(0){\rm{CAT}}\left(0)space (see [1, §7.2]). Moreover, it follows from the result of Eskenazis, Mendel, and Naor [8] that there exists a Cycl4(0){{\rm{Cycl}}}_{4}\left(0)space that does not admit a coarse embedding into any CAT(0){\rm{CAT}}\left(0)space (see [17, p. 116]).The following theorem is our main result, which can be viewed as an analogue of Reshetnyak’s majorization theorem for Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )spaces.Theorem 1.5Let κ∈R\kappa \in {\mathbb{R}}. If X is a Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space, then for any integer n≥3n\ge 3, and for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xthat satisfies∑i∈Z/nZdX(f(i),f(i+[1]n))<2Dκ,f(j)≠f(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}f\left(j)\ne f\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, there exists a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}that satisfies the following two conditions: (1)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, we havedκ(g(i),g(i+[1]n))=dX(f(i),f(i+[1]n)),dκ(g(i),g(j))≥dX(f(i),f(j)).{d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa }\left(g\left(i),g\left(j))\ge {d}_{X}(f\left(i),f\left(j)).(2)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with i≠ji\ne j, we have [g(i),g(j)]∩[g(i−[1]n),g(i+[1]n)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n}),g\left(i+{\left[1]}_{n})]\ne \varnothing , where we denote by [a,b]\left[a,b]the line segment in Mκ2{M}_{\kappa }^{2}with endpoints a and b.Note that when the polygon with vertices g(i)g\left(i), i∈Z/nZi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}is non-degenerate, the condition (2) in Theorem 1.5 means that this polygon is convex.1.1Gromov’s remark about the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditionsTheorem 1.4 tells us that the Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )condition implies the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions for all integers n≥5n\ge 5under the assumption that the metric space is Dκ{D}_{\kappa }-geodesic. In the study of upper curvature bound for general metric spaces, it is natural to ask whether this implication is true without assuming that the metric space is Dκ{D}_{\kappa }-geodesic. Concerning this question, Gromov [10, §15, Remarks (b)] stated “We shall see later on that Cycl4⇒Cyclk{{\rm{Cycl}}}_{4}\Rightarrow {{\rm{Cycl}}}_{k}for all k≥5k\ge 5in the geodesic case but this is apparently not so in general.” However, we prove as a direct consequence of Theorem 1.5 that this implication actually holds true without assuming that the metric space is Dκ{D}_{\kappa }-geodesic:Theorem 1.6Let κ∈R\kappa \in {\mathbb{R}}. Every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )for all integers n≥5n\ge 5.1.2Gromov’s question about the Wirtinger inequalitiesIn [10, §6], Gromov also introduced the following conditions on a metric space.Definition 1.7Fix an integer n≥4n\ge 4. We say that a metric space (X,dX)\left(X,{d}_{X})is a Wirn{{\rm{Wir}}}_{n}space if any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xsatisfies (1.2)0≤sin2jπn∑i∈Z/nZdX(f(i),f(i+[1]n))2−sin2πn∑i∈Z/nZdX(f(i),f(i+[j]n))20\le {\sin }^{2}\frac{j\pi }{n}\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}{(f\left(i),f\left(i+{\left[1]}_{n}))}^{2}-{\sin }^{2}\frac{\pi }{n}\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}{(f\left(i),f\left(i+{\left[j]}_{n}))}^{2}for every j∈Z∩[2,n−2]j\in {\mathbb{Z}}\cap \left[2,n-2].The family of inequalities (1.2) can be thought of as a discrete and nonlinear analogue of classical Wirtinger’s inequality for functions on S1{S}^{1}. Every Euclidean space is Wirn{{\rm{Wir}}}_{n}for all integers n≥4n\ge 4, which was first proved by Pech [13] before Gromov introduced the notion of Wirn{{\rm{Wir}}}_{n}spaces. Therefore, it follows from the definition of Cycln(0){{\rm{Cycl}}}_{n}\left(0)spaces that every Cycln(0){{\rm{Cycl}}}_{n}\left(0)space is Wirn{{\rm{Wir}}}_{n}for each integer n≥4n\ge 4. Thus, for general metric spaces, the following implications are true for each integer n≥4n\ge 4: CAT(0)⇒Cycln(0)⇒Wirn.{\rm{CAT}}\left(0)\Rightarrow {{\rm{Cycl}}}_{n}\left(0)\Rightarrow {{\rm{Wir}}}_{n}.In [10, p. 133, §25, Question], Gromov posed the question of whether the implication Cycl4(0)⇒Wirn{{\rm{Cycl}}}_{4}\left(0)\Rightarrow {{\rm{Wir}}}_{n}holds true for every integer n≥5n\ge 5without assuming that the metric space is geodesic. Kondo, Toyoda, and Uehara [11] answered this question affirmatively:Theorem 1.8[11] Every Cycl4(0){{\rm{Cycl}}}_{4}\left(0)space is Wirn{{\rm{Wir}}}_{n}for all integers n≥4n\ge 4.Since Theorem 1.6 implies the stronger implication Cycl4(0)⇒Cycln(0){{\rm{Cycl}}}_{4}\left(0)\Rightarrow {{\rm{Cycl}}}_{n}\left(0)for every integer n≥4n\ge 4, Theorem 1.6 strengthens Theorem 1.8 and gives another proof of it.1.3The ⊠\boxtimes -inequalitiesIt was remarked in [10, §7] that the Cycl4(0){{\rm{Cycl}}}_{4}\left(0)condition is equivalent to the validity of a certain family of inequalities, defined as follows.Definition 1.9We say that a metric space (X,dX)\left(X,{d}_{X})satisfies the ⊠\boxtimes -inequalities if for any t,s∈[0,1]t,s\in \left[0,1]and any x,y,z,w∈Xx,y,z,w\in X, we have0≤(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2.0\le \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}.Gromov [10] and Sturm [16] proved independently that every CAT(0){\rm{CAT}}\left(0)space satisfies the ⊠\boxtimes -inequalities. The name “⊠\boxtimes -inequalities” is based on a notation used by Gromov [10] and was used in [11] and [17]. Sturm [16] called these inequalities the weighted quadruple inequalities. Gromov stated the following fact in [10, §7].Theorem 1.10[10] A metric space is Cycl4(0){{\rm{Cycl}}}_{4}\left(0)if and only if it satisfies the ⊠\boxtimes -inequalities.For a proof of Theorem 1.10, see Section 6. The following two corollaries follow from Theorems 1.5 and 1.6 immediately.Corollary 1.11If a metric space X satisfies the ⊠\boxtimes -inequalities, then for any integer n≥3n\ge 3, and for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xthat satisfies f(j)≠f(j+[1]n)f\left(j)\ne f\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, there exists a map g:Z/nZ→R2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {{\mathbb{R}}}^{2}that satisfies the following two conditions: (1)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, we have‖g(i)−g(i+[1]n)‖=dX(f(i),f(i+[1]n)),‖g(i)−g(j)‖≥dX(f(i),f(j)).\Vert g\left(i)-g\left(i+{\left[1]}_{n})\Vert ={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}\Vert g\left(i)-g\left(j)\Vert \ge {d}_{X}(f\left(i),f\left(j)).(2)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with i≠ji\ne j, we have [g(i),g(j)]∩[g(i−[1]n),g(i+[1]n)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n}),g\left(i+{\left[1]}_{n})]\ne \varnothing , where we denote by [a,b]\left[a,b]the line segment in R2{{\mathbb{R}}}^{2}with endpoints a and b.Corollary 1.12If a metric space satisfies the ⊠\boxtimes -inequalities, then it is Cycln(0){{\rm{Cycl}}}_{n}\left(0)for every integer n≥4n\ge 4.1.4Graph comparisonWe can say that the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )condition is defined by comparing embeddings of the cycle graph with nnvertices into a given metric space with embeddings of the same graph into Mκ2{M}_{\kappa }^{2}(see Definition 1.4 in [17]). By replacing the cycle graph with another graph, and Mκ2{M}_{\kappa }^{2}with another space, we can define a large number of new conditions. In recent years, such graph comparison conditions have been used by Lebedeva, Petrunin, and Zolotov [12] and the present author [17], among others, and play an interesting role in the study of the geometry of metric spaces.1.5Outline of the proof of Theorem 1.5Our proof of Theorem 1.5 is based on the idea used by Ballmann in his lecture note [4] for proving Reshetnyak’s majorization theorem. We also recommend [2, Chapter 9, L] for the proof of Reshetnyak’s majorization theorem by using Ballmann’s idea. To use Ballmann’s idea in our setting, we need the following lemma, which we prove in Section 5.Lemma 1.13Let κ∈R\kappa \in {\mathbb{R}}, and let (X,dX)\left(X,{d}_{X})be a Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space. Suppose x,y,z,w∈Xx,y,z,w\in Xand x′,y′,z′,w′∈Mκ2x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x′,y′)+dκ(y′,z′)+dκ(z′,w′)+dκ(w′,x′)<2Dκ,dX(x,y)≤dκ(x′,y′),dX(y,z)≤dκ(y′,z′),dX(z,w)≤dκ(z′,w′),dX(w,x)≤dκ(w′,x′),dκ(x′,z′)≤dX(x,z).\begin{array}{l}{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )+{d}_{\kappa }\left(w^{\prime} ,x^{\prime} )\lt 2{D}_{\kappa },\\ {d}_{X}\left(x,y)\le {d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{X}(y,z)\le {d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{X}\left(z,w)\le {d}_{\kappa }\left(z^{\prime} ,w^{\prime} ),\\ {d}_{X}\left(w,x)\le {d}_{\kappa }\left(w^{\prime} ,x^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z).\end{array}Then we have dX(y,w)≤dκ(y′,p)+dκ(p,w′){d}_{X}(y,w)\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} )for every p∈[x′,z′]p\in \left[x^{\prime} ,z^{\prime} ].Once we have proved Lemma 1.13, we can prove Theorem 1.5 by just following the idea of Ballmann mentioned above. We show this in Section 3.Remark 1.14For the use of Lemma 1.13 to prove Theorem 1.5 by following Ballmann’s idea, it is no problem to replace the assumption dκ(x′,z′)≤dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z)in Lemma 1.13 with the equality dκ(x′,z′)=dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )={d}_{X}\left(x,z). However, proving the dκ(x′,z′)=dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )={d}_{X}\left(x,z)version of Lemma 1.13 still requires a similar argument as we do in this article for proving Lemma 1.13.For κ≤0\kappa \le 0, Lemma 1.13 can be proved by a straightforward computation (see Remark 2.7). To prove Lemma 1.13 for a general real number κ\kappa , we prove two lemmas in Section 4, which generalize Alexandrov’s lemma [6, p. 25]. By using those lemmas, we prove Lemma 1.13 in Section 5.Remark 1.15It was pointed out by an anonymous referee that the dκ(x′,z′)=dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )={d}_{X}\left(x,z)version of Lemma 1.13 can also be proved by using Zalgaller’s arm lemma [18] instead of the generalized Alexandrov’s lemma.1.6Organization of the paperThis article is organized as follows. In Section 2, we recall some definitions and results from metric geometry and the geometry of Mκ2{M}_{\kappa }^{2}. In Section 3, we prove Theorems 1.5 and 1.6 by using Lemma 1.13. In Section 4, we prove two lemmas, which generalize Alexandrov’s lemma [6, p. 25]. In Section 5, we prove Lemma 1.13 by using those lemmas proved in Section 4. In Section 6, we present a proof of Theorem 1.10 for completeness.2PreliminariesIn this section, we recall some definitions and results from metric geometry and the geometry of Mκ2{M}_{\kappa }^{2}.2.1GeodesicsLet (X,dX)\left(X,{d}_{X})be a metric space. A geodesic in XXis an isometric embedding of an interval of the real line into XX. For x,y∈Xx,y\in X, a geodesic segment with endpoints x and y is the image of a geodesic γ:[0,dX(x,y)]→X\gamma :\left[0,{d}_{X}\left(x,y)]\to Xwith γ(0)=x\gamma \left(0)=x, γ(dX(x,y))=y\gamma \left({d}_{X}\left(x,y))=y. If there exists a unique geodesic segment with endpoints xxand yy, we denote it by [x,y]\left[x,y]. We also denote the sets [x,y]⧹{x,y}\left[x,y]\setminus \left\{x,y\right\}, [x,y]⧹{x}\left[x,y]\setminus \left\{x\right\}and [x,y]⧹{y}\left[x,y]\setminus \{y\}by (x,y)\left(x,y), (x,y]\left(x,y]and [x,y)\left[x,y), respectively. A metric space XXis called geodesic if for any x,y∈Xx,y\in X, there exists a geodesic segment with endpoints xxand yy. Let D∈(0,∞)D\in \left(0,\infty ). We say that XXis D-geodesic if for any x,y∈Xx,y\in Xwith dX(x,y)<D{d}_{X}\left(x,y)\lt D, there exists a geodesic segment with endpoints xxand yy. A subset SSof XXis called convex if for any x,y∈Sx,y\in S, every geodesic segment in XXwith endpoints xxand yyis contained in SS. If this condition holds for any x,y∈Sx,y\in Swith dX(x,y)<D{d}_{X}\left(x,y)\lt D, then SSis called D-convex.2.2The geometry of Mκ2{M}_{\kappa }^{2}Let κ∈R\kappa \in {\mathbb{R}}. For any x,y∈Mκ2x,y\in {M}_{\kappa }^{2}with dκ(x,y)<Dκ{d}_{\kappa }\left(x,y)\lt {D}_{\kappa }, there exists a unique geodesic segment [x,y]\left[x,y]with endpoints xxand yy. We mean by a line in Mκ2{M}_{\kappa }^{2}the image of an isometric embedding of R{\mathbb{R}}into Mκ2{M}_{\kappa }^{2}if κ≤0\kappa \le 0, and a great circle in Mκ2{M}_{\kappa }^{2}if κ>0\kappa \gt 0. Then for any two distinct points x,y∈Mκ2x,y\in {M}_{\kappa }^{2}with dκ(x,y)<Dκ{d}_{\kappa }\left(x,y)\lt {D}_{\kappa }, there exits a unique line through xxand yy, which we denote by ℓ(x,y)\ell \left(x,y). For any line ℓ\ell in Mκ2{M}_{\kappa }^{2}, Mκ2⧹ℓ{M}_{\kappa }^{2}\setminus \ell consists of exactly two connected components. We call each connected component of Mκ2⧹ℓ{M}_{\kappa }^{2}\setminus \ell a side of ℓ\ell . One side of ℓ\ell is called the opposite side of the other. If x,y∈Mκ2x,y\in {M}_{\kappa }^{2}lie on the same side of a line, then dκ(x,y)<Dκ{d}_{\kappa }\left(x,y)\lt {D}_{\kappa }. Each side of a line is a convex subset of Mκ2{M}_{\kappa }^{2}. For a subset SSof a side of some line in Mκ2{M}_{\kappa }^{2}, the convex hull of S is the intersection of all convex subsets of Mκ2{M}_{\kappa }^{2}containing SS, or equivalently, the minimal convex subset of Mκ2{M}_{\kappa }^{2}containing SS. We denote the convex hull of SSby conv(S){\rm{conv}}\left(S). We recall the following well-known fact, which is trivial when κ≤0\kappa \le 0.Proposition 2.1Let κ∈R\kappa \in {\mathbb{R}}, and let n≥3n\ge 3be an integer. Suppose f:Z/nZ→Mκ2f:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}is a map such that ∑i∈Z/nZdκ(f(i),f(i+[1]n))<2Dκ{\sum }_{i\in {\mathbb{Z}}\text{/}n{\mathbb{Z}}}{d}_{\kappa }(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa }. Then there exists a line L in Mκ2{M}_{\kappa }^{2}such that all f(i)f\left(i), i∈Z/nZi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}lie on one side of L, and thus, conv(f(Z/nZ)){\rm{conv}}(f\left({\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}))is contained in one side of L.For x,y,z∈Mκ2x,y,z\in {M}_{\kappa }^{2}with 0<dκ(x,y)<Dκ0\lt {d}_{\kappa }\left(x,y)\lt {D}_{\kappa }and 0<dκ(y,z)<Dκ0\lt {d}_{\kappa }(y,z)\lt {D}_{\kappa }, we denote by ∠xyz∈[0,π]\angle xyz\in \left[0,\pi ]the interior angle measure at yyof the (possibly degenerate) triangle with vertices xx, yy, and zz. By the law of cosines for Mκ2{M}_{\kappa }^{2}(see [6, p. 24]), ∠xyz∈[0,π]\angle xyz\in \left[0,\pi ]satisfies the following formula: cos∠xyz=dκ(x,y)2+dκ(y,z)2−dκ(z,x)22dκ(x,y)dκ(y,z),ifκ=0,cosh(−κdκ(x,y))cosh(−κdκ(y,z))−cosh(−κdκ(z,x))sinh(−κdκ(x,y))sinh(−κdκ(y,z)),ifκ<0,cos(κdκ(z,x))−cos(κdκ(x,y))cos(κdκ(y,z))sin(κdκ(x,y))sin(κdκ(y,z)),ifκ>0.\cos \angle xyz=\left\{\begin{array}{ll}\frac{{{\rm{d}}}_{\kappa }{\left(x,y)}^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(z,x)}^{2}}{2{d}_{\kappa }\left(x,y){d}_{\kappa }(y,z)},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa =0,\\ \frac{\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))-\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(z,x))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\sinh (\sqrt{-\kappa }{d}_{\kappa }(y,z))},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \lt 0,\\ \frac{\cos (\sqrt{\kappa }{d}_{\kappa }\left(z,x))-\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\sin (\sqrt{\kappa }{d}_{\kappa }(y,z))},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}\kappa \gt 0.\end{array}\right.The following proposition follows immediately from the law of cosines.Proposition 2.2Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,x′,y′,z′∈Mκ2x,y,z,x^{\prime} ,y^{\prime} ,z^{\prime} \in {M}_{\kappa }^{2}are points such that0<dκ(x,y)=dκ(x′,y′)<Dκ,0<dκ(y,z)=dκ(y′,z′)<Dκ.0\lt {d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )\lt {D}_{\kappa },\hspace{1.0em}0\lt {d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} )\lt {D}_{\kappa }.Then dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )if and only if ∠xyz≤∠x′y′z′\angle xyz\le \angle x^{\prime} y^{\prime} z^{\prime} . Moreover, dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} )if and only if ∠xyz=∠x′y′z′\angle xyz=\angle x^{\prime} y^{\prime} z^{\prime} .The following proposition also follows from the law of cosines.Proposition 2.3Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z∈Mκ2x,y,z\in {M}_{\kappa }^{2}are three distinct points such thatdκ(x,y)+dκ(y,z)<Dκ,dκ(x,y)≤dκ(y,z).{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }\left(x,y)\le {d}_{\kappa }(y,z).Suppose x′,y′,z′∈Mκ2x^{\prime} ,y^{\prime} ,z^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x,y)=dκ(x′,y′),dκ(y,z)=dκ(y′,z′),dκ(z,x)≤dκ(z′,x′).{d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(z,x)\le {d}_{\kappa }\left(z^{\prime} ,x^{\prime} ).Then ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxz.ProofWe consider three cases.CASE 1: κ=0\kappa =0. In this case, we have ∠yzx≤∠yxz\angle yzx\le \angle yxzand ∠y′z′x′≤∠y′x′z′\angle y^{\prime} z^{\prime} x^{\prime} \le \angle y^{\prime} x^{\prime} z^{\prime} by hypothesis, and therefore, ∠yzx≤π/2\angle yzx\le \pi \hspace{0.1em}\text{/}\hspace{0.1em}2and ∠y′z′x′≤π/2\angle y^{\prime} z^{\prime} x^{\prime} \le \pi \hspace{0.1em}\text{/}\hspace{0.1em}2. It follows that 0≤dκ(x,z)2+dκ(y,z)2−dκ(x,y)2,0≤dκ(x′,z′)2+dκ(y′,z′)2−dκ(x′,y′)2=dκ(x′,z′)2+dκ(y,z)2−dκ(x,y)2,\begin{array}{rcl}0& \le & {d}_{\kappa }{\left(x,z)}^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2},\\ 0& \le & {d}_{\kappa }{\left(x^{\prime} ,z^{\prime} )}^{2}+{d}_{\kappa }{(y^{\prime} ,z^{\prime} )}^{2}-{d}_{\kappa }{\left(x^{\prime} ,y^{\prime} )}^{2}={d}_{\kappa }{\left(x^{\prime} ,z^{\prime} )}^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2},\end{array}and therefore, we have (2.1)0≤α2+dκ(y,z)2−dκ(x,y)20\le {\alpha }^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2}for any α∈[dκ(x,z),dκ(x′,z′)]\alpha \in \left[{d}_{\kappa }\left(x,z),{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )]. Define a function f1:(0,∞)→R{f}_{1}:\left(0,\infty )\to {\mathbb{R}}by f1(α)=dκ(x,y)2+α2−dκ(y,z)22dκ(x,y)α.{f}_{1}\left(\alpha )=\frac{{{\rm{d}}}_{\kappa }{\left(x,y)}^{2}+{\alpha }^{2}-{d}_{\kappa }{(y,z)}^{2}}{2{d}_{\kappa }\left(x,y)\alpha }.Then we have ddαf1(α)=12dκ(x,y)α2(α2+dκ(y,z)2−dκ(x,y)2)≥0\frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{1}\left(\alpha )=\frac{1}{2{d}_{\kappa }\left(x,y){\alpha }^{2}}({\alpha }^{2}+{d}_{\kappa }{(y,z)}^{2}-{d}_{\kappa }{\left(x,y)}^{2})\ge 0for any α∈[dκ(x,z),dκ(x′,z′)]\alpha \in \left[{d}_{\kappa }\left(x,z),{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )]by (2.1), which implies that ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxzbecause f1(dκ(x,z))=cos∠yxz,f1(dκ(x′,z′))=cos∠y′x′z′{f}_{1}\left({d}_{\kappa }\left(x,z))=\cos \angle yxz,\hspace{1.0em}{f}_{1}\left({d}_{\kappa }\left(x^{\prime} ,z^{\prime} ))=\cos \angle y^{\prime} x^{\prime} z^{\prime} by the law of cosines.CASE 2: κ<0\kappa \lt 0. In this case, we define a function f2:(0,∞)→R{f}_{2}:\left(0,\infty )\to {\mathbb{R}}by f2(α)=cosh(−κdκ(x,y))cosh(−κα)−cosh(−κdκ(y,z))sinh(−κdκ(x,y))sinh(−κα).{f}_{2}\left(\alpha )=\frac{\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\cosh (\sqrt{-\kappa }\alpha )-\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))\sinh (\sqrt{-\kappa }\alpha )}.Then we have ddαf2(α)=−κ(cosh(−κdκ(y,z))cosh(−κα)−cosh(−κdκ(x,y)))sinh(−κdκ(x,y))sinh2(−κα)≥−κ(cosh(−κdκ(y,z))−cosh(−κdκ(x,y)))sinh(−κdκ(x,y))sinh2(−κα)≥0\begin{array}{rcl}\frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{2}\left(\alpha )& =& \frac{\sqrt{-\kappa }(\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))\cosh (\sqrt{-\kappa }\alpha )-\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)){\sinh }^{2}(\sqrt{-\kappa }\alpha )}\\ & \ge & \frac{\sqrt{-\kappa }(\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))-\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)))}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y)){\sinh }^{2}(\sqrt{-\kappa }\alpha )}\ge 0\end{array}for any α∈(0,∞)\alpha \in \left(0,\infty )by hypothesis. This implies that ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxzbecause f2(dκ(x,z))=cos∠yxz,f2(dκ(x′,z′))=cos∠y′x′z′{f}_{2}\left({d}_{\kappa }\left(x,z))=\cos \angle yxz,\hspace{1.0em}{f}_{2}\left({d}_{\kappa }\left(x^{\prime} ,z^{\prime} ))=\cos \angle y^{\prime} x^{\prime} z^{\prime} by the law of cosines.CASE 3: κ>0\kappa \gt 0. In this case, we define a function f3:(0,Dκ)→R{f}_{3}:\left(0,{D}_{\kappa })\to {\mathbb{R}}by f3(α)=cos(κdκ(y,z))−cos(κdκ(x,y))cos(κα)sin(κdκ(x,y))sin(κα).{f}_{3}\left(\alpha )=\frac{\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))-\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\cos (\sqrt{\kappa }\alpha )}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,y))\sin (\sqrt{\kappa }\alpha )}.Then ddαf3(α)=κ(cos(κdκ(x,y))−cos(κdκ(y,z))cos(κα))sin(κdκ(x,y))sin2(κα).\frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{3}\left(\alpha )=\frac{\sqrt{\kappa }(\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\cos (\sqrt{\kappa }\alpha ))}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,y)){\sin }^{2}(\sqrt{\kappa }\alpha )}.If dκ(y,z)≤Dκ/2{d}_{\kappa }(y,z)\le {D}_{\kappa }\hspace{-0.08em}\text{/}\hspace{-0.02em}2, then 0<κdκ(x,y)≤κdκ(y,z)≤π20\lt \sqrt{\kappa }{d}_{\kappa }\left(x,y)\le \sqrt{\kappa }{d}_{\kappa }(y,z)\le \frac{\pi }{2}by hypothesis, and therefore, cos(κdκ(x,y))−cos(κdκ(y,z))cos(κα)≥cos(κdκ(x,y))−cos(κdκ(y,z))≥0\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\cos (\sqrt{\kappa }\alpha )\ge \cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\ge 0for any α∈(0,Dκ)\alpha \in \left(0,{D}_{\kappa }). If dκ(y,z)>Dκ/2{d}_{\kappa }(y,z)\gt {D}_{\kappa }\hspace{0.1em}\text{/}\hspace{0.1em}2, then 0<κdκ(x,y)<π2<κdκ(y,z)<π,κdκ(x,y)<π−κdκ(y,z)0\lt \sqrt{\kappa }{d}_{\kappa }\left(x,y)\lt \frac{\pi }{2}\lt \sqrt{\kappa }{d}_{\kappa }(y,z)\lt \pi ,\hspace{1.0em}\sqrt{\kappa }{d}_{\kappa }\left(x,y)\lt \pi -\sqrt{\kappa }{d}_{\kappa }(y,z)by hypothesis, and therefore, cos(κdκ(x,y))−cos(κdκ(y,z))cos(κα)≥cos(κdκ(x,y))+cos(κdκ(y,z))=cos(κdκ(x,y))−cos(π−κdκ(y,z))>cos(κdκ(x,y))−cos(κdκ(x,y))=0\begin{array}{l}\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\cos (\sqrt{\kappa }\alpha )\ge \cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))+\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))\\ \hspace{1.0em}=\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\pi -\sqrt{\kappa }{d}_{\kappa }(y,z))\gt \cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))-\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))=0\end{array}for any α∈(0,Dκ)\alpha \in \left(0,{D}_{\kappa }). Thus, we always have 0≤ddαf3(α)0\le \frac{{\rm{d}}}{{\rm{d}}\alpha }{f}_{3}\left(\alpha )for any α∈(0,Dκ)\alpha \in \left(0,{D}_{\kappa }). This implies that ∠y′x′z′≤∠yxz\angle y^{\prime} x^{\prime} z^{\prime} \le \angle yxzbecause f3(dκ(x,z))=cos∠yxz,f3(dκ(x′,z′))=cos∠y′x′z′{f}_{3}\left({d}_{\kappa }\left(x,z))=\cos \angle yxz,\hspace{1.0em}{f}_{3}\left({d}_{\kappa }\left(x^{\prime} ,z^{\prime} ))=\cos \angle y^{\prime} x^{\prime} z^{\prime} by the law of cosines.The aforementioned three cases exhaust all possibilities.□The following formulas follow from straightforward computation.Proposition 2.4Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z∈Mκ2x,y,z\in {M}_{\kappa }^{2}are points such that x≠zx\ne zand dκ(a,b)<Dκ{d}_{\kappa }\left(a,b)\lt {D}_{\kappa }for any a,b∈{x,y,z}a,b\in \left\{x,y,z\right\}. Suppose γ:[0,dκ(x,z)]→Mκ2\gamma :\left[0,{d}_{\kappa }\left(x,z)]\to {M}_{\kappa }^{2}is the geodesic such that γ(0)=x\gamma \left(0)=xand γ(dκ(x,z))=z\gamma \left({d}_{\kappa }\left(x,z))=z. Let t∈[0,1]t\in \left[0,1], and let p=γ(tdκ(x,z))p=\gamma (t{d}_{\kappa }\left(x,z)). If κ=0\kappa =0, thendκ(y,p)2=(1−t)dκ(x,y)2+tdκ(y,z)2−t(1−t)dκ(x,z)2.{d}_{\kappa }{(y,p)}^{2}=\left(1-t){d}_{\kappa }{\left(x,y)}^{2}+t{d}_{\kappa }{(y,z)}^{2}-t\left(1-t){d}_{\kappa }{\left(x,z)}^{2}.If κ<0\kappa \lt 0, thencosh(−κdκ(y,p))=1sinh(−κdκ(x,z))(sinh(−κ(1−t)dκ(x,z))cosh(−κdκ(x,y))+sinh(−κtdκ(x,z))cosh(−κdκ(y,z))).\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,p))=\frac{1}{\sinh (\sqrt{-\kappa }{d}_{\kappa }\left(x,z))}(\sinh (\sqrt{-\kappa }\left(1-t){d}_{\kappa }\left(x,z))\cosh (\sqrt{-\kappa }{d}_{\kappa }\left(x,y))+\sinh (\sqrt{-\kappa }t{d}_{\kappa }\left(x,z))\cosh (\sqrt{-\kappa }{d}_{\kappa }(y,z))).If κ>0\kappa \gt 0, thencos(κdκ(y,p))=sin(κ(1−t)dκ(x,z))cos(κdκ(x,y))+sin(κtdκ(x,z))cos(κdκ(y,z))sin(κdκ(x,z)).\cos (\sqrt{\kappa }{d}_{\kappa }(y,p))=\frac{\sin (\sqrt{\kappa }\left(1-t){d}_{\kappa }\left(x,z))\cos (\sqrt{\kappa }{d}_{\kappa }\left(x,y))+\sin (\sqrt{\kappa }t{d}_{\kappa }\left(x,z))\cos (\sqrt{\kappa }{d}_{\kappa }(y,z))}{\sin (\sqrt{\kappa }{d}_{\kappa }\left(x,z))}.The next corollary follows immediately from Proposition 2.4Corollary 2.5Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,y˜∈Mκ2x,y,z,\tilde{y}\in {M}_{\kappa }^{2}are points such thatdκ(x,y)≤dκ(x,y˜)<Dκ,dκ(y,z)≤dκ(y˜,z)<Dκ,0<dκ(x,z)<Dκ.{d}_{\kappa }\left(x,y)\le {d}_{\kappa }\left(x,\tilde{y})\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }(y,z)\le {d}_{\kappa }(\tilde{y},z)\lt {D}_{\kappa },\hspace{1.0em}0\lt {d}_{\kappa }\left(x,z)\lt {D}_{\kappa }.Then we have dκ(y,p)≤dκ(y˜,p){d}_{\kappa }(y,p)\le {d}_{\kappa }(\tilde{y},p)for any p∈[x,z]p\in \left[x,z].Although it is not necessary for our purpose, it is worth noting that for κ∈(−∞,0]\kappa \in \left(-\infty ,0], Proposition 2.4 also implies the following corollary.Corollary 2.6Let κ∈(−∞,0]\kappa \in \left(-\infty ,0]. Suppose x,y,z,x˜,y˜,z˜∈Mκ2x,y,z,\tilde{x},\tilde{y},\tilde{z}\in {M}_{\kappa }^{2}are points such thatdκ(x,y)≤dκ(x˜,y˜)<Dκ,dκ(y,z)≤dκ(y˜,z˜)<Dκ,0<dκ(x˜,z˜)≤dκ(x,z)<Dκ.{d}_{\kappa }\left(x,y)\le {d}_{\kappa }\left(\tilde{x},\tilde{y})\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }(y,z)\le {d}_{\kappa }(\tilde{y},\tilde{z})\lt {D}_{\kappa },\hspace{1.0em}0\lt {d}_{\kappa }\left(\tilde{x},\tilde{z})\le {d}_{\kappa }\left(x,z)\lt {D}_{\kappa }.Let γ:[0,dκ(x,z)]→Mκ2\gamma :\left[0,{d}_{\kappa }\left(x,z)]\to {M}_{\kappa }^{2}and γ˜:[0,dκ(x˜,z˜)]→Mκ2\tilde{\gamma }:\left[0,{d}_{\kappa }\left(\tilde{x},\tilde{z})]\to {M}_{\kappa }^{2}be the geodesics such thatγ(0)=x,γ(dκ(x,z))=z,γ˜(0)=x˜,γ˜(dκ(x˜,z˜))=z˜.\gamma \left(0)=x,\hspace{1.0em}\gamma \left({d}_{\kappa }\left(x,z))=z,\hspace{1.0em}\tilde{\gamma }\left(0)=\tilde{x},\hspace{1.0em}\tilde{\gamma }\left({d}_{\kappa }\left(\tilde{x},\tilde{z}))=\tilde{z}.Fix t∈[0,1]t\in \left[0,1], and set p=γ(tdκ(x,z))p=\gamma \left(t{d}_{\kappa }\left(x,z)), p˜=γ˜(tdκ(x˜,z˜))\tilde{p}=\tilde{\gamma }\left(t{d}_{\kappa }\left(\tilde{x},\tilde{z})). Then dκ(y,p)≤dκ(y˜,p˜){d}_{\kappa }(y,p)\le {d}_{\kappa }(\tilde{y},\tilde{p}).Remark 2.7For the case in which κ∈(−∞,0]\kappa \in \left(-\infty ,0], we can prove Lemma 1.13 easily by using Corollary 2.6. To prove Lemma 1.13 for general κ∈R\kappa \in {\mathbb{R}}, we will prove a generalization of Alexandrov’s lemma [6, p. 25] in Section 4.2.3CAT(κ){\rm{CAT}}\left(\kappa )spacesA geodesic triangle in a metric space XXis a triple △=(γ1,γ2,γ3)\bigtriangleup =\left({\gamma }_{1},{\gamma }_{2},{\gamma }_{3})of geodesics γi:[ai,bi]→X{\gamma }_{i}:\left[{a}_{i},{b}_{i}]\to Xsuch that γ1(b1)=γ2(a2){\gamma }_{1}\left({b}_{1})={\gamma }_{2}\left({a}_{2}), γ2(b2)=γ3(a3){\gamma }_{2}\left({b}_{2})={\gamma }_{3}\left({a}_{3})and γ3(b3)=γ1(a1){\gamma }_{3}\left({b}_{3})={\gamma }_{1}\left({a}_{1}). Let κ∈R\kappa \in {\mathbb{R}}. If the perimeter ∑i=13∣bi−ai∣{\sum }_{i=1}^{3}| {b}_{i}-{a}_{i}| of the geodesic triangle △\bigtriangleup is less than 2Dκ2{D}_{\kappa }, then there exists a geodesic triangle △κ=(γ1κ,γ2κ,γ3κ){\bigtriangleup }^{\kappa }=\left({\gamma }_{1}^{\kappa },{\gamma }_{2}^{\kappa },{\gamma }_{3}^{\kappa }), γiκ:[ai,bi]→Mκ2{\gamma }_{i}^{\kappa }:\left[{a}_{i},{b}_{i}]\to {M}_{\kappa }^{2}in Mκ2{M}_{\kappa }^{2}. Such a geodesic triangle △κ{\bigtriangleup }^{\kappa }is unique up to isometry of Mκ2{M}_{\kappa }^{2}. The geodesic triangle △\bigtriangleup is said to be κ\kappa -thin if dX(γi(s),γj(t))≤dκ(γiκ(s),γjκ(t)){d}_{X}\left({\gamma }_{i}\left(s),{\gamma }_{j}\left(t))\le {d}_{\kappa }\left({\gamma }_{i}^{\kappa }\left(s),{\gamma }_{j}^{\kappa }\left(t))for any i,j∈{1,2,3}i,j\in \left\{1,2,3\right\}, any s∈[ai,bi]s\in \left[{a}_{i},{b}_{i}], and any t∈[aj,bj]t\in \left[{a}_{j},{b}_{j}].Definition 2.8Let κ∈R\kappa \in {\mathbb{R}}. A metric space XXis called a CAT(κ){\rm{CAT}}\left(\kappa )space if XXis Dκ{D}_{\kappa }-geodesic, and any geodesic triangle in XXwith perimeter <2Dκ\lt 2{D}_{\kappa }is κ\kappa -thin.By definition, Mκ2{M}_{\kappa }^{2}is a CAT(κ){\rm{CAT}}\left(\kappa )space. It is easily observed that if (X,dX)\left(X,{d}_{X})is a CAT(κ){\rm{CAT}}\left(\kappa )space, then for any x,y∈Xx,y\in Xwith dX(x,y)<Dκ{d}_{X}\left(x,y)\lt {D}_{\kappa }, there exists the unique geodesic segment with endpoints xxand yy. Every Dκ{D}_{\kappa }-convex subset of a CAT(κ){\rm{CAT}}\left(\kappa )space equipped with the induced metric is a CAT(κ){\rm{CAT}}\left(\kappa )space. For a detailed exposition of CAT(κ){\rm{CAT}}\left(\kappa )spaces, see [2,6,7]. We note that in [2, p. 96, Definition 9.3], the term CAT(κ){\rm{CAT}}\left(\kappa )space is used with somewhat different meaning.3Ballmann’s argument for our settingIn this section, we prove Theorems 1.5 and 1.6 by using Lemma 1.13. As we mentioned in Section 1, our proof is based on the idea used by Ballmann in his lecture note [4] for proving Reshetnyak’s majorization theorem. We will prove Lemma 1.13 later in Section 5. First, we define the following conditions by slightly modifying the definition of the Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )conditions.Definition 3.1Fix κ∈R\kappa \in {\mathbb{R}}and a positive integer nn. We say that a metric space (X,dX)\left(X,{d}_{X})is a Cycln′(κ){{\rm{Cycl}}}_{n}^{^{\prime} }\left(\kappa )space if for any map f:Z/nZ→Xf:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xthat satisfies ∑i∈Z/nZdX(f(i),f(i+[1]n))<2Dκ,f(j)≠f(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}f\left(j)\ne f\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, there exists a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}that satisfies the following two conditions: (1)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, dκ(g(i),g(i+[1]n))=dX(f(i),f(i+[1]n)),dκ(g(i),g(j))≥dX(f(i),f(j)).{d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n})),\hspace{1.0em}{d}_{\kappa }\left(g\left(i),g\left(j))\ge {d}_{X}(f\left(i),f\left(j)).(2)For any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}with i≠ji\ne j, [g(i),g(j)]∩[g(i−[1]n),g(i+[1]n)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n}),g\left(i+{\left[1]}_{n})]\ne \varnothing .We call such a map g:Z/nZ→Mκ2g:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}that satisfies the aforementioned two conditions a comparison map for ff.Note that when the polygon with vertices g(i)g\left(i), i∈Z/nZi\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}is non-degenerate, the condition (2) in Definition 3.1 means that this polygon is convex. By definition, every metric space is Cycl1′(κ){{\rm{Cycl}}}_{1}^{^{\prime} }\left(\kappa ), Cycl2′(κ){{\rm{Cycl}}}_{2}^{^{\prime} }\left(\kappa ), and Cycl3′(κ){{\rm{Cycl}}}_{3}^{^{\prime} }\left(\kappa )for all κ∈R\kappa \in {\mathbb{R}}. It is also easily seen that for any κ∈R\kappa \in {\mathbb{R}}and any integer n≥4n\ge 4, if a metric space is Cyclm′(κ){{\rm{Cycl}}}_{m}^{^{\prime} }\left(\kappa )for all m∈Z∩[1,n]m\in {\mathbb{Z}}\cap \left[1,n], then it is Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa ). To prove Theorem 1.5, it clearly suffices to prove that every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycln′(κ){{\rm{Cycl}}}_{n}^{^{\prime} }\left(\kappa )for all positive integers nn.Proof of Theorem 1.5 by using Lemma 1.13Fix κ∈R\kappa \in {\mathbb{R}}. We will prove that every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycln′(κ){{\rm{Cycl}}}_{n}^{^{\prime} }\left(\kappa )for all positive integers nnby induction on nn. As we mentioned above, every metric space is Cycl1′(κ){{\rm{Cycl}}}_{1}^{^{\prime} }\left(\kappa ), Cycl2′(κ){{\rm{Cycl}}}_{2}^{^{\prime} }\left(\kappa ), and Cycl3′(κ){{\rm{Cycl}}}_{3}^{^{\prime} }\left(\kappa )trivially. Fix an integer n≥3n\ge 3, and assume that every Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space is Cycll′(κ){{\rm{Cycl}}}_{l}^{^{\prime} }\left(\kappa )for all l∈Z∩[1,n]l\in {\mathbb{Z}}\cap \left[1,n]. Let (X,dX)\left(X,{d}_{X})be an arbitrary Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space, and let f:Z/(n+1)Z→Xf:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}\to Xbe an arbitrary map that satisfies (3.1)∑i∈Z/(n+1)ZdX(f(i),f(i+[1]n+1))<2Dκ,f(j)≠f(j+[1]n+1)\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}}{d}_{X}(f\left(i),f\left(i+{\left[1]}_{n+1}))\lt 2{D}_{\kappa },\hspace{1.0em}f\left(j)\ne f\left(j+{\left[1]}_{n+1})for every j∈Z/(n+1)Zj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n+1){\mathbb{Z}}. Then what we need to prove is the existence of a comparison map for ff. Define a map f0:Z/nZ→X{f}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Xby f0([m]n)=f([m]n+1),m∈Z∩[0,n−1].{f}_{0}\left({\left[m]}_{n})=f\left({\left[m]}_{n+1}),\hspace{1.0em}m\in {\mathbb{Z}}\cap \left[0,n-1].We consider two cases.CASE 1: The map f satisfies f([n−1]n+1)≠f([0]n+1)f\left({\left[n-1]}_{n+1})\ne f\left({\left[0]}_{n+1}). In this case, it follows from (3.1) that ∑i∈Z/nZdX(f0(i),f0(i+[1]n))<2Dκ,f0(j)≠f0(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}}}{d}_{X}({f}_{0}\left(i),{f}_{0}\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}{f}_{0}\left(j)\ne {f}_{0}\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, and so there exists a comparison map g0:Z/nZ→Mκ2{g}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}for f0{f}_{0}by the inductive hypothesis. Since dκ(g0([n−1]n),g0([0]n))=dX(f([n−1]n+1),f([0]n+1)),{d}_{\kappa }\left({g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n}))={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[0]}_{n+1})),there exists p∈Mκ2p\in {M}_{\kappa }^{2}such that dκ(g0([n−1]n),p)=dX(f([n−1]n+1),f([n]n+1)),dκ(p,g0([0]n))=dX(f([n]n+1),f([0]n+1)).{d}_{\kappa }\left({g}_{0}\left({\left[n-1]}_{n}),p)={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1})),\hspace{1em}{d}_{\kappa }\left(p,{g}_{0}\left({\left[0]}_{n}))={d}_{X}(f\left({\left[n]}_{n+1}),f\left({\left[0]}_{n+1})).Because g0{g}_{0}is a comparison map for f0{f}_{0}, the condition (2) in Definition 3.1 guarantees that for any i,j∈Z/nZi,j\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}, g0(i){g}_{0}\left(i)and g0(j){g}_{0}\left(j)do not lie on opposite sides of ℓ(g0([n−1]n),g0([0]n))\ell \left({g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n})). So we may assume that ppis not on the same side of ℓ(g0([n−1]n),g0([0]n))\ell \left({g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n}))as g0(i){g}_{0}\left(i)for every i∈Z/nZi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}. Define a map g:Z/(n+1)Z→Mκ2g:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n+1){\mathbb{Z}}\to {M}_{\kappa }^{2}by g([m]n+1)=g0([m]n),ifm∈Z∩[0,n−1],p,ifm=n.g\left({\left[m]}_{n+1})=\left\{\phantom{\rule[-1.33em]{}{0ex}}\begin{array}{ll}{g}_{0}\left({\left[m]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-1],\\ p,\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n.\end{array}\right.Then it is straightforward to see that (3.2)dκ(g([l]n+1),g([m]n+1))≥dX(f([l]n+1),f([m]n+1)){d}_{\kappa }(g\left({\left[l]}_{n+1}),g\left({\left[m]}_{n+1}))\ge {d}_{X}(f\left({\left[l]}_{n+1}),f\left({\left[m]}_{n+1}))for any l,m∈Z∩[0,n−1]l,m\in {\mathbb{Z}}\cap \left[0,n-1], (3.3)dκ(g(i),g(i+[1]n+1))=dX(f(i),f(i+[1]n+1)){d}_{\kappa }\left(g\left(i),g\left(i+{\left[1]}_{n+1}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n+1}))for any i∈Z/(n+1)Zi\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n+1){\mathbb{Z}}, and (3.4)dκ(g([n−1]n+1),g([0]n+1))=dX(f([n−1]n+1),f([0]n+1)).{d}_{\kappa }\left(g\left({\left[n-1]}_{n+1}),g\left({\left[0]}_{n+1}))={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[0]}_{n+1})).We divide CASE 1 into the following two subcases.SUBCASE 1A: The map g satisfies∠g([n−2]n+1)g([n−1]n+1)g([0]n+1)+∠g([0]n+1)g([n−1]n+1)g([n]n+1)≤π,∠g([1]n+1)g([0]n+1)g([n−1]n+1)+∠g([n−1]n+1)g([0]n+1)g([n]n+1)≤π.\begin{array}{rcl}\angle g\left({\left[n-2]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})+\angle g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[n]}_{n+1})& \le & \pi ,\\ \angle g\left({\left[1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})+\angle g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n]}_{n+1})& \le & \pi .\end{array}In this subcase, since g0{g}_{0}is a comparison map for f0{f}_{0}, it follows from elementary geometry that we have [g(i),g(j)]∩[g(i−[1]n+1),g(i+[1]n+1)]≠∅\left[g\left(i),g\left(j)]\cap \left[g\left(i-{\left[1]}_{n+1}),g\left(i+{\left[1]}_{n+1})]\ne \varnothing for any i,j∈Z/(n+1)Zi,j\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}with i≠ji\ne j. In particular, we have [g([n−1]n+1),g([0]n+1)]∩[g([m]n+1),g([n]n+1)]≠∅\left[g\left({\left[n-1]}_{n+1}),g\left({\left[0]}_{n+1})]\cap \left[g\left({\left[m]}_{n+1}),g\left({\left[n]}_{n+1})]\ne \varnothing for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]. Clearly, we also have dκ(g([0]n+1),g([m]n+1))+dκ(g([m]n+1),g([n−1]n+1))+dκ(g([n−1]n+1),g([n]n+1))+dκ(g([n]n+1),g([0]n+1))<2Dκ.{d}_{\kappa }\left(g\left({\left[0]}_{n+1}),g\left({\left[m]}_{n+1}))+{d}_{\kappa }\left(g\left({\left[m]}_{n+1}),g\left({\left[n-1]}_{n+1}))+{d}_{\kappa }\left(g\left({\left[n-1]}_{n+1}),g\left({\left[n]}_{n+1}))+{d}_{\kappa }\left(g\left({\left[n]}_{n+1}),g\left({\left[0]}_{n+1}))\lt 2{D}_{\kappa }.Therefore, Lemma 1.13 implies dκ(g([m]n+1),g([n]n+1))≥dX(f([m]n+1),f([n]n+1)){d}_{\kappa }\left(g\left({\left[m]}_{n+1}),g\left({\left[n]}_{n+1}))\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n]}_{n+1}))for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]because XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), and we have dX(f([0]n+1),f([m]n+1))≤dκ(g([0]n+1),g([m]n+1)),dX(f([m]n+1),f([n−1]n+1))≤dκ(g([m]n+1),g([n−1]n+1)),dX(f([n−1]n+1),f([n]n+1))=dκ(g([n−1]n+1),g([n]n+1)),dX(f([n]n+1),f([0]n+1))=dκ(g([n]n+1),g([0]n+1)),dX(f([0]n+1),f([n−1]n+1))=dκ(g([0]n+1),g([n−1]n+1))\begin{array}{rcl}{d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[m]}_{n+1}))& \le & {d}_{\kappa }\left(g\left({\left[0]}_{n+1}),g\left({\left[m]}_{n+1})),\\ {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n-1]}_{n+1}))& \le & {d}_{\kappa }\left(g\left({\left[m]}_{n+1}),g\left({\left[n-1]}_{n+1})),\\ {d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1}))& =& {d}_{\kappa }\left(g\left({\left[n-1]}_{n+1}),g\left({\left[n]}_{n+1})),\\ {d}_{X}(f\left({\left[n]}_{n+1}),f\left({\left[0]}_{n+1}))& =& {d}_{\kappa }\left(g\left({\left[n]}_{n+1}),g\left({\left[0]}_{n+1})),\\ {d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[n-1]}_{n+1}))& =& {d}_{\kappa }\left(g\left({\left[0]}_{n+1}),g\left({\left[n-1]}_{n+1}))\end{array}by (3.2), (3.3), and (3.4). Thus, ggis a comparison map for ff.SUBCASE 1B: The map g satisfies(3.5)π<∠g([n−2]n+1)g([n−1]n+1)g([0]n+1)+∠g([0]n+1)g([n−1]n+1)g([n]n+1)\pi \lt \angle g\left({\left[n-2]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})+\angle g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})g\left({\left[n]}_{n+1})or π<∠g([1]n+1)g([0]n+1)g([n−1]n+1)+∠g([n−1]n+1)g([0]n+1)g([n]n+1).\pi \lt \angle g\left({\left[1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n-1]}_{n+1})+\angle g\left({\left[n-1]}_{n+1})g\left({\left[0]}_{n+1})g\left({\left[n]}_{n+1}).In this subcase, we may assume without loss of generality that we have (3.5). Let S=conv(g0(Z/nZ)),T=conv({g0([n−1]n),p,g0([0]n)}).S={\rm{conv}}\left({g}_{0}\left({\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}n{\mathbb{Z}})),\hspace{1.0em}T={\rm{conv}}\left(\left\{{g}_{0}\left({\left[n-1]}_{n}),p,{g}_{0}\left({\left[0]}_{n})\right\}).Equip the subsets SSand TTof Mκ2{M}_{\kappa }^{2}with the induced metrics, and regard them as disjoint metric spaces. Define (R,dR)\left(R,{d}_{R})to be the metric space obtained by gluing SSand TTby identifying [g0([n−1]n),g0([0]n)]⊆S\left[{g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n})]\subseteq Swith [g0([n−1]n),g0([0]n)]⊆T\left[{g}_{0}\left({\left[n-1]}_{n}),{g}_{0}\left({\left[0]}_{n})]\subseteq T. Then RRis a CAT(κ){\rm{CAT}}\left(\kappa )space by Reshetnyak’s gluing theorem (see [14] or [6, Chapter II.11, Theorem 11.1]). We denote by rm{r}_{m}the point in RRrepresented by g0([m]n)∈S{g}_{0}\left({\left[m]}_{n})\in Sfor each m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1], and by rn{r}_{n}the point in RRrepresented by p∈Tp\in T(Figure 1). Define a map f1:Z/nZ→R{f}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to Rby f1([m]n)=rm,ifm∈Z∩[0,n−2],rn,ifm=n−1.{f}_{1}\left({\left[m]}_{n})=\left\{\phantom{\rule[-1.33em]{}{0ex}}\begin{array}{ll}{r}_{m},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ {r}_{n},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1.\end{array}\right.Then it follows from (3.1) and the definition of f1{f}_{1}that ∑i∈Z/nZdR(f1(i),f1(i+[1]n))<2Dκ,f1(j)≠f1(j+[1]n)\sum _{i\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}}{d}_{R}({f}_{1}\left(i),{f}_{1}\left(i+{\left[1]}_{n}))\lt 2{D}_{\kappa },\hspace{1.0em}{f}_{1}\left(j)\ne {f}_{1}\left(j+{\left[1]}_{n})for every j∈Z/nZj\in {\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}. Because RRis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )by Theorem 1.4, there exists a comparison map g1:Z/nZ→Mκ2{g}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}for f1{f}_{1}by the inductive hypothesis. It follows from (3.5) and Alexandrov’s lemma [6, p. 25] that dκ(g1([n−2]n),g1([n−1]n))=dR(rn−2,rn)=dR(rn−2,rn−1)+dR(rn−1,rn).{d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-2},{r}_{n})={d}_{R}\left({r}_{n-2},{r}_{n-1})+{d}_{R}\left({r}_{n-1},{r}_{n}).Therefore, there exists a point q∈[g1([n−2]n),g1([n−1]n)]q\in \left[{g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n})]such that dκ(g1([n−2]n),q)=dR(rn−2,rn−1),dκ(q,g1([n−1]n))=dR(rn−1,rn).{d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),q)={d}_{R}\left({r}_{n-2},{r}_{n-1}),\hspace{1.0em}{d}_{\kappa }\left(q,{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-1},{r}_{n}).Define a map g2:Z/(n+1)Z→Mκ2{g}_{2}:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}\to {M}_{\kappa }^{2}by g2([m]n+1)=g1([m]n),ifm∈Z∩[0,n−2],q,ifm=n−1,g1([n−1]n),ifm=n.{g}_{2}\left({\left[m]}_{n+1})=\left\{\begin{array}{ll}{g}_{1}\left({\left[m]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ q,\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1,\\ {g}_{1}\left({\left[n-1]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n.\end{array}\right.Then, we clearly have [g2(i),g2(j)]∩[g2(i−[1]n+1),g2(i+[1]n+1)]≠∅\left[{g}_{2}\left(i),{g}_{2}\left(j)]\cap \left[{g}_{2}\left(i-{\left[1]}_{n+1}),{g}_{2}\left(i+{\left[1]}_{n+1})]\ne \varnothing for any i,j∈Z/(n+1)Zi,j\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}with i≠ji\ne j. It is straightforward to see that dκ(g2([l]n+1),g2([m]n+1))≥dX(f([l]n+1),f([m]n+1)){d}_{\kappa }\left({g}_{2}\left({\left[l]}_{n+1}),{g}_{2}\left({\left[m]}_{n+1}))\ge {d}_{X}(f\left({\left[l]}_{n+1}),f\left({\left[m]}_{n+1}))for any l,m∈Z∩[0,n−2]l,m\in {\mathbb{Z}}\cap \left[0,n-2], and dκ(g2(i),g2(i+[1]n+1))=dX(f(i),f(i+[1]n+1)){d}_{\kappa }\left({g}_{2}\left(i),{g}_{2}\left(i+{\left[1]}_{n+1}))={d}_{X}(f\left(i),f\left(i+{\left[1]}_{n+1}))for any i∈Z/(n+1)Zi\in {\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}. Lemma 1.13 implies (3.6)dκ(g1([m]n),q)≥dR(rm,rn−1){d}_{\kappa }\left({g}_{1}\left({\left[m]}_{n}),q)\ge {d}_{R}\left({r}_{m},{r}_{n-1})for every m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2]because RRis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), and we have dκ(g1([n−1]n),g1([m]n))≥dR(rn,rm),dκ(g1([m]n),g1([n−2]n))≥dR(rm,rn−2),dκ(g1([n−2]n),q)=dR(rn−2,rn−1),dκ(q,g1([n−1]n))=dR(rn−1,rn),dκ(g1([n−2]n),g1([n−1]n))=dR(rn−2,rn),q∈[g1([n−2]n),g1([n−1]n)]\begin{array}{l}{d}_{\kappa }\left({g}_{1}\left({\left[n-1]}_{n}),{g}_{1}\left({\left[m]}_{n}))\ge {d}_{R}\left({r}_{n},{r}_{m}),\hspace{1.0em}{d}_{\kappa }\left({g}_{1}\left({\left[m]}_{n}),{g}_{1}\left({\left[n-2]}_{n}))\ge {d}_{R}\left({r}_{m},{r}_{n-2}),\\ {d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),q)={d}_{R}\left({r}_{n-2},{r}_{n-1}),\hspace{1.0em}{d}_{\kappa }\left(q,{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-1},{r}_{n}),\\ {d}_{\kappa }\left({g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n}))={d}_{R}\left({r}_{n-2},{r}_{n}),\hspace{1.0em}q\in \left[{g}_{1}\left({\left[n-2]}_{n}),{g}_{1}\left({\left[n-1]}_{n})]\end{array}by definition of g1{g}_{1}. It follows from (3.6) and the definition of g2{g}_{2}that dκ(g2([m]n+1),g2([n−1]n+1))≥dX(f([m]n+1),f([n−1]n+1)){d}_{\kappa }\left({g}_{2}\left({\left[m]}_{n+1}),{g}_{2}\left({\left[n-1]}_{n+1}))\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n-1]}_{n+1}))for every m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2]. Together with the definition of RR, Lemma 1.13 also implies (3.7)dR(rm,rn)≥dX(f([m]n+1),f([n]n+1)){d}_{R}\left({r}_{m},{r}_{n})\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n]}_{n+1}))for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]because XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), and we have dR(r0,rm)≥dX(f([0]n+1),f([m]n+1)),dR(rm,rn−1)≥dX(f([m]n+1),f([n−1]n+1)),dR(rn−1,rn)=dX(f([n−1]n+1),f([n]n+1)),dR(rn,r0)=dX(f([n]n+1),f([0]n+1)),dR(r0,rn−1)=dX(f([0]n+1),f([n−1]n+1)).\begin{array}{l}{d}_{R}\left({r}_{0},{r}_{m})\ge {d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[m]}_{n+1})),\hspace{1.0em}{d}_{R}\left({r}_{m},{r}_{n-1})\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n-1]}_{n+1})),\\ {d}_{R}\left({r}_{n-1},{r}_{n})={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1})),\hspace{1.0em}{d}_{R}\left({r}_{n},{r}_{0})={d}_{X}(f\left({\left[n]}_{n+1}),f\left({\left[0]}_{n+1})),\\ {d}_{R}\left({r}_{0},{r}_{n-1})={d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[n-1]}_{n+1})).\end{array}It follows from (3.7) and the definition of g2{g}_{2}that dκ(g2([m]n+1),g2([n]n+1))≥dX(f([m]n+1),f([n]n+1)){d}_{\kappa }\left({g}_{2}\left({\left[m]}_{n+1}),{g}_{2}\left({\left[n]}_{n+1}))\ge {d}_{X}(f\left({\left[m]}_{n+1}),f\left({\left[n]}_{n+1}))for every m∈Z∩[0,n−1]m\in {\mathbb{Z}}\cap \left[0,n-1]. Thus, g2{g}_{2}is a comparison map for ff.CASE 2: The map f satisfies f([n−1]n+1)=f([0]n+1)f\left({\left[n-1]}_{n+1})=f\left({\left[0]}_{n+1}). In this case, we set d=dX(f([n−1]n+1),f([n]n+1))=dX(f([0]n+1),f([n]n+1)).d={d}_{X}(f\left({\left[n-1]}_{n+1}),f\left({\left[n]}_{n+1}))={d}_{X}(f\left({\left[0]}_{n+1}),f\left({\left[n]}_{n+1})).Define a map f˜0:Z/(n−1)Z→X{\tilde{f}}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n-1){\mathbb{Z}}\to Xby f˜0([m]n−1)=f([m]n+1){\tilde{f}}_{0}\left({\left[m]}_{n-1})=f\left({\left[m]}_{n+1}), m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2]. Then there exists a comparison map g˜0:Z/(n−1)Z→Mκ2{\tilde{g}}_{0}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n-1){\mathbb{Z}}\to {M}_{\kappa }^{2}for f˜0{\tilde{f}}_{0}by the inductive hypothesis. Let S˜=conv(g˜0(Z/(n−1)Z)),T˜=[0,d].\tilde{S}={\rm{conv}}\left({\tilde{g}}_{0}\left({\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}\left(n-1){\mathbb{Z}})),\hspace{1.0em}\tilde{T}=\left[0,d].Equip S˜⊆Mκ2\tilde{S}\subseteq {M}_{\kappa }^{2}and T˜⊆R\tilde{T}\subseteq {\mathbb{R}}the induced metrics, and regard them as metric spaces in their own right. Define (R˜,dR˜)\left(\tilde{R},{d}_{\tilde{R}})to be the metric space obtained by gluing S˜\tilde{S}and T˜\tilde{T}by identifying {g˜0([0]n−1)}⊆S˜\left\{{\tilde{g}}_{0}\left({\left[0]}_{n-1})\right\}\subseteq \tilde{S}with {0}⊆T˜\left\{0\right\}\subseteq \tilde{T}. Then R˜\tilde{R}is a CAT(κ){\rm{CAT}}\left(\kappa )space by Reshetnyak’s gluing theorem. We denote by r˜m{\tilde{r}}_{m}the point in R˜\tilde{R}represented by g˜0([m]n−1)∈S˜{\tilde{g}}_{0}\left({\left[m]}_{n-1})\in \tilde{S}for each m∈Z∩[0,n−2]m\in {\mathbb{Z}}\cap \left[0,n-2], by r˜n−1{\tilde{r}}_{n-1}the point in R˜\tilde{R}represented by g˜0([0]n−1)∈S˜{\tilde{g}}_{0}\left({\left[0]}_{n-1})\in \tilde{S}, and by r˜n{\tilde{r}}_{n}the point in R˜\tilde{R}represented by d∈T˜d\in \tilde{T}. In particular, we have r˜0=r˜n−1{\tilde{r}}_{0}={\tilde{r}}_{n-1}. Define a map f˜1:Z/nZ→R˜{\tilde{f}}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to \tilde{R}by f˜1([m]n)=r˜m,ifm∈Z∩[0,n−2],r˜n,ifm=n−1.{\tilde{f}}_{1}\left({\left[m]}_{n})=\left\{\phantom{\rule[-1.33em]{}{0ex}}\begin{array}{ll}{\tilde{r}}_{m},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ {\tilde{r}}_{n},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1.\end{array}\right.Then since R˜\tilde{R}is Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )by Theorem 1.4, there exists a comparison map g˜1:Z/nZ→Mκ2{\tilde{g}}_{1}:{\mathbb{Z}}\hspace{-0.08em}\text{/}\hspace{-0.08em}n{\mathbb{Z}}\to {M}_{\kappa }^{2}of f˜1{\tilde{f}}_{1}by the inductive hypothesis. By definition of the gluing of metric spaces, we have dκ(g˜1([n−2]n),g˜1([n−1]n))=dR˜(r˜n−2,r˜n)=dR˜(r˜n−2,r˜n−1)+dR˜(r˜n−1,r˜n),{d}_{\kappa }\left({\tilde{g}}_{1}\left({\left[n-2]}_{n}),{\tilde{g}}_{1}\left({\left[n-1]}_{n}))={d}_{\tilde{R}}\left({\tilde{r}}_{n-2},{\tilde{r}}_{n})={d}_{\tilde{R}}\left({\tilde{r}}_{n-2},{\tilde{r}}_{n-1})+{d}_{\tilde{R}}\left({\tilde{r}}_{n-1},{\tilde{r}}_{n}),and therefore, there exists a point q˜∈[g˜1([n−2]n),g˜1([n−1]n)]\tilde{q}\in \left[{\tilde{g}}_{1}\left({\left[n-2]}_{n}),{\tilde{g}}_{1}\left({\left[n-1]}_{n})]such that dκ(g˜1([n−2]n),q˜)=dR˜(r˜n−2,r˜n−1),dκ(q˜,g˜1([n−1]n))=dR˜(r˜n−1,r˜n).{d}_{\kappa }\left({\tilde{g}}_{1}\left({\left[n-2]}_{n}),\tilde{q})={d}_{\tilde{R}}\left({\tilde{r}}_{n-2},{\tilde{r}}_{n-1}),\hspace{1.0em}{d}_{\kappa }\left(\tilde{q},{\tilde{g}}_{1}\left({\left[n-1]}_{n}))={d}_{\tilde{R}}\left({\tilde{r}}_{n-1},{\tilde{r}}_{n}).Define a map g˜2:Z/(n+1)Z→Mκ2{\tilde{g}}_{2}:{\mathbb{Z}}\hspace{0.1em}\text{/}\hspace{0.1em}\left(n+1){\mathbb{Z}}\to {M}_{\kappa }^{2}by g˜2([m]n+1)=g˜1([m]n),ifm∈Z∩[0,n−2],q˜,ifm=n−1,g˜1([n−1]n),ifm=n.{\tilde{g}}_{2}\left({\left[m]}_{n+1})=\left\{\begin{array}{ll}{\tilde{g}}_{1}\left({\left[m]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m\in {\mathbb{Z}}\cap \left[0,n-2],\\ \tilde{q},\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n-1,\\ {\tilde{g}}_{1}\left({\left[n-1]}_{n}),\hspace{1.0em}& {\rm{if}}\hspace{0.33em}m=n.\end{array}\right.Then the same argument as in SUBCASE 1B shows that g˜2{\tilde{g}}_{2}is a comparison map for ff, which completes the proof.□Theorem 1.6 follows immediately from Theorem 1.5.Figure 1The CAT(κ){\rm{CAT}}\left(\kappa )space RR.Proof of Theorem 1.6Let XXbe a Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa )space. Then XXis Cyclm′(κ){{\rm{Cycl}}}_{m}^{^{\prime} }\left(\kappa )for every positive integer mmas shown in the proof of Theorem 1.5, which implies that XXis Cycln(κ){{\rm{Cycl}}}_{n}\left(\kappa )for every integer n≥4n\ge 4.□4A generalization of Alexandrov’s lemmaIn this section, we prove two lemmas, Lemmas 4.6 and 4.10, which generalize Alexandrov’s lemma [6, p. 25]. We use these lemmas to prove Lemma 1.13 in Section 5.Before proving the first lemma, we recall some elementary facts about angle measure in Mκ2{M}_{\kappa }^{2}. We first recall the following three basic propositions about the sum of two adjacent angle measures in Mκ2{M}_{\kappa }^{2}.Proposition 4.1Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Then ∠xoz≤∠xoy+∠yoz\angle xoz\le \angle xoy+\angle yoz.Proposition 4.2Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Assume that ∠xoz=∠xoy+∠yoz\angle xoz=\angle xoy+\angle yoz. Then all of the following conditions are true: y and z do not lie on opposite sides of ℓ(o,x)\ell \left(o,x);x and y do not lie on opposite sides of ℓ(o,z)\ell \left(o,z);x and z do not lie on the same side of ℓ(o,y)\ell \left(o,y).Proposition 4.3Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Then the identity ∠xoz=∠xoy+∠yoz\angle xoz=\angle xoy+\angle yozholds if and only if yyand zzdo not lie on opposite sides of ℓ(o,x)\ell \left(o,x), and ∠xoy≤∠xoz\angle xoy\le \angle xoz.We recall two more elementary propositions.Proposition 4.4Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}. Then the following conditions are equivalent: (1)∠xoy+∠yoz+∠zox=2π\angle xoy+\angle yoz+\angle zox=2\pi ;(2)π≤∠yox+∠xoz\pi \le \angle yox+\angle xoz, and yyand zzdo not lie on the same side of ℓ(o,x)\ell \left(o,x);(3)π≤∠zoy+∠yox\pi \le \angle zoy+\angle yox, and zzand xxdo not lie on the same side of ℓ(o,y)\ell \left(o,y);(4)π≤∠xoz+∠zoy\pi \le \angle xoz+\angle zoy, and xxand yydo not lie on the same side of ℓ(o,z)\ell \left(o,z).When {o,x,y,z}\left\{o,x,y,z\right\}is contained in one side of some line, each condition in the statement of Proposition 4.4 means that oolies on the solid triangle with vertices xx, yy, and zz. Therefore, it is intuitively obvious that the following proposition holds true.Proposition 4.5Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are four distinct points such that {o,x,y,z}\left\{o,x,y,z\right\}is contained in one side of some line. Assume that one of (hence all of) the conditions (1), (2), (3), and (4) in the statement of Proposition 4.4holds true. Then all of the following conditions are true: o and x do not lie on opposite sides of ℓ(y,z)\ell (y,z);∠zxo+∠oxy=∠zxy\angle zxo+\angle oxy=\angle zxy;If o∈ℓ(x,y)∪ℓ(y,z)∪ℓ(z,x)o\in \ell \left(x,y)\cup \ell (y,z)\cup \ell \left(z,x), then o∈[x,y]∪[y,z]∪[z,x]o\in \left[x,y]\cup [y,z]\cup \left[z,x].We now start to prove the first lemma of this section, which was proved for κ=0\kappa =0in [17, Lemma 8.4].Lemma 4.6Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x,y)+dκ(y,z)+dκ(z,w)+dκ(w,x)<2Dκ,dκ(x,y)=dκ(x′,y′),dκ(y,z)=dκ(y′,z′),dκ(z,w)=dκ(z′,w′),dκ(w,x)=dκ(w′,x′),dκ(y,w)≤dκ(y′,w′).\begin{array}{l}{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)+{d}_{\kappa }\left(w,x)\lt 2{D}_{\kappa },\\ {d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(z,w)={d}_{\kappa }\left(z^{\prime} ,w^{\prime} ),\\ {d}_{\kappa }\left(w,x)={d}_{\kappa }\left(w^{\prime} ,x^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} ).\end{array}Whenever x, y, z, and w are distinct except the possibility that y=wy=w, we assume in addition that π≤∠yzx+∠xzw\pi \le \angle yzx+\angle xzw, and that y and w do not lie on the same side of ℓ(x,z)\ell \left(x,z). Thendκ(x,z)≤dκ(x′,z′).{d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).ProofIf one of the identities x=yx=y, x=zx=z, x=wx=w, y=zy=z, or z=wz=wholds, then it is straightforward to check the desired inequality holds true. So we assume that none of these identities hold. Suppose that y=wy=w. Then y∈ℓ(x,z)y\in \ell \left(x,z)because yyand wwdo not lie on the same side of ℓ(x,z)\ell \left(x,z)by hypothesis, and therefore, we have ∠xzy=∠xzw=0\angle xzy=\angle xzw=0or ∠xzy=∠xzw=π\angle xzy=\angle xzw=\pi . Because π≤∠yzx+∠xzw\pi \le \angle yzx+\angle xzwby hypothesis, it follows that (4.1)∠xzy=∠xzw=π.\angle xzy=\angle xzw=\pi .We also have (4.2)dκ(x,z)+dκ(z,y)<Dκ{d}_{\kappa }\left(x,z)+{d}_{\kappa }\left(z,y)\lt {D}_{\kappa }because otherwise κ\kappa would be greater than 0, and dκ(x,z)+dκ(z,y)+dκ(y,x){d}_{\kappa }\left(x,z)+{d}_{\kappa }\left(z,y)+{d}_{\kappa }(y,x)would be equal to 2Dκ2{D}_{\kappa }, which would imply that dκ(x,y)+dκ(y,z)+dκ(z,w)+dκ(w,x)≥dκ(x,y)+dκ(y,z)+dκ(z,x)=2Dκ,{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)+{d}_{\kappa }\left(w,x)\ge {d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,x)=2{D}_{\kappa },contradicting the hypothesis. It follows from (4.1) and (4.2) that z∈[x,y]z\in \left[x,y], and thus, dκ(x,z)=dκ(x,y)−dκ(y,z)=dκ(x′,y′)−dκ(y′,z′)≤dκ(x′,z′).{d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x,y)-{d}_{\kappa }(y,z)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).So henceforth we assume that xx, yy, zz, and wware distinct. We consider two cases.CASE 1: z∉ℓ(x,y)∪ℓ(y,w)∪ℓ(w,x)z\notin \ell \left(x,y)\cup \ell (y,w)\cup \ell \left(w,x). Let w˜∈Mκ2\tilde{w}\in {M}_{\kappa }^{2}be the point such that dκ(z,w˜)=dκ(z,w),∠yzw˜=∠y′z′w′,{d}_{\kappa }\left(z,\tilde{w})={d}_{\kappa }\left(z,w),\hspace{1.0em}\angle yz\tilde{w}=\angle y^{\prime} z^{\prime} w^{\prime} ,and w˜\tilde{w}is not on the opposite side of ℓ(y,z)\ell (y,z)from ww, as shown in Figure 2. Then Proposition 2.2 implies dκ(y,w˜)=dκ(y′,w′),{d}_{\kappa }(y,\tilde{w})={d}_{\kappa }(y^{\prime} ,w^{\prime} ),and therefore, by using Proposition 2.2 again, we obtain (4.3)∠w˜yz=∠w′y′z′.\angle \tilde{w}yz=\angle w^{\prime} y^{\prime} z^{\prime} .Since dκ(y,w)≤dκ(y′,w′){d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} )by hypothesis, Proposition 2.2 implies that ∠yzw≤∠y′z′w′=∠yzw˜,\angle yzw\le \angle y^{\prime} z^{\prime} w^{\prime} =\angle yz\tilde{w},and therefore, Proposition 4.3 implies that (4.4)∠yzw+∠wzw˜=∠yzw˜≤π.\angle yzw+\angle wz\tilde{w}=\angle yz\tilde{w}\le \pi .Because π≤∠yzx+∠xzw\pi \le \angle yzx+\angle xzw, and yyand wware not on the same side of ℓ(x,z)\ell \left(x,z), Proposition 4.4 implies that π≤∠yzw+∠wzx\pi \le \angle yzw+\angle wzx. Combining this with (4.4) yields ∠wzw˜≤∠wzx\angle wz\tilde{w}\le \angle wzx. Furthermore, w˜\tilde{w}and xxare not on opposite sides of ℓ(z,w)\ell \left(z,w)because (4.4) implies that w˜\tilde{w}is not on the same side of ℓ(z,w)\ell \left(z,w)as yyby Proposition 4.2, the hypothesis implies that xxis not on the same side of ℓ(z,w)\ell \left(z,w)as yyby Proposition 4.4, and y∉ℓ(z,w)y\notin \ell \left(z,w)by the assumption of CASE 1. Therefore, Proposition 4.3 implies that (4.5)∠wzx=∠wzw˜+∠w˜zx.\angle wzx=\angle wz\tilde{w}+\angle \tilde{w}zx.We have ∠w˜zx≤∠wzx\angle \tilde{w}zx\le \angle wzxby (4.5), and therefore, Proposition 2.2 implies that dκ(w˜,x)≤dκ(w,x)=dκ(w′,x′).{d}_{\kappa }\left(\tilde{w},x)\le {d}_{\kappa }\left(w,x)={d}_{\kappa }\left(w^{\prime} ,x^{\prime} ).Using Proposition 2.2 again, this implies that (4.6)∠w˜yx≤∠w′y′x′.\angle \tilde{w}yx\le \angle w^{\prime} y^{\prime} x^{\prime} .Because the hypothesis implies ∠yzw+∠wzx+∠xzy=2π\angle yzw+\angle wzx+\angle xzy=2\pi by Proposition 4.4, it follows from (4.4) and (4.5) that ∠w˜zx+∠xzy=2π−∠yzw˜≥π.\angle \tilde{w}zx+\angle xzy=2\pi -\angle yz\tilde{w}\ge \pi .Furthermore, yyand w˜\tilde{w}are not on the same side of ℓ(x,z)\ell \left(x,z)because yyis not on the same side of ℓ(x,z)\ell \left(x,z)as wwby hypothesis, (4.5) implies that w˜\tilde{w}is not on the opposite side of ℓ(x,z)\ell \left(x,z)from wwby Proposition 4.2, and w∉ℓ(x,z)w\notin \ell \left(x,z)by the assumption of CASE 1. Therefore, Proposition 4.5 implies that (4.7)∠w˜yx=∠w˜yz+∠zyx.\angle \tilde{w}yx=\angle \tilde{w}yz+\angle zyx.We also have (4.8)∠w′y′x′≤∠w′y′z′+∠z′y′x′\angle w^{\prime} y^{\prime} x^{\prime} \le \angle w^{\prime} y^{\prime} z^{\prime} +\angle z^{\prime} y^{\prime} x^{\prime} by Proposition 4.1. By combining (4.3), (4.6), (4.7), and (4.8), we obtain ∠zyx=∠w˜yx−∠w˜yz≤∠w′y′x′−∠w˜yz=∠w′y′x′−∠w′y′z′≤∠z′y′x′,\angle zyx=\angle \tilde{w}yx-\angle \tilde{w}yz\le \angle w^{\prime} y^{\prime} x^{\prime} -\angle \tilde{w}yz=\angle w^{\prime} y^{\prime} x^{\prime} -\angle w^{\prime} y^{\prime} z^{\prime} \le \angle z^{\prime} y^{\prime} x^{\prime} ,and therefore, Proposition 2.2 implies that dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).CASE 2: z∈ℓ(x,y)∪ℓ(y,w)∪ℓ(w,x)z\in \ell \left(x,y)\cup \ell (y,w)\cup \ell \left(w,x). In this case, z∈[x,y]∪[y,w]∪[w,x]z\in \left[x,y]\cup [y,w]\cup \left[w,x]by Proposition 4.5. If z∈[x,y]z\in \left[x,y], then dκ(x,z)=dκ(x,y)−dκ(y,z)=dκ(x′,y′)−dκ(y′,z′)≤dκ(x′,z′).{d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x,y)-{d}_{\kappa }(y,z)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).If z∈[w,x]z\in \left[w,x], then we obtain dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )similarly. So assume that z∈[y,w]z\in [y,w]. Then we have dκ(y,w)≤dκ(y′,w′)≤dκ(y′,z′)+dκ(z′,w′)=dκ(y,z)+dκ(z,w)=dκ(y,w),{d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} )\le {d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )={d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)={d}_{\kappa }(y,w),and thus, dκ(y,w)=dκ(y′,w′){d}_{\kappa }(y,w)={d}_{\kappa }(y^{\prime} ,w^{\prime} )and z′∈[y′,w′]z^{\prime} \in [y^{\prime} ,w^{\prime} ]. Hence, Proposition 2.2 implies ∠xyz=∠xyw=∠x′y′w′=∠x′y′z′,\angle xyz=\angle xyw=\angle x^{\prime} y^{\prime} w^{\prime} =\angle x^{\prime} y^{\prime} z^{\prime} ,and therefore, by using Proposition 2.2 again, we obtain dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).□Figure 2Proof of Lemma 4.6.Remark 4.7Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points that satisfy the hypothesis of Lemma 4.6. Alexandrov’s lemma [6, p. 25] states that if the identity dκ(y′,w′)=dκ(y′,z′)+dκ(z′,w′){d}_{\kappa }(y^{\prime} ,w^{\prime} )={d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )holds in addition, then we have dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} ).Before proving the second lemma of this section, we recall two additional elementary facts about the sum of two adjacent angle measures in Mκ2{M}_{\kappa }^{2}.Proposition 4.8Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z∈Mκ2o,x,y,z\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z}a\in \left\{x,y,z\right\}, and dκ(x,z)<Dκ{d}_{\kappa }\left(x,z)\lt {D}_{\kappa }. If [x,z]∩[o,y]≠∅\left[x,z]\cap \left[o,y]\ne \varnothing , then∠xoz=∠xoy+∠yoz.\angle xoz=\angle xoy+\angle yoz.Proposition 4.9Let κ∈R\kappa \in {\mathbb{R}}. Suppose o,x,y,z,w∈Mκ2o,x,y,z,w\in {M}_{\kappa }^{2}are points such that 0<dκ(o,a)<Dκ0\lt {d}_{\kappa }\left(o,a)\lt {D}_{\kappa }for every a∈{x,y,z,w}a\in \left\{x,y,z,w\right\}. If we have∠xoz=∠xoy+∠yoz,∠xow=∠xoz+∠zow,\angle xoz=\angle xoy+\angle yoz,\hspace{1.0em}\angle xow=\angle xoz+\angle zow,then we have∠yow=∠yoz+∠zow,∠xow=∠xoy+∠yow.\angle yow=\angle yoz+\angle zow,\hspace{1.0em}\angle xow=\angle xoy+\angle yow.We now prove the second lemma of this section.Lemma 4.10Let κ∈R\kappa \in {\mathbb{R}}. Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points such thatdκ(x′,y′)+dκ(y′,z′)+dκ(z′,w′)+dκ(w′,x′)<2Dκ,dκ(x,y)=dκ(x′,y′),dκ(y,z)=dκ(y′,z′),dκ(z,w)=dκ(z′,w′),dκ(w,x)=dκ(w′,x′),dκ(x′,z′)≤dκ(x,z),\begin{array}{l}{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )+{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )+{d}_{\kappa }\left(w^{\prime} ,x^{\prime} )\lt 2{D}_{\kappa },\\ {d}_{\kappa }\left(x,y)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(y,z)={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(z,w)={d}_{\kappa }\left(z^{\prime} ,w^{\prime} ),\\ {d}_{\kappa }\left(w,x)={d}_{\kappa }\left(w^{\prime} ,x^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z),\end{array}and [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing . Thendκ(y,w)≤dκ(y′,w′).{d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} ).ProofIf x′=z′x^{\prime} =z^{\prime} , then it follows from the hypothesis that x′∈[y′,w′]x^{\prime} \in [y^{\prime} ,w^{\prime} ], which implies the desired inequality by the triangle inequality. So we assume that x′≠z′x^{\prime} \ne z^{\prime} . We next consider the case in which y′∈ℓ(x′,z′)y^{\prime} \in \ell \left(x^{\prime} ,z^{\prime} ). If y′∈[x′,z′]y^{\prime} \in \left[x^{\prime} ,z^{\prime} ], then we have dκ(x,z)≤dκ(x,y)+dκ(y,z)=dκ(x′,y′)+dκ(y′,z′)=dκ(x′,z′)≤dκ(x,z),{d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)={d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )={d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z),and thus, dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} )and y∈[x,z]y\in \left[x,z]. Hence, Proposition 2.2 implies ∠yxw=∠zxw=∠z′x′w′=∠y′x′w′,\angle yxw=\angle zxw=\angle z^{\prime} x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} w^{\prime} ,and therefore, by using Proposition 2.2 again, we obtain dκ(y,w)=dκ(y′,w′).{d}_{\kappa }(y,w)={d}_{\kappa }(y^{\prime} ,w^{\prime} ).If y′∈ℓ(x′,z′)⧹[x′,z′]y^{\prime} \in \ell \left(x^{\prime} ,z^{\prime} )\setminus \left[x^{\prime} ,z^{\prime} ], then it follows from the hypothesis that x′x^{\prime} or z′z^{\prime} is on [y′,w′][y^{\prime} ,w^{\prime} ], which implies the desired inequality by the triangle inequality. Similarly, the desired inequality holds true in the case in which w′∈ℓ(x′,z′)w^{\prime} \in \ell \left(x^{\prime} ,z^{\prime} )as well. So henceforth we assume in addition that y′∉ℓ(x′,z′)y^{\prime} \notin \ell \left(x^{\prime} ,z^{\prime} )and w′∉ℓ(x′,z′)w^{\prime} \notin \ell \left(x^{\prime} ,z^{\prime} )In particular, this requires that x′x^{\prime} , y′y^{\prime} , z′z^{\prime} , and w′w^{\prime} are distinct.By hypothesis, one of the inequalities dκ(x′,y′)+dκ(y′,z′)<Dκ{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )\lt {D}_{\kappa }or dκ(z′,w′)+dκ(w′,x′)<Dκ{d}_{\kappa }\left(z^{\prime} ,w^{\prime} )+{d}_{\kappa }\left(w^{\prime} ,x^{\prime} )\lt {D}_{\kappa }holds true. We may assume without loss of generality that dκ(x′,y′)+dκ(y′,z′)<Dκ,dκ(x′,y′)≤dκ(y′,z′).{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} )\lt {D}_{\kappa },\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,y^{\prime} )\le {d}_{\kappa }(y^{\prime} ,z^{\prime} ).Then we have (4.9)∠yxz≤∠y′x′z′\angle yxz\le \angle y^{\prime} x^{\prime} z^{\prime} by Proposition 2.3. Let z˜∈Mκ2\tilde{z}\in {M}_{\kappa }^{2}be the point such that dκ(y′,z˜)=dκ(y′,z′),∠x′y′z˜=∠xyz,{d}_{\kappa }(y^{\prime} ,\tilde{z})={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}\angle x^{\prime} y^{\prime} \tilde{z}=\angle xyz,and z˜\tilde{z}is not on the opposite side of ℓ(x′,y′)\ell \left(x^{\prime} ,y^{\prime} )from z′z^{\prime} , as shown in Figure 3. Then Proposition 2.2 implies dκ(x′,z˜)=dκ(x,z),{d}_{\kappa }\left(x^{\prime} ,\tilde{z})={d}_{\kappa }\left(x,z),and therefore, by using Proposition 2.2 again, we obtain (4.10)∠y′x′z˜=∠yxz.\angle y^{\prime} x^{\prime} \tilde{z}=\angle yxz.Since dκ(x′,z′)≤dκ(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z), Proposition 2.2 implies (4.11)∠x′y′z′≤∠xyz=∠x′y′z˜.\angle x^{\prime} y^{\prime} z^{\prime} \le \angle xyz=\angle x^{\prime} y^{\prime} \tilde{z}.Because z˜\tilde{z}is not on the opposite side of ℓ(x′,y′)\ell \left(x^{\prime} ,y^{\prime} )from z′z^{\prime} , (4.11) implies (4.12)∠x′y′z˜=∠x′y′z′+∠z′y′z˜\angle x^{\prime} y^{\prime} \tilde{z}=\angle x^{\prime} y^{\prime} z^{\prime} +\angle z^{\prime} y^{\prime} \tilde{z}by Proposition 4.3. The hypothesis that [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing implies (4.13)∠x′y′z′=∠x′y′w′+∠w′y′z′\angle x^{\prime} y^{\prime} z^{\prime} =\angle x^{\prime} y^{\prime} w^{\prime} +\angle w^{\prime} y^{\prime} z^{\prime} by Proposition 4.8. By Proposition 4.9, (4.12) and (4.13) imply (4.14)∠x′y′z˜=∠x′y′w′+∠w′y′z˜.\angle x^{\prime} y^{\prime} \tilde{z}=\angle x^{\prime} y^{\prime} w^{\prime} +\angle w^{\prime} y^{\prime} \tilde{z}.By combining (4.11), (4.13), and (4.14), we obtain ∠w′y′z′≤∠w′y′z˜\angle w^{\prime} y^{\prime} z^{\prime} \le \angle w^{\prime} y^{\prime} \tilde{z}, and therefore, Proposition 2.2 implies dκ(w,z)=dκ(w′,z′)≤dκ(w′,z˜).{d}_{\kappa }\left(w,z)={d}_{\kappa }\left(w^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(w^{\prime} ,\tilde{z}).Using Proposition 2.2 again, this implies (4.15)∠zxw≤∠z˜x′w′.\angle zxw\le \angle \tilde{z}x^{\prime} w^{\prime} .The hypothesis that [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing implies (4.16)∠y′x′w′=∠y′x′z′+∠z′x′w′\angle y^{\prime} x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} w^{\prime} by Proposition 4.8. It follows from (4.9) and (4.10) that ∠y′x′z˜≤∠y′x′z′\angle y^{\prime} x^{\prime} \tilde{z}\le \angle y^{\prime} x^{\prime} z^{\prime} , which implies (4.17)∠y′x′z′=∠y′x′z˜+∠z˜x′z′\angle y^{\prime} x^{\prime} z^{\prime} =\angle y^{\prime} x^{\prime} \tilde{z}+\angle \tilde{z}x^{\prime} z^{\prime} by Proposition 4.3 because z˜\tilde{z}is not on the opposite side of ℓ(x′,y′)\ell \left(x^{\prime} ,y^{\prime} )from z′z^{\prime} . By Proposition 4.9, (4.16), and (4.17) imply (4.18)∠y′x′w′=∠y′x′z˜+∠z˜x′w′.\angle y^{\prime} x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} \tilde{z}+\angle \tilde{z}x^{\prime} w^{\prime} .We also have (4.19)∠yxw≤∠yxz+∠zxw\angle yxw\le \angle yxz+\angle zxwby Proposition 4.1. By combining (4.10), (4.15), (4.18), and (4.19), we obtain ∠yxw≤∠yxz+∠zxw≤∠y′x′z˜+∠z˜x′w′=∠y′x′w′,\angle yxw\le \angle yxz+\angle zxw\le \angle y^{\prime} x^{\prime} \tilde{z}+\angle \tilde{z}x^{\prime} w^{\prime} =\angle y^{\prime} x^{\prime} w^{\prime} ,and therefore, Proposition 2.2 implies dκ(y,w)≤dκ(y′,w′){d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} ).□Figure 3Proof of Lemma 4.10.Recall that for any x,y,z,w∈Mκ2x,y,z,w\in {M}_{\kappa }^{2}such that dκ(x,y)+dκ(y,z)+dκ(z,w)+dκ(w,x)<2Dκ{d}_{\kappa }\left(x,y)+{d}_{\kappa }(y,z)+{d}_{\kappa }\left(z,w)+{d}_{\kappa }\left(w,x)\lt 2{D}_{\kappa }, and xx, yy, zz, and wware distinct except the possibility that y=wy=w, we have [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing if and only if ∠yxz+∠zxw≤π\angle yxz+\angle zxw\le \pi , ∠yzx+∠xzw≤π\angle yzx+\angle xzw\le \pi , and yyand wware not on the same side of ℓ(x,z)\ell \left(x,z).Lemmas 4.6 and 4.10 imply the following corollary.Corollary 4.11Suppose x,y,z,w,x′,y′,z′,w′∈Mκ2x,y,z,w,x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {M}_{\kappa }^{2}are points that satisfy the hypothesis of Lemma 4.10. Assume in addition that x≠zx\ne zand that yyand wwdo not lie on the same side of ℓ(x,z)\ell \left(x,z). Then [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing .ProofWe have (4.20)dκ(y,w)≤dκ(y′,w′){d}_{\kappa }(y,w)\le {d}_{\kappa }(y^{\prime} ,w^{\prime} )by Lemma 4.10. If one of the identities x=yx=y, x=wx=w, y=zy=z, or z=wz=wholds, then we have [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing clearly. So we assume that xx, yy, zz, and wware distinct except the possibility that y=wy=w. Suppose for the sake of contradiction that [x,z]∩[y,w]=∅\left[x,z]\cap [y,w]=\varnothing . Since yyand wwdo not lie on the same side of ℓ(x,z)\ell \left(x,z), this requires that we have π<∠yxz+∠zxw\pi \lt \angle yxz+\angle zxwor π<∠yzx+∠xzw\pi \lt \angle yzx+\angle xzw. Therefore, (4.20) implies dκ(x,z)≤dκ(x′,z′){d}_{\kappa }\left(x,z)\le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )by Lemma 4.6. Combining this with the hypothesis that dκ(x′,z′)≤dκ(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{\kappa }\left(x,z)yields dκ(x,z)=dκ(x′,z′){d}_{\kappa }\left(x,z)={d}_{\kappa }\left(x^{\prime} ,z^{\prime} ). Therefore, it follows from Proposition 2.2 that we have π<∠yxz+∠zxw=∠y′x′z′+∠z′x′w′\pi \lt \angle yxz+\angle zxw=\angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} w^{\prime} or π<∠yzx+∠xzw=∠y′z′x′+∠x′z′w′,\pi \lt \angle yzx+\angle xzw=\angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} w^{\prime} ,which implies [x′,z′]∩[y′,w′]=∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]=\varnothing , contradicting the hypothesis that [x′,z′]∩[y′,w′]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]\ne \varnothing . Thus, we have [x,z]∩[y,w]≠∅\left[x,z]\cap [y,w]\ne \varnothing .□5Proof of Lemma 1.13In this section, we prove Lemma 1.13.Proof of Lemma 1.13Let κ\kappa , XX, xx, yy, zz, ww, x′x^{\prime} , y′y^{\prime} , z′z^{\prime} , and w′w^{\prime} be as in the hypothesis, and fix p∈[x′,z′]p\in \left[x^{\prime} ,z^{\prime} ]. If x′=z′x^{\prime} =z^{\prime} , then p=x′p=x^{\prime} , and therefore, dX(y,w)≤dX(y,x)+dX(x,w)≤dκ(y′,x′)+dκ(x′,w′)=dκ(y′,p)+dκ(p,w′).{d}_{X}(y,w)\le {d}_{X}(y,x)+{d}_{X}\left(x,w)\le {d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,w^{\prime} )={d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ).So henceforth we assume that x′≠z′x^{\prime} \ne z^{\prime} . Then we also have x≠zx\ne zsince dκ(x′,z′)≤dX(x,z){d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z). Let w¯∈Mκ2\overline{w}\in {M}_{\kappa }^{2}be the point such that dκ(x′,w¯)=dκ(x′,w′),dκ(w¯,z′)=dκ(w′,z′),{d}_{\kappa }\left(x^{\prime} ,\overline{w})={d}_{\kappa }\left(x^{\prime} ,w^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(\overline{w},z^{\prime} )={d}_{\kappa }\left(w^{\prime} ,z^{\prime} ),and w¯\overline{w}is not on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} )as y′y^{\prime} . (If w′w^{\prime} is not on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} )as y′y^{\prime} , then w¯\overline{w}is w′w^{\prime} itself.) Then it is easily seen that dκ(p,w¯)=dκ(p,w′){d}_{\kappa }\left(p,\overline{w})={d}_{\kappa }\left(p,w^{\prime} ), and therefore, (5.1)dκ(y′,w¯)≤dκ(y′,p)+dκ(p,w¯)=dκ(y′,p)+dκ(p,w′).{d}_{\kappa }(y^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,\overline{w})={d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ).Because XXis Cycl4(κ){{\rm{Cycl}}}_{4}\left(\kappa ), there exist x0,y0,z0,w0∈Mκ2{x}_{0},{y}_{0},{z}_{0},{w}_{0}\in {M}_{\kappa }^{2}such that dκ(x0,y0)≤dX(x,y),dκ(y0,z0)≤dX(y,z),dκ(z0,w0)≤dX(z,w),dκ(w0,x0)≤dX(w,x),dX(x,z)≤dκ(x0,z0),dX(y,w)≤dκ(y0,w0).\begin{array}{l}{d}_{\kappa }\left({x}_{0},{y}_{0})\le {d}_{X}\left(x,y),\hspace{1.0em}{d}_{\kappa }({y}_{0},{z}_{0})\le {d}_{X}(y,z),\hspace{1.0em}{d}_{\kappa }\left({z}_{0},{w}_{0})\le {d}_{X}\left(z,w),\\ {d}_{\kappa }\left({w}_{0},{x}_{0})\le {d}_{X}\left(w,x),\hspace{1.0em}{d}_{X}\left(x,z)\le {d}_{\kappa }\left({x}_{0},{z}_{0}),\hspace{1.0em}{d}_{X}(y,w)\le {d}_{\kappa }({y}_{0},{w}_{0}).\end{array}These points satisfy ∣dκ(x′,y′)−dκ(y′,z′)∣≤dκ(x′,z′)≤dX(x,z)≤dκ(x0,z0)≤dκ(x0,y0)+dκ(y0,z0)≤dX(x,y)+dX(y,z)≤dκ(x′,y′)+dκ(y′,z′),| {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )| \le {d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z)\le {d}_{\kappa }\left({x}_{0},{z}_{0})\le {d}_{\kappa }\left({x}_{0},{y}_{0})+{d}_{\kappa }({y}_{0},{z}_{0})\le {d}_{X}\left(x,y)+{d}_{X}(y,z)\le {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} ),and thus, ∣dκ(x′,y′)−dκ(y′,z′)∣≤dκ(x0,z0)≤dκ(x′,y′)+dκ(y′,z′).| {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )-{d}_{\kappa }(y^{\prime} ,z^{\prime} )| \le {d}_{\kappa }\left({x}_{0},{z}_{0})\le {d}_{\kappa }\left(x^{\prime} ,y^{\prime} )+{d}_{\kappa }(y^{\prime} ,z^{\prime} ).This guarantees that there exists a point y˜∈Mκ2\tilde{y}\in {M}_{\kappa }^{2}such that dκ(x0,y˜)=dκ(x′,y′),dκ(y˜,z0)=dκ(y′,z′).{d}_{\kappa }\left({x}_{0},\tilde{y})={d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(\tilde{y},{z}_{0})={d}_{\kappa }(y^{\prime} ,z^{\prime} ).Similarly, there also exists a point w˜∈Mκ2\tilde{w}\in {M}_{\kappa }^{2}such that dκ(x0,w˜)=dκ(x′,w′),dκ(w˜,z0)=dκ(w′,z′).{d}_{\kappa }\left({x}_{0},\tilde{w})={d}_{\kappa }\left(x^{\prime} ,w^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(\tilde{w},{z}_{0})={d}_{\kappa }\left(w^{\prime} ,z^{\prime} ).Clearly, we may assume that w˜\tilde{w}does not lie on the same side of ℓ(x0,z0)\ell \left({x}_{0},{z}_{0})as y˜\tilde{y}. We consider two cases.CASE 1: [x′,z′]∩[y′,w¯]≠∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,\overline{w}]\ne \varnothing . In this case, it follows from Lemma 4.10 that (5.2)dκ(y˜,w˜)≤dκ(y′,w¯){d}_{\kappa }(\tilde{y},\tilde{w})\le {d}_{\kappa }(y^{\prime} ,\overline{w})because dκ(x0,y˜)=dκ(x′,y′),dκ(y˜,z0)=dκ(y′,z′),dκ(z0,w˜)=dκ(z′,w¯),dκ(w˜,x0)=dκ(w¯,x′),dκ(x′,z′)≤dX(x,z)≤dκ(x0,z0).\begin{array}{rcl}{d}_{\kappa }\left({x}_{0},\tilde{y})& =& {d}_{\kappa }\left(x^{\prime} ,y^{\prime} ),\hspace{1.0em}{d}_{\kappa }(\tilde{y},{z}_{0})={d}_{\kappa }(y^{\prime} ,z^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left({z}_{0},\tilde{w})={d}_{\kappa }\left(z^{\prime} ,\overline{w}),\\ {d}_{\kappa }\left(\tilde{w},{x}_{0})& =& {d}_{\kappa }\left(\overline{w},x^{\prime} ),\hspace{1.0em}{d}_{\kappa }\left(x^{\prime} ,z^{\prime} )\le {d}_{X}\left(x,z)\le {d}_{\kappa }\left({x}_{0},{z}_{0}).\end{array}We also have [x0,z0]∩[y˜,w˜]≠∅\left[{x}_{0},{z}_{0}]\cap [\tilde{y},\tilde{w}]\ne \varnothing by Corollary 4.11. Choose p0∈[x0,z0]∩[y˜,w˜]{p}_{0}\in \left[{x}_{0},{z}_{0}]\cap [\tilde{y},\tilde{w}]. Then dκ(y0,p0)≤dκ(y˜,p0),dκ(p0,w0)≤dκ(p0,w˜){d}_{\kappa }({y}_{0},{p}_{0})\le {d}_{\kappa }(\tilde{y},{p}_{0}),\hspace{1.0em}{d}_{\kappa }\left({p}_{0},{w}_{0})\le {d}_{\kappa }\left({p}_{0},\tilde{w})by Corollary 2.5, and therefore, (5.3)dκ(y0,w0)≤dκ(y0,p0)+dκ(p0,w0)≤dκ(y˜,p0)+dκ(p0,w˜)=dκ(y˜,w˜).{d}_{\kappa }({y}_{0},{w}_{0})\le {d}_{\kappa }({y}_{0},{p}_{0})+{d}_{\kappa }\left({p}_{0},{w}_{0})\le {d}_{\kappa }(\tilde{y},{p}_{0})+{d}_{\kappa }\left({p}_{0},\tilde{w})={d}_{\kappa }(\tilde{y},\tilde{w}).It follows from (5.1), (5.2), and (5.3) that dX(y,w)≤dκ(y0,w0)≤dκ(y˜,w˜)≤dκ(y′,w¯)≤dκ(y′,p)+dκ(p,w′).{d}_{X}(y,w)\le {d}_{\kappa }({y}_{0},{w}_{0})\le {d}_{\kappa }(\tilde{y},\tilde{w})\le {d}_{\kappa }(y^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ).CASE 2: [x′,z′]∩[y′,w¯]=∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,\overline{w}]=\varnothing . In this case, we have x′≠y′x^{\prime} \ne y^{\prime} , x′≠w′x^{\prime} \ne w^{\prime} , y′≠z′y^{\prime} \ne z^{\prime} and z′≠w¯z^{\prime} \ne \overline{w}, and one of the inequalities π<∠y′x′z′+∠z′x′w¯\pi \lt \angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} \overline{w}or π<∠y′z′x′+∠x′z′w¯\pi \lt \angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} \overline{w}holds since y′y^{\prime} and w¯\overline{w}are not on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} ). We may assume without loss of generality that π<∠y′x′z′+∠z′x′w¯.\pi \lt \angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} \overline{w}.Then we have dκ(y′,x′)+dκ(x′,w¯)≤dκ(y′,p)+dκ(p,w¯){d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,\overline{w})by Alexandrov’s lemma [6, p. 25], and therefore, dX(y,w)≤dX(y,x)+dX(x,w)≤dκ(y′,x′)+dκ(x′,w′)=dκ(y′,x′)+dκ(x′,w¯)≤dκ(y′,p)+dκ(p,w¯)=dκ(y′,p)+dκ(p,w′),{d}_{X}(y,w)\le {d}_{X}(y,x)+{d}_{X}\left(x,w)\le {d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,w^{\prime} )={d}_{\kappa }(y^{\prime} ,x^{\prime} )+{d}_{\kappa }\left(x^{\prime} ,\overline{w})\le {d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,\overline{w})={d}_{\kappa }(y^{\prime} ,p)+{d}_{\kappa }\left(p,w^{\prime} ),which completes the proof.□6Proof of Theorem 1.10In this section, we present a proof of Theorem 1.10 for completeness (see Remark 6.2). First, we recall the following fact, which was established by Sturm when he proved in [16, Theorem 4.9] that if a geodesic metric space satisfies the ⊠\boxtimes -inequalities, then it is CAT(0){\rm{CAT}}\left(0).Proposition 6.1Let (X,dX)\left(X,{d}_{X})be a metric space that satisfies the ⊠\boxtimes -inequalities. Suppose x,y,z∈Xx,y,z\in Xare points such that x≠zx\ne z, and(6.1)dX(x,z)=dX(x,y)+dX(y,z).{d}_{X}\left(x,z)={d}_{X}\left(x,y)+{d}_{X}(y,z).Set t=dX(x,y)/dX(x,z)t={d}_{X}\left(x,y)\hspace{0.1em}\text{/}\hspace{0.1em}{d}_{X}\left(x,z). Then we havedX(y,w)2≤(1−t)dX(x,w)2+tdX(z,w)2−t(1−t)dX(x,z)2.{d}_{X}{(y,w)}^{2}\le \left(1-t){d}_{X}{\left(x,w)}^{2}+t{d}_{X}{\left(z,w)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}.for any w∈Xw\in X.ProofBy the hypothesis (6.1), we compute (1−t)dX(x,y)2+tdX(y,z)2=dX(y,z)dX(x,z)dX(x,y)2+dX(x,y)dX(x,z)dX(y,z)2=dX(x,y)dX(y,z)dX(x,z)(dX(x,y)+dX(y,z))=dX(x,y)dX(y,z)=t(1−t)dX(x,z)2.\begin{array}{rcl}\left(1-t){d}_{X}{\left(x,y)}^{2}+t{d}_{X}{(y,z)}^{2}& =& \frac{{{\rm{d}}}_{X}(y,z)}{{d}_{X}\left(x,z)}{d}_{X}{\left(x,y)}^{2}+\frac{{{\rm{d}}}_{X}\left(x,y)}{{d}_{X}\left(x,z)}{d}_{X}{(y,z)}^{2}\\ & =& \frac{{{\rm{d}}}_{X}\left(x,y){d}_{X}(y,z)}{{d}_{X}\left(x,z)}({d}_{X}\left(x,y)+{d}_{X}(y,z))\\ & =& {d}_{X}\left(x,y){d}_{X}(y,z)\\ & =& t\left(1-t){d}_{X}{\left(x,z)}^{2}.\end{array}Combining this with the ⊠\boxtimes -inequality in XXyields 0≤(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+s(1−t)dX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2=(1−s)((1−t)dX(x,y)2+tdX(y,z)2)+tsdX(z,w)2+s(1−t)dX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2=(1−s)t(1−t)dX(x,z)2+tsdX(z,w)2+s(1−t)dX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)2=tsdX(z,w)2+s(1−t)dX(w,x)2−st(1−t)dX(x,z)2−s(1−s)dX(y,w)2\begin{array}{rcl}0& \le & \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\\ & =& \left(1-s)(\left(1-t){d}_{X}{\left(x,y)}^{2}+t{d}_{X}{(y,z)}^{2})+ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\\ & =& \left(1-s)t\left(1-t){d}_{X}{\left(x,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\\ & =& ts{d}_{X}{\left(z,w)}^{2}+s\left(1-t){d}_{X}{\left(w,x)}^{2}-st\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}\end{array}for every s∈[0,1]s\in \left[0,1]. For any s∈(0,1]s\in \left(0,1], dividing this by ss, we obtain (1−s)dX(y,w)2≤tdX(z,w)2+(1−t)dX(w,x)2−t(1−t)dX(x,z)2.\left(1-s){d}_{X}{(y,w)}^{2}\le t{d}_{X}{\left(z,w)}^{2}+\left(1-t){d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}.Letting s→0s\to 0in this inequality yields the desired inequality.□We now prove Theorem 1.10.Proof of Theorem 1.10First, assume that a metric space (X,dX)\left(X,{d}_{X})is Cycl4(0){{\rm{Cycl}}}_{4}\left(0). Then for any x,y,z,w∈Xx,y,z,w\in X, there exist x0,y0,z0,w0∈R2{x}_{0},{y}_{0},{z}_{0},{w}_{0}\in {{\mathbb{R}}}^{2}such that ‖x0−y0‖≤dX(x,y),‖y0−z0‖≤dX(y,z),‖z0−w0‖≤dX(z,w),‖w0−x0‖≤dX(w,x),‖x0−z0‖≥dX(x,z),‖y0−w0‖≥dX(y,w).\begin{array}{l}\Vert {x}_{0}-{y}_{0}\Vert \le {d}_{X}\left(x,y),\hspace{1.0em}\Vert {y}_{0}-{z}_{0}\Vert \le {d}_{X}(y,z),\hspace{1.0em}\Vert {z}_{0}-{w}_{0}\Vert \le {d}_{X}\left(z,w),\\ \Vert {w}_{0}-{x}_{0}\Vert \le {d}_{X}\left(w,x),\hspace{1.0em}\Vert {x}_{0}-{z}_{0}\Vert \ge {d}_{X}\left(x,z),\hspace{1.0em}\Vert {y}_{0}-{w}_{0}\Vert \ge {d}_{X}(y,w).\end{array}Therefore, for any s,t∈[0,1]s,t\in \left[0,1], we have (1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)d(y,w)2≥(1−t)(1−s)‖x0−y0‖2+t(1−s)‖y0−z0‖2+ts‖z0−w0‖2+(1−t)s‖w0−x0‖2−t(1−t)‖x0−z0‖2−s(1−s)‖y0−w0‖2=‖((1−t)x′+tz′)−((1−s)y′+sw′)‖2≥0,\begin{array}{l}\left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s)d{(y,w)}^{2}\\ \hspace{1.0em}\ge \left(1-t)\left(1-s)\Vert {x}_{0}-{y}_{0}{\Vert }^{2}+t\left(1-s)\Vert {y}_{0}-{z}_{0}{\Vert }^{2}+ts\Vert {z}_{0}-{w}_{0}{\Vert }^{2}+\left(1-t)s\Vert {w}_{0}-{x}_{0}{\Vert }^{2}-t\left(1-t)\Vert {x}_{0}-{z}_{0}{\Vert }^{2}-s\left(1-s)\Vert {y}_{0}-{w}_{0}{\Vert }^{2}\\ \hspace{1.0em}=\Vert \left(\left(1-t)x^{\prime} +tz^{\prime} )-\left(\left(1-s)y^{\prime} +sw^{\prime} ){\Vert }^{2}\ge 0,\end{array}which means that XXsatisfies the ⊠\boxtimes -inequalities.For the converse, assume that (X,dX)\left(X,{d}_{X})satisfies the ⊠\boxtimes -inequalities. Fix x,y,z,w∈Xx,y,z,w\in X. If xx, yy, zz, and wware not distinct, then we can embed {x,y,z,w}\left\{x,y,z,w\right\}isometrically into R2{{\mathbb{R}}}^{2}. So we assume that xx, yy, zz, and wware distinct. Then there exist x′,y′,z′,w′∈R2x^{\prime} ,y^{\prime} ,z^{\prime} ,w^{\prime} \in {{\mathbb{R}}}^{2}such that ‖x′−y′‖=dX(x,y),‖y′−z′‖=dX(y,z),‖z′−w′‖=dX(z,x),‖x′−w′‖=dX(x,w),‖w′−z′‖=dX(w,z),\begin{array}{l}\Vert x^{\prime} -y^{\prime} \Vert ={d}_{X}\left(x,y),\hspace{1.0em}\Vert y^{\prime} -z^{\prime} \Vert ={d}_{X}(y,z),\hspace{1.0em}\Vert z^{\prime} -w^{\prime} \Vert ={d}_{X}\left(z,x),\\ \Vert x^{\prime} -w^{\prime} \Vert ={d}_{X}\left(x,w),\hspace{1.0em}\Vert w^{\prime} -z^{\prime} \Vert ={d}_{X}\left(w,z),\end{array}and y′y^{\prime} and w′w^{\prime} do not lie on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} ). We consider three cases.CASE 1: [x′,z′]∩(y′,w′)≠∅\left[x^{\prime} ,z^{\prime} ]\cap (y^{\prime} ,w^{\prime} )\ne \varnothing . In this case, there exist s∈(0,1)s\in \left(0,1)and t∈[0,1]t\in \left[0,1]such that (1−t)x′+tz′=(1−s)y′+sw′.\left(1-t)x^{\prime} +tz^{\prime} =\left(1-s)y^{\prime} +sw^{\prime} .It follows that 0=∥((1−t)x′+tz′)−((1−s)y′+sw′)∥2=(1−t)(1−s)‖x′−y′‖2+t(1−s)‖y′−z′‖2+ts‖z′−w′‖2+(1−t)s‖w′−x′‖2−t(1−t)‖x′−z′‖2−s(1−s)‖y′−w′‖2=(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)‖y′−w′‖2.\begin{array}{rcl}0& =& {\parallel (\left(1-t)x^{\prime} +tz^{\prime} )-(\left(1-s)y^{\prime} +sw^{\prime} )\parallel }^{2}\\ & =& \left(1-t)\left(1-s)\Vert x^{\prime} -y^{\prime} {\Vert }^{2}+t\left(1-s)\Vert y^{\prime} -z^{\prime} {\Vert }^{2}+ts\Vert z^{\prime} -w^{\prime} {\Vert }^{2}+\left(1-t)s\Vert w^{\prime} -x^{\prime} {\Vert }^{2}-t\left(1-t)\Vert x^{\prime} -z^{\prime} {\Vert }^{2}-s\left(1-s)\Vert y^{\prime} -w^{\prime} {\Vert }^{2}\\ & =& \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s)\Vert y^{\prime} -w^{\prime} {\Vert }^{2}.\end{array}On the other hand, we have 0≤(1−t)(1−s)dX(x,y)2+t(1−s)dX(y,z)2+tsdX(z,w)2+(1−t)sdX(w,x)2−t(1−t)dX(x,z)2−s(1−s)dX(y,w)20\le \left(1-t)\left(1-s){d}_{X}{\left(x,y)}^{2}+t\left(1-s){d}_{X}{(y,z)}^{2}+ts{d}_{X}{\left(z,w)}^{2}+\left(1-t)s{d}_{X}{\left(w,x)}^{2}-t\left(1-t){d}_{X}{\left(x,z)}^{2}-s\left(1-s){d}_{X}{(y,w)}^{2}because XXsatisfies the ⊠\boxtimes -inequalities. Comparing these yields dX(y,w)≤‖y′−w′‖{d}_{X}(y,w)\le \Vert y^{\prime} -w^{\prime} \Vert .CASE 2: [x′,z′]∩{y′,w′}≠∅\left[x^{\prime} ,z^{\prime} ]\cap \{y^{\prime} ,w^{\prime} \}\ne \varnothing . In this case, we may assume without loss of generality that y′∈[x′,z′]y^{\prime} \in \left[x^{\prime} ,z^{\prime} ]. Then we can write y′=(1−c)x′+cz′,y^{\prime} =\left(1-c)x^{\prime} +cz^{\prime} ,where (6.2)c=‖x′−y′‖‖x′−z′‖=dX(x,y)dX(x,z)∈(0,1).c=\frac{\Vert x^{\prime} -y^{\prime} \Vert }{\Vert x^{\prime} -z^{\prime} \Vert }=\frac{{{\rm{d}}}_{X}\left(x,y)}{{d}_{X}\left(x,z)}\in \left(0,1).It follows that ‖y′−w′‖2=‖(1−c)x′+cz′−w′‖2=(1−c)‖x′−w′‖2+c‖z′−w′‖2−c(1−c)‖x′−z′‖2=(1−c)dX(x,w)2+cdX(z,w)2−c(1−c)dX(x,z)2.\begin{array}{rcl}\Vert y^{\prime} -w^{\prime} {\Vert }^{2}& =& \Vert \left(1-c)x^{\prime} +cz^{\prime} -w^{\prime} {\Vert }^{2}\\ & =& \left(1-c)\Vert x^{\prime} -w^{\prime} {\Vert }^{2}+c\Vert z^{\prime} -w^{\prime} {\Vert }^{2}-c\left(1-c)\Vert x^{\prime} -z^{\prime} {\Vert }^{2}\\ & =& \left(1-c){d}_{X}{\left(x,w)}^{2}+c{d}_{X}{\left(z,w)}^{2}-c\left(1-c){d}_{X}{\left(x,z)}^{2}.\end{array}On the other hand, it follows from (6.2) and Proposition 6.1 that dX(y,w)2≤(1−c)dX(x,w)2+cdX(z,w)2−c(1−c)dX(x,z)2{d}_{X}{(y,w)}^{2}\le \left(1-c){d}_{X}{\left(x,w)}^{2}+c{d}_{X}{\left(z,w)}^{2}-c\left(1-c){d}_{X}{\left(x,z)}^{2}because we have dX(x,z)=‖x′−z′‖=‖x′−y′‖+‖y′−z′‖=dX(x,y)+dX(y,z).{d}_{X}\left(x,z)=\Vert x^{\prime} -z^{\prime} \Vert =\Vert x^{\prime} -y^{\prime} \Vert +\Vert y^{\prime} -z^{\prime} \Vert ={d}_{X}\left(x,y)+{d}_{X}(y,z).Combining these yields dX(y,w)≤‖y′−w′‖{d}_{X}(y,w)\le \Vert y^{\prime} -w^{\prime} \Vert .CASE 3: [x′,z′]∩[y′,w′]=∅\left[x^{\prime} ,z^{\prime} ]\cap [y^{\prime} ,w^{\prime} ]=\varnothing . Since y′y^{\prime} and w′w^{\prime} do not lie on the same side of ℓ(x′,z′)\ell \left(x^{\prime} ,z^{\prime} ), this requires that we have ∠y′z′x′+∠x′z′w′>π\angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} w^{\prime} \gt \pi or ∠y′x′z′+∠z′x′w′>π\angle y^{\prime} x^{\prime} z^{\prime} +\angle z^{\prime} x^{\prime} w^{\prime} \gt \pi . We may assume without loss of generality that ∠y′z′x′+∠x′z′w′>π\angle y^{\prime} z^{\prime} x^{\prime} +\angle x^{\prime} z^{\prime} w^{\prime} \gt \pi . Therefore, it follows from Alexandrov’s lemma [6, p. 25] that there exist x˜,y˜,w˜∈R2\tilde{x},\tilde{y},\tilde{w}\in {{\mathbb{R}}}^{2}and z˜∈[y˜,w˜]\tilde{z}\in [\tilde{y},\tilde{w}]such that ‖x˜−y˜‖=dX(x,y),‖x˜−w˜‖=dX(x,w),‖y˜−w˜‖=dX(y,z)+dX(z,w),‖y˜−z˜‖=dX(y,z),‖z˜−w˜‖=dX(z,w),\begin{array}{l}\Vert \tilde{x}-\tilde{y}\Vert ={d}_{X}\left(x,y),\hspace{1.0em}\Vert \tilde{x}-\tilde{w}\Vert ={d}_{X}\left(x,w),\hspace{1.0em}\Vert \tilde{y}-\tilde{w}\Vert ={d}_{X}(y,z)+{d}_{X}\left(z,w),\\ \Vert \tilde{y}-\tilde{z}\Vert ={d}_{X}(y,z),\hspace{1.0em}\Vert \tilde{z}-\tilde{w}\Vert ={d}_{X}\left(z,w),\end{array}and dX(x,z)=‖x′−z′‖≤‖x˜−z˜‖.{d}_{X}\left(x,z)=\Vert x^{\prime} -z^{\prime} \Vert \le \Vert \tilde{x}-\tilde{z}\Vert .The points x˜,y˜,z˜,w˜∈R2\tilde{x},\tilde{y},\tilde{z},\tilde{w}\in {{\mathbb{R}}}^{2}have the desired property.The aforementioned three cases exhaust all possibilities, and thus, XXis Cycl4(0){{\rm{Cycl}}}_{4}\left(0).□Remark 6.2A proof of Theorem 1.10 can also be found in [11, Lemma 2.6]. However, the case corresponding to CASE 2 in the aforementioned proof is omitted in the proof in [11].

Journal

Analysis and Geometry in Metric Spacesde Gruyter

Published: Jan 1, 2023

Keywords: Reshetnyak’s majorization theorem; CAT(κ) space; Cycl n (κ) space; ⊠-inequalities; weighted quadruple inequalities; Primary 53C23; Secondary 51F99

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