Cat(0) polygonal complexes are 2-median

Median spaces are spaces in which for every three points the three intervals between them intersect at a single point. It is well known that rank-1 affine buildings are median spaces, but by a result of Haettel, higher rank buildings are not even coarse median. We define the notion of “2-median space”, which roughly says that for every four points the minimal discs filling the four geodesic triangles they span intersect in a point or a geodesic segment. We show that CAT(0) Euclidean polygonal complexes, and in particular rank-2 affine buildings, are 2-median. In the appendix, we recover a special case of a result of Stadler of a Fary–Milnor type theorem and show in elementary tools that a minimal disc filling a geodesic triangle is injective.


Introduction
It is a well known fact that R and metric trees have the property that for any three points, the intersection of the geodesic segments between them is a single point.That is, metric trees are median in the following sense: Definition 1.1.Let (X, d) be a metric space • For x, y ∈ X we define the interval between x, y to be • The space X is median if for any x, y, z ∈ X, we have that [x, y] ∩ [x, z] ∩ [y, z] = {m(x, y, z)} is a unique point.
Median spaces have been extensively studied: Niblo and Reeves [15] proved that any action of a Property (T) group on a finite-dimensional CAT(0) cube complex has a global fixed point.In fact, Chatterji, Drutu and Haglund [5] prove that a locally compact second countable group has Property (T) if and only if any continuous action of it on a median space has bounded orbits.Sapir [19] proved that acting on a median space implies the Rapid Decay Property.
Median algebras, which are spaces equipped with a map satisfying algebraic properties that the geometric median satisfies, are studied as well, see the work of Fiovaranti [8], [7] and Bader and Taller [2].In 2013 Bowditch defined a generalisation of a median space, called coarse median, and showed that the mapping class group of a surface is a coarse median space.Niblo, Wright and Zhang [16] further simplified this definition and gave a definition of a coarse median algebra.
In view of the above, we would like to have a higher-rank analogue.However, Haettel [10] proved that higher rank symmetric spaces and thick affine buildings which are not products of trees are not even coarse median spaces.This shows that one needs to modify the definition of median to fit the higher rank setting.We will focus on dimension two.
As a first step, let us examine the case of the Euclidean plane E 2 .Clearly, E 2 is not a median space as for a triple of non-collinear points the geodesic segments between them do not intersect.Instead, let us consider four points x 1 , x 2 , x 3 , x 4 ∈ E 2 and the four full triangles they span {▲ (x i , x j , x k )} 1≤i<j<k≤4 .
Note that if x 1 , x 2 , x 3 , x 4 are non-collinear then the intersection of those triangles is a single point, as in Fig. 1, where in the figure we see the two different possible arrangements (the top and bottom of the figure) of the points in the plane and the four full triangles are coloured.We can see that o is the unique point in the intersection of the four full triangles.We note that the Helly property for the intersection of convex sets gives us that the intersection exists in this case.In the degenerate case, in which all four points are on a common line, we have that the intersection is an interval, it is exactly the interval x i x j ∩ x k x l for {i, j, k, l} = {1, 2, 3, 4} for which that intersection is non-empty.
We prove that a similar phenomenon, which we call 2-median, holds in 2dimensional CAT(0) Euclidean polygonal complexes.It is not clear to us what should ultimately be the general definition of a 2-median space, and we therefore refrain from giving such.Instead, we give an ad hoc definition in our restricted setting of 2-dimensional CAT(0) Euclidean polygonal complexes, and show that it always holds.
Let X be a 2-dimensional CAT(0) Euclidean polygonal complexes, and let x, y, z ∈ X.In Definition 4.1, we define the full triangle ▲ (x, y, z) as the minimal subset of X necessary in order to fill the geodesic triangle △(x, y, z) by a disc.If △(x, y, z) is a Jordan curve, the full triangle ▲ (x, y, z) coincides with the image of the (unique) minimal disc in the sense of Lytchak and Wenger [13].
In this case, it was shown by Stadler [20] that ▲ (x, y, z) is homeomorphic to a disc.
Definition 1.2.The space X is a 2-median space if for any x 1 , x 2 , x 3 , x 4 ∈ X either there exists {i, j, k, l} = {1, 2, 3, 4} such that x i x j ∩ x k x l is an interval, in which case ⋂ 1≤i<j<k≤4 ▲ (x i , x j , x k ) = x i x j ∩ x k x l , or ⋂ 1≤i<j<k≤4 ▲ (x i , x j , x k ) is a single point.
Our main result is the following: Theorem A. 2-dimensional CAT(0) polygonal complexes are 2-median.
In [12], Kleiner and Lang define and prove higher rank hyperbolicity properties in a general setting.In particular, they prove that 2-dimensional CAT(0) polygonal complexes have slim simplices; that is, if f maps the boundary of the 3-simplex to a 2-dimensional CAT(0) polygonal complex X, such that f is a (L, a) quasi isometry on each facet, then the image of each facet is contained in the D = D(X, L, a) neighbourhood of the image of the other facets.
As we said, we refrain from giving a general definition of 2-median spaces.However, speculating that such a definition will resemble Definition 1.2 (after the appropriate notions of intervals and full triangles are defined), we expect the following: the 2-median property should essentially be a rank-two phenomenon -e.g. the intersection of the triangles spanned by four generic points in the 3dimensional Euclidean space E 3 is empty; it should be a non-positive curvature phenomenon -e.g. the intersection of the triangles of the regular tetrahedron in the sphere S 2 is empty; full triangles should be minimal discs and not convex hulls -cf.Example 4.2.
The outline of the paper is as follows: §2 and §3 are introductory: in §2 we review some properties of CAT(0) Euclidean polygonal complexes and their links and in §3 we define a generalized disc and a generalized annulus which we will use throughout the text.
In §4 we give an ad hoc topological definition of ▲ (x, y, z) for a CAT(0) polygonal complex and characterize it.
In §5 we prove using a special case of a result by Stadler [20], that in this setting our ad hoc definition coincides with the definition above.
In §7 we discuss the discreteness of the intersection.We show that the intersection of the four full triangles is discrete while the intersection of two of the triangles does not have isolated points.
In §8 we prove the theorem in the degenerate case.
In §9 we prove that a near-immersion from a disc filling an embedded polygonal curve is injective: Theorem 1.4.Let X be a CAT(0) polygonal complex, and let Γ be an embedded polygonal curve with ordered vertices x 1 , ..., x n such that κ(Γ) < 4π.Assume f ∶ ∆ → X is a near-immersion, where ∆ is a finite ∆-complex homeomorphic to a disc and f | ∂∆ ∶ ∂∆ → Γ is a homeomorphism, then f is injective.
In §10 we prove the theorem in the general case.The idea is to obtain a near-immersion from a deflated tetrahedron (see Fig. 2) to our space and show that it is actually injective.We obtain a near-immersion by taking any cellular map from a deflated tetrahedron to our space and folding, and showing that when the procedure terminates we remain with a deflated tetrahedron.In §11 we discuss further directions and in §A we give an elementary proof of Stadler's result for CAT(0) polygonal complexes.
We assume that the reader is familiar with CAT(0) spaces.Specifically, we will focus on Euclidean CAT(0) polygonal complexes i.e.CAT(0) polygonal complexes whose cells are convex Euclidean polygons (not assumed to be regular).The relevant definitions can be found in parts I and II of [4].
• If ξ, ζ are points in lk(p), we denote by ∠ p (ξ, ζ) the angle between ξ and ζ in X (as defined in [4,Definition I.1.12]).This defines a metric ∠ p on lk(x).We denote the length of a path α in lk(p) by ∡ p α, and by ∡ p (ξ, ζ) the induced length metric, i.e. the length of the shortest path connecting them in lk(p).Note that ∠ p does not necessarily equal the metric ∡ p .
We will abuse notation and for x, y ∈ X and γ a path in X we denote by ∠ p (x, y), ∡ p (α), ∡ p (x, y) the angle, length and distance of the projections of x, y and γ to lk(p).
In cases in which the point referred to is clear, we will sometimes omit the subscript p, • Let Y be a geodesic space, x, y ∈ Y .When there exists a unique geodesic between x, y (as is the case in CAT(0) spaces and when x, y ∈ lk(q) and ∠ q (x, y) < π for q in a CAT(0) space), we denote it by xy.
The CAT(0) condition for CAT(0) polygonal complexes takes a simple form: Claim 2.2 (Link condition [4, Theorem II.5.2]).A Euclidean polygonal complex is CAT(0) if and only if it is simply connected and for each 0-cell v we have that the length of every injective loop in lk(v) (i.e. the girth of lk(v)) has length at least 2π.
In the remainder of this section, we discuss the connection between the link of point p ∈ X and X − p.
Recall that there exists ϵ > 0 such that the sphere around p of radius ϵ, S, can be identified with lk(p), and the closest point projection, π p ∶ X − p → S, is a deformation retraction of X − p (see [4,Proposition II.2.4]).
When it is clear which point we are referring to, we will denote π p (x) by x.
Proof.By [4,Theorem I.7.16] there exists r > 0 such that S, the sphere or radius r around p, is homeomorphic to lk(p) and the ball or radius r around p is isometric to C(lk(p)); where C(lk(p)) is the cone over lk(p) with the metric defined such that for u, v ∈ lk(p) the angle between pu, pv in lk(p) is min {∡ p (x, y), π}.
As the next example shows, in general if X is a CAT(0) polygonal complex, p ∈ X and q, r ∈ X such that p ∉ qr, it might be that π p (qr) ≠ π p (q)π p (r).However, we will see in Claim 2.5 that π p (qr), π p (q)π p (r) are homotopic relative to endpoints in lk(p).
Proof.Let S be as in Proposition 2.3 and identify lk(p) with S. Let α q ∶ [0, 1] → X be a parametrisation of q q where q ∈ S = lk(p), such that α q (0) = q and α q (1) = q.Similarly consider α r .Let We have that G is continuous and well defined into X − p, so π p ○ G gives a homotopy between π p ○ γ 0 and π p ○ γ 1 .
As in Proposition 2.3, the ball bounded by S is isometric to the cone over S. The cone over qr is convex and contains γ 1 , so π p ○ γ 1 = qr.We get that π p ○ G is a homotopy relative to {q, r} between π p (qr) and qr = π p (q)π p (r).
As an immediate corollary, we get the following.
Corollary 2.6.Let X be a CAT(0) polygonal complex, let ∆ be a geodesic triangle with vertices x, y, z ∈ X, and let p ∈ X.
where all homotopies are relative to {x, ŷ, ẑ} when they are defined.See Fig. 4 for an illustration of the four cases.In what follows we will need to discuss complexes which are similar to discs and annuli.We will record in this section the technicalities of the definitions.
• A generalized disc D is a contractible finite ∆-complex (see [11] Chapter 2 for definition of ∆-complex) with a piecewise linear embedding to R 2 , see Fig. 5.A boundary cycle is a map α ∶ C → D (1) such that α(C) is contained in the topological boundary, contains all 1-cells that are contained in the topological boundary and does not cross itself, that is, if e i , e i+1 are consecutive 1-cells in α then e −1  i , e i+1 are consecutive in the cyclic order coming from the planar embedding of all 1-cells emanating from t(e i ), and where C is cycle graph which is of minimal length among all such graphs for which there exists such a map.The boundary cycle can be thought of as the boundary of an epsilon-neighbourhood of the embedded generalized disk.
As complexes, they are just a connected and simply connected union of disks and trees.But it is important to note that generalized disks come with their embedding into R 2 .
Any two parametrized boundary cycles of D differ by precomposition by an automorphism of C. By abuse of language, we call α the boundary cycle of D.  A generalized annulus is a finite ∆-complex with a piecewise linear embedding to R 2 which is homotopy equivalent to S 1 and has no separating points.For example see Fig. 6.Given a generalized annulus A, it separates R 2 into a bounded and an unbounded component where the bounded component is a generalized disc D. Let the inner boundary of A be the boundary cycle of D and similarly A ∪ D is a generalized disc and we define the outer boundary of A to be the boundary cycle of A ∪ D.
Note that the inner boundary of a generalized annulus is a circle, as it has no separating points.Definition 3.3.Let A be a generalized annulus and let α ∶ C → A (1) be the inner boundary of A where C is a cycle graph.The equivalence relation ∼ on α(C) is a good gluing if the relation ∼ identifies α(e i ) with α(j i ) where e 1 , . . ., e k , j −1 k , ..., j −1 1 are the edges of C in this order, and at most one of int(e i ), int(j i ) is on the outer boundary of A. For example see Fig. 6.Proof.Assume A is a generalized annulus and ∼ is a good gluing of A. The quotient A/ ∼ is a surface: indeed, consider the links of vertices in A -they are subsets of the circle that have at most two components (as otherwise either the vertex would be a cut vertex or A would not be homotopy equivalent to S 1 ), the good gluing ensures that they glue up to have one component which is either a circle or a non-trivial arc.In fact, we see that the image of the outer boundary of A is the only boundary component of the surface A/ ∼.
The generalized annulus A is homotopic to a circle and so χ(A) = 0.The CW complex A/ ∼ has the same number of 2-cells as A, k fewer 1-cells, and k−1 fewer 0-cells where k is as in Definition 3.3.It follows that χ(A/ ∼) = χ(A) + 1 = 1.Since the compact surface A/ ∼ has one boundary component and χ(A/ ∼) = 1 it must be a disc.

Triangles in CAT(0) polygonal complexes
Throughout let X be a CAT(0) polygonal complex.E.g. a rank two affine building.Note that a polygonal complex can be subdivided to be a triangle complex, so without loss of generality, we assume that X is a CAT(0) triangle complex, i.e. each 2-cell is a Euclidean triangle, which is not assumed to be regular.
2. Define the full triangle ▲ (x, y, z) to be the union of △(x, y, z) and the set of points p in X − △(x, y, z) for which △(x, y, z) is not trivial in π 1 (X − p).
Example 4.2.In Fig. 3 we see △(p, q, r) in green in the space T × [0, 1] and its corresponding full triangle in yellow.Note that ▲ (p, q, r) is not convex -the geodesic between s and p does not lie inside of it.
Our uniqueness result would be false if we had considered the convex hull of the three points to be the full triangle, as one could see from Fig. 7: the purple line is not in the intersection of the four triangles, but is in the intersection of the four convex hulls, as it is not in ▲ (x, y, w), but is in conv(x, y, w), and it is in ▲ (x, z, w) = conv(x, z, w) and in zy ⊆ ▲ (x, y, z) ∩ ▲ (y, z, w).So the intersection of the four convex hulls is not a single point.: Intersection of the convex hulls is not a single point.Top left: ▲ (x, y, z) = conv(x, y, z) and ▲ (x, z, w) = conv(x, z, w); bottom left: ▲ (x, y, w) , ▲ (z, y, w); right: conv(x, y, w) In Fig. 3 the full triangle is not a disc, but rather a disc with a "spike" attached to it.This leads us to the following definition.Proof.The intersection of any two geodesics in a CAT(0) space is either empty or a (possibly degenerate) segment.Consider xy∩xz, this is a (possibly degenerate) segment, as it is non-empty.Let x ′ be the point furthest away from x on that interval and define y ′ , z ′ similarly.We get △(x, y, z) = △(x ′ , y ′ , z ′ )∪xx ′ ∪yy ′ ∪zz ′ , as in Fig. 8.
It is easy to see that unless △(x, y, z) is a tripod, A similar phenomenon occurs in the link of a point in X.That is Lemma 4.5.Let G be a metric graph with girth at least 2π and let x, y, z ∈ X be such that d(x, y), d(y, z), d(x, z) < π, then the geodesic triangle △(x, y, z) (which is well defined by assumption on the distances between the points and the girth) is a spiky triangle.
The proof of this lemma is the same as the proof of Lemma 4.4 where the sentence "The intersection of any two geodesics in a CAT(0) space is either empty or a (possibly degenerate) segment" is replaced with "The intersection of any two geodesics of length < π in G is either empty or a (possibly degenerate) segment by assumption on the girth".
We want to understand how ▲ (x, y, z) looks locally.Consider x, y, z ∈ X and q ∈ ▲ (x, y, z).Let B be a small enough ball around q.We want to describe B ∩ ▲ (x, y, z).
We start by analyzing points q in ▲ (x, y, z) − △(x, y, z).By Proposition 2.3, q ∈ ▲ (x, y, z) − △(x, y, z) if and only if c = π q (△(x, y, z)) is a homotopically non-trivial cycle in lk(q) or q ∈ △(x, y, z).We claim that for a small enough ball B around q the points of B ∩ ▲ (x, y, z) are the cone over the "essential part" C of c which is defined below: Definition 4.6.Let G be a graph.Given a cycle c in G which is not trivial in homotopy (resp.a path), the essential part of c is the unique closed nonbacktracking cycle (resp.path) which is homotopic (resp.homotopic relative to end points) to c.Given c a disjoint union of cycles and paths in G, the essential part of c is the union of the essential parts of the connected components of c. Claim 4.7.Let q ∈ ▲ (x, y, z) − △(x, y, z) and let B be a small enough ball around q such that ∂B ≅ lk(q) and B ∩ △(x, y, z) = ∅.Let C be the essential part of π q (△(x, y, z)).Then for p ∈ B, p ∈ C if and only if p ∈ ▲ (x, y, z).
Let Ĉ be the cone over C with respect to q.It is a convex disc and p ∈ C if and only if p is in the cone.If p ∉ Ĉ then C is filled with a disk Ĉ in X − p, and so p ∉ ▲ (x, y, z).

Similarly one proves the following claim
Claim 4.8.Let q ∈ △(x, y, z) and let B be a small enough ball around q such that ∂B ≅ lk(q) and B does not intersect any side of △(x, y, z) not containing q.Let C be the essential part of π q (△(x, y, z) − q).Then for p ∈ B, p ∈ C if and only if p ∈ ▲ (x, y, z).
Proof.We argue similarly to the proof of Claim 4.7, by analysing Ĉ (in the notations of the proof of Claim 4.7) for the different cases in Corollary 2.6.
We remind that Corollary 2.6 characterizes the connected components of π q (△(x, y, z) − q).
In Theorem 5.1 we will show that full triangles in X are homeomorphic to full spiky triangles.Definition 4.9.We say x 1 , x 2 , x 3 ∈ X are strongly non-collinear if they are distinct and for any {i, j, k} = {1, 2, 3} we have Definition 4.10.The points x 1 , x 2 , x 3 , x 4 ∈ X form a lanky quadrilateral if x i x j ∩ x k x l is a non-degenerate segment for some {i, j, k, l} = {1, 2, 3, 4}.E.g. see Fig. 9.Note that this does not imply that the four points lie on a geodesic.In the definition of 2-median, Definition 1.2, there are two cases to consider: either the four points form a lanky quadrilateral, in which case the intersection of the four full triangles is an interval, or not and then the intersection is a point.

Minimal triangles
In Lemma 4.4 we showed that geodesic triangles in X are homeomorphic to spiky triangles.In this section, we prove that full triangles in X are homeomorphic to full spiky triangles.
Theorem 5.1.Let x, y, z ∈ X and let T be an (abstract) full spiky triangle with ∂T ≅ △(x, y, z).Then there exists an embedding f ∶ T ↪ X such that f | ∂T maps ∂T homeomorphically onto △(x, y, z).
Any such map has image ▲ (x, y, z).
We will prove the theorem first in the strongly non-collinear case and then deduce it to the general case.The existence of an embedding in the strongly non-collinear case is a special case of Theorem 78 in [20].We give a more elementary proof in our case in the appendix, see Theorem A.1.
The uniqueness of the image in the strongly non-collinear case follows from the following lemma.Lemma 5.2.Assume Γ is a piecewise linear curve in a CAT(0) polygonal complex X and assume there exists a ∆-complex homeomorphic to D 2 and a cellular and injective map f ∶ D → X such that f (∂D) = Γ, then f (D) equals the union of Γ with the set of points p ∈ X − Γ such that Γ is not trivial in π 1 (X − p).
In particular, for any g ∶ D 2 → X cellular with respect to some ∆-complex structure on D 2 and injective such that g(∂D) = Γ, we have g(D 2 ) = f (D).
Proof.Denote by M the union of Γ with the set of points p ∈ X such that Γ is not trivial in π 1 (X − p).Clearly M ⊆ f (D), as if p ∉ f (D) then f provides a null homotopy of Γ in X − p and thus p ∉ M .
For the other direction, let p ∈ D. There are projections, as in Claim 2.5, Now for the general case.
proof of Theorem 5.1.The existence of such a map follows from the strongly non-collinear case using the following lemma.
Lemma 5.3.If f is a map from an abstract spiky triangle T to X such that it maps ∂T homeomorphically onto △(x, y, z) and the triangle part of T is mapped injectively, then f is injective.
Proof.Let x, y, z ∈ X and denote by x ′ ∈ X the point in which xy, xz diverge and similarly denote y ′ , z ′ , as in Fig. 8.By Lemma 4.4, △(x, y, z) is homeomorphic to a spiky triangle and △(x ′ , y ′ , z ′ ) is homeomorphic to a triangle.
By Theorem 78 in [20], there exists a full triangle D and Glue to D the three spikes, i.e. let I 1 , I 2 , I 3 be three intervals homeomorphic to xx ′ , yy ′ , zz ′ respectively and let a j ∈ I j be one of their endpoints.Consider ) and consider the map f ∶ T → X, induced from f and the homeomorphisms of I j to the geodesic segments.
In order to show f is injective it is enough to show that for p ∈ ∂T − D we have f (p) ∉ f (D) = ▲ (x ′ , y ′ , z ′ ).Indeed as f | ∂T is a homeomoprhism, we have that f (p) ∉ △(x ′ , y ′ , z ′ ), hence, by Claim 4.7, f (p) ∈ ▲ (x ′ , y ′ , z ′ ) if and only if the concatenation of x′ ŷ′ , ŷ′ ẑ′ , ẑ′ x′ in lk(p) is not null homotopic.But two out of the three segments are constant in lk(p), as they are continuations of geodesics emanating from p, so the union of the three is null homotopic.
In conclusion, we found an injective map from an abstract full spiky triangle to X mapping the boundary of the triangle homeomorphically onto △(x, y, z).We are left to prove that any such map f ∶ T → X has image ▲ (x, y, z).
Let D be the triangle part of T and let x ′ , y ′ , z ′ as above.We have that f maps ∂D homeomorphically onto △(x, y, z), hence f (D) = ▲ (x ′ , y ′ , z ′ ) by Lemma 5.2.We get Remark 5.4.We have shown that in the strongly non-collinear case, our ad hoc definition coincides with the definition of the image of a minimal disc as in Lytchak and Wenger [13].

Existence of intersection
For x 1 , x 2 , x 3 , x 4 ∈ X let W 1 , W 2 , W 3 , W 4 be the four full triangles they define, where W i , 1 ≤ i ≤ 4, is the full triangle of the three points {x 1 , x 2 , x 3 , x 4 } − {x i }.In Proposition 6.2 we will show ⋂ 4 i=1 W i ≠ ∅.If one of the W i 's is a tripod or interval (i.e. the triangle part of the spiky triangle is degenerate), the claim is clear -the branching point of the tripod would be in the intersection.So we may assume this is not the case.
By Theorem 5.1 there exists F 1 , F 2 , F 3 , F 4 full spiky triangles, f i ∶ F i → W i homeomorphism such that each side of F i maps to a side of W i .The maps glue along the boundaries of the F i 's and descend to a map f .That is, consider We will make use of Sperner's lemma, see [9] for the proof.Lemma 6.3 (Sperner's lemma).Consider a triangulation of a triangle with vertices p, q, r.Let there be a colouring of the 0-cells by three colours such that each of the three vertices p, q, r has a distinct colour and the 0-cells that lie on any side of the triangle have only two colours -the two colours at the endpoints of that side.Then there exists at least one "rainbow triangle" -a 2-cell with vertices coloured with all three colours.More precisely, there must be an odd number of rainbow triangles.
Figure 10: Colouring of the sides of F 4 Proof of Proposition 6.2.Consider a sequence of triangulations of the triangle part of F 4 such that the supremum of diameters of the 2-cells goes to 0. We define a colouring of the 0-cells of the triangulation by three colours 1, 2, 3. On ∂F 4 we define it as in Fig. 10, i.e. for a 0-cell v ∈ f −1 (x 2 x 3 ) − f −1 (x 3 ) we let its colour be 1, the colours of 0-cells in are 2 and 3 respectively.For v a 0-cell in the interior of F 4 we define its colour to be min {j | f (v) ∈ W j } which is well defined from the previous claim.By Sperner's lemma, Lemma 6.3, in each coloured triangulation there exists a rainbow triangle T n .Let p ′ n be the barycenter of T n , by compactness (up to passing to a subsequence) p ′ n converges to p ′ ∈ F 4 .We now claim that Note that for j ∈ {1, 2, 3} the sequence of vertices of T n coloured j also converges to p ′ , as the bound on the diameter of 2-cells goes to 0, hence p ′ is arbitrarily close to W j for all j ∈ {1, 2, 3}, hence p ∈ W j , and of course p ∈ W 4 , as p ′ ∈ F 4 , so we are done.

Discreteness of intersection
In this section we will prove the following theorem.
four full triangles they define, then • If x 1 , x 2 , x 3 , x 4 do not form a lanky quadrilateral, then ⋂ 4 i=1 W i is a finite set of points.
• If x 1 , x 2 , x 3 , x 4 form a lanky quadrilateral, then ⋂ 4 i=1 W i is the union of a finite set and ⋃ {i,j,k,l}={1,2,3,4} (x i x j ∩x k x l ), in particular it is 1-dimensional.
Let x 1 , x 2 , x 3 , x 4 ∈ X, we will refer to them as vertices and to the geodesic segments between them as sides.By opposite sides we mean x i x j , x k x l for {i, j, k, l} = {1, 2, 3, 4}.Denote W 1 , W 2 , W 3 , W 4 the four full triangles in X. Assume throughout this section that p ∈ ⋂ 4 i=1 W i .We will study the link of p and show that the case p ∈ ⋂ 4 i=1 W i is very restricted.Indeed, denoting x1 , x2 , x3 , x4 the image of x 1 , x 2 , x 3 , x 4 in lk(p), we have from Claim 4.7 and Claim 4.8 that either two are antipodal, that is p is on a side; or △(x i , xj , xk ) ⊆ lk(p) are well defined for all i, j, k ∈ {1, 2, 3, 4}, distinct and, by Corollary 2.6, not trivial in homotopy.
, then p is in the intersection of two opposite sides.
Proof.The link of p is a circle of length 2π, as in Fig. 11.Assume by contradiction that p is not on a side containing x i , that is, the antipodal point to xi is not xj for all j ∈ {1, 2, 3, 4}.The points xi and its antipodal separate lk(p) into two arcs.By the pigeonhole principle, one of the arcs contains at least two out of {x j } j≠i , say xk , xl , so △(x i , xk , xl ) is null homotopic in lk(p) which is a contradiction to p ∈ ▲ (x i , x k , x l ) and Corollary 2.6, which states that In conclusion, p is on a side emanating from each vertex.If p is not in the intersection of two opposite sides, then up to permutation of the vertices, p ∈ ⋂ j=2,3,4 x 1 x j and in particular x2 = x3 = x4 .That cannot be, as then △(x 2 , x3 , x4 ) is null homotopic in lk(p) which is again a contradiction to Corollary 2.6.i=1 W i and p ∈ X (1) − X (0) is contained in the interior of a 1-cell e, then either p is on the intersection of two opposite sides or p is on a side that intersects e transversely.
Proof.The link lk(p) is a graph with two vertices, which we will refer to as poles and denote S, N , and no loops.For the purpose of this proof, we will refer to the 1-cells in lk(p) as longitudes in order to differentiate from the 1-cells in X.We may consider only the longitudes on which x1 , x2 , x3 , x4 lie, as the geodesics between them will be included in the union of those longitudes.There are at most four such longitudes.Note that if there are only two such longitudes, then the same proof as in the previous claim will give us that p is in the intersection of two opposite sides.Also note that if p is on a side that does not intersect e transversely, then there are only two such longitudes and we are done.
Remember that each longitude is of length π and if p ∉ x i x j then by Proposi- We arrive to a contradiction, as the geodesic between x3 , x4 does not pass through one of the poles, say S, hence the triangle △(x 2 , x3 , x4 ) is null homotopic -does not pass through S.
Case 2. x1 , x2 , x3 , x4 lie on four longitudes.It cannot be that in lk(p) all geodesics emanating from a vertex pass through the same pole.Indeed, assume towards contradiction and without loss of generality that all geodesics emanating from x1 pass through S. We get that all other geodesics pass through N because △(x 1 , xi , xj ) is not null homotopic for 2 ≤ i < j ≤ 4. But then △(x 2 , x3 , x4 ) is null homotopic, as all the geodesic segments in it pass through N , so do not pass through S.This is a contradiction to p ∈ ▲ (x 2 , x 3 , x 4 ).
Assume without loss of generality that x1 is the closest to S and x4 is the closest to N (need not be unique), as in Fig. 13.If the geodesic between x1 , x4 passes through S (respectively N ) then all geodesics emanating from x1 (respectively x4 ) pass through S (respectively N ).So all geodesics emanating from x1 or x4 pass through the same pole which is a contradiction by the previous paragraph.i=1 W i is either in X (0) , in the transverse intersection of X (1) with sides, or in the intersection of sides.
We have that each side crosses transversely a finite amount of 1-cells in X, X (0) ∩ ⋃ 4 i=1 W i is finite, and if x 1 , x 2 , x 3 , x 4 do not form a lanky quadrilateral, two sides intersect at most at a point, so ⋂ 4 i=1 W i is finite.If they do, we have that ⋂ 4 i=1 W i is a union of a finite set of points with all intersections of opposite sides.
This theorem is essential to the proof, it will be used together with the following lemma, regarding the intersection of two triangles sharing a side.Lemma 7.4.Assume x 1 , x 2 , x 3 , x 4 ∈ X, x 1 ≠ x 2 and q ∈ ▲ (x 1 , x 2 , x 3 ) ∩ ▲ (x 1 , x 2 , x 4 ), then q is not an isolated point in the intersection.
Proof.Let x1 , x2 , x3 , x4 be the images of x 1 , x 2 , x 3 , x 4 in lk(q).Note that if q ∈ x 1 x 2 we are done, as x 1 x 2 ⊆ ▲ (x 1 , x 2 , x 3 ) ∩ ▲ (x 1 , x 2 , x 4 ) so q is not an isolated point in the intersection.Assume this is not the case.
By Claim 4.7 and Claim 4.8 it is enough to show that C 3 , C 4 , the essential parts of π q (△(x 1 , x 2 , x 3 )−q), π q (△(x 1 , x 2 , x 4 )−q) in lk(q) respectively, intersect.We will actually show that they intersect on x1 x2 . Clearly Note that: This is because: -If q ∈ x 2 x 4 we can a path π of length π between x4 , x2 and the concatenation of x2 ŵ1 , ŵ1 x4 , π either contains a cycle and then, as in the previous case, we get what we want or is fully backtracking and then α + d( ŵ1 , x4 ) ≥ π.
This is because the concatenation of x1 ŵ1 , ŵ1 x4 is exactly x1 x4 , by choice of ŵ1 , so we are done by Proposition 2.3.
From the second point we have ), and then from the first point β < α.But of course, this argument was symmetric so we also have α < β which is a contradiction.
By the triangle inequality, we have the following Summing up these inequalities we get so all inequalities are equalities.
Geodesics are unique in a CAT(0) space so we get In particular q ∈ x 1 x 2 ∩ x 3 x 4 .

Near-immersions
Recall Definition 1.3.We rewrite the definition for 2-dimensional complexes.Definition 9.1.A cellular map f ∶ A → B between 2-dimensional polygonal complexes is a near-immersion if it is injective on each cell and there are no twin cells -different 2-cells t, t ′ sharing a 1-cell such that f (t) = f (t ′ ).Theorem 9.2.Let X be a CAT(0) polygonal complex, and let Γ be an embedded polygonal curve with ordered vertices x 1 , ..., x n such that κ(Γ) < 4π.Assume f ∶ ∆ → X is a near-immersion, where ∆ is a finite ∆-complex homeomorphic to a disc and f | ∂∆ ∶ ∂∆ → Γ is a homeomorphism, then f is injective.
In particular, Γ = △(x, y, z) satisfies the conclusion of the theorem when x, y, z ∈ X are strongly non-collinear, as in this case κ(Γ) < 4π.
We get the following corollary using Lemma 5.3.
Corollary 9.3.Let X be a CAT(0) polygonal complex, x, y, z ∈ X and f ∶ T → X a near-immersion such that T is a full (abstract) spiky triangle and f | ∂T ∶ ∂T → △(x, y, z) is a homeomorphism.Then f is injective and its image is ▲ (x, y, z).
Proof.By Theorem 9.2, f is injective on the triangle part of T , hence by Lemma 5.3 f is injective.By Theorem 5.1 f (T ) = ▲ (x, y, z).
We will prove Theorem 9.2 by arguing that if f is not injective we get a contradiction to Gauss-Bonnet Theorem.
We will remind the reader of the statement of Gauss-Bonnet.
Definition 9.4.A pseudo-surface with boundary is a 2-dimensional simplicial complex S such that every 1-cell is contained in one or two 2-cells.Its boundary, ∂S, is the union of the 1-cells which are incident to only one 2-cell.Its interior, int(S), is defined to be S − ∂S.
where χ(S) is the Euler characteristic of S.
Note that the sums are finite, as for v ∉ S (0) we have that the appropriate summand is 0.
In the rest of this section we will prove Theorem 9.2.Assume ∆ is a finite ∆-complex homeomorphic to a disc and f ∶ ∆ → X is a near-immersion mapping the boundary of ∆ homeomorphically onto Γ.We endow ∆ with a piecewise Euclidean structure by endowing each 2-cell σ of ∆ with the Euclidean metric of its image in X under f .Let p ∈ ∆, since f is a near-immersion, each arc α in lk(p) is sent to an arc f p (α) in lk(f (p)) which has the same length.It follows that the girth of lk(p) is at least that of lk(f (p)), which is at least 2π since X is CAT(0).By the link condition, Claim 2.2, ∆ is CAT(0).
Denote abusively by x i ∈ ∆ the unique point in ∂∆ with image x i ∈ X.
We collect a few important facts about ∆ in the following claims.
Claim 9.7.We have, for all v ∈ ∂∆ − {x 1 , ..., x n }, k v ≤ 0, Proof.Proof of (1): Since ∆ is CAT(0), the girth of lk(v) is at least 2π.If v ∈ int ∆ then lk(v) is a circle, and is on the geodesic segment connecting x i , x i+1 or x n , x 1 .By the triangle inequality for angles, the sum of the angles of v in 2-cells containing v is at least Let γ be a geodesic in ∆ connecting the points p, q.For v ∈ int(γ) define Proof.To prove (i), we will use the following lemma.
Lemma 9.9.Let P be a geodesic polygon in a CAT(0) space with ordered vertices v 1 , . . ., v n then ∑ n i=1 (π − ∠ vi P ) ≥ 2π.The proof of Lemma 9.9 is standard and is obtained by cutting the polygon into triangles by erecting geodesic segments between v i for 2 ≤ i ≤ n − 2 and v n and using the fact that the sum of angles in a triangle in a CAT(0) space is less than or equal to π ([4] Proposition II.1.7).See for example Theorem 2.4 in [3] and Corollary 2.4 in [1].
Consider the geodesic polygon P which is the image of f ○ γ.We have where the last inequality follows from Lemma 9.9.This proves (i).
To prove (ii), let v ∈ int(γ) ∩ int(∆).If T v = 0 then by Claim 9.7(1) K v ≤ 0 = T v , therefore we may assume that T v > 0, i.e. the angle that the path f ○ γ makes at f (v) in X is strictly less than π.
Let γ + and γ − be the two points in lk(v) corresponding to the geodesic γ passing through v.The points γ + , γ − break the circle lk(v) into two arcs α 1 , α 2 , as in the left hand side of Fig. 18.As the girth of lk(v) is greater than 2π, it follows that ∡α 1 + ∡α 2 ≥ 2π.Since f is a near-immersion, the paths From Proposition 2.3 there is a path β in lk(f (v)) whose length realizes the angle that f ○ γ makes at f (v).In lk(f (v)), all three paths β, f v ○ α 1 , f v ○ α 2 are non-backtracking and connect f v (γ + ), f v (γ − ), as in the right hand side of Fig. 18.
The links of v (on the left) and f (v) (on the right) In conclusion as claimed.
To prove (iii), let v ∈ int(γ) ∩ {x 1 , . . ., x n }.We must have k v ≤ 0, because otherwise, we could find a shortcut of γ not passing through v, in contradiction to γ being a geodesic.Hence if T v = 0, then k v ≤ 0 = T v , as claimed.Therefore, we may assume that T v > 0, i.e. that the angle that the path f ○ γ makes at f (v) in X is less than π.
As in the proof of Claim 9.8 (iii), let γ + , γ − be the two points in lk(v) corresponding to γ.They break the arc lk(v) into three arcs α 1 , α 2 , α 3 where α 2 is the middle arc and α 1 , α 3 may be degenerate, as in the left-hand side of Fig. 19.Since f is a near-immersion, the paths From Proposition 2.3 there is a path β in lk(f (v)) whose length realizes the angle that the path f ○ γ makes at f (v).We have that β, ) respectively, as in the right hand side of Fig. 19.
We have ∡α 2 ≥ ∠(γ − , γ + ) = π where the last equality follows, since γ is a geodesic, hence in particular The links of v (on the left) and f (v) (on the right) So where the last inequality is from the triangle inequality for angles.This completes the proof of the claim.
We are now ready to prove Theorem 9.2.

Minimal tetrahedron
The idea of the proof of the main theorem is to show that there is an embedding of a "deflated tetrahedron", as in Fig. 20, to X, such that the sides are mapped to x i x j , because then the "faces" of the deflated tetrahedron must be mapped to ▲ (x i , x j , x k ), and they intersect in a unique point.In our minds, we start with a finite ∆-complex, as in Fig. 20, and a map to X, and we modify those until we get an embedding.This leads us to our next definition, keeping in mind that in Fig. 20 we have six spiky triangles, {▲ (y i , y j , ω)} i≤i<j≤4 , glued on their boundaries.Definition 10.1.
• A deflated tetrahedron is a collection of: six generalized triangles (which are allowed to be degenerate, recall Definition 3.1) {D ij } 1≤i<j≤4 with the marked points on D ij denoted y i , y j , ω, (denoting bijections between y i ω in D ij and y i ω in D ik , such that the complex T ijk , obtained by gluing D ij , D jk , D ik along their boundaries via the bijections above, is homeomorphic to a spiky triangle.
For a deflated tetrahedron {D ij } 1≤i<j≤4 let Z be the complex obtained by gluing {D ij } 1≤i<j≤4 along their boundaries, that is, where ∼ is generated by the given bijections.
f is a cellular map, up to a refinement of the cellular structure on X, and injective on every cell, from y i y j to the geodesic segment x i x j in X is a homeomorphism and descends to a map from Z.
• A pair ({D ij } 1≤i<j≤4 , f ) of a deflated tetrahedron and a filling map of x 1 , x 2 , x 3 , x 4 through o will be referred to as a deflated tetrahedron filling x 1 , x 2 , x 3 , x 4 through o.
We note that we do not require f | Dij to be injective, but we will see that, under an appropriate minimality assumption (Definition 10.3), f is injective on D ij and so is its induced map on Z (Proposition 10.2), and that Z is a deformation retract of what one will imagine a deflated tetrahedron to be -a shape like in Fig. 20.
Proof.By Theorem 5.1, {▲ (x i , x j , o)} 1≤i<j≤4 are spiky triangles.As in the proof of Theorem 5.1, by [21] for each (i, j) we may refine the cellular structure of X and get finite ∆-complexes D ij ≅ ▲ (x i , x j , o) and cellular, injective maps is near-immersion and hence from Corollary 9.3 it is injective.In particular {T ijk } 1≤I<j<k≤4 embed in Z.
Proof.Assume towards contradiction that f is not near-immersion.Then, as f is injective on 2-cells, there are twin cells in T ijk , i.e. 2-cells t, t ′ that share a 1-cell and such that f (t) = f (t ′ ).
If t, t ′ ⊆ D ij for some i, j we will commence as in the proof of Theorem 5.1, that is, define C ij the connected component of ω in D ij −int(t∪t ′ )/ ∼, where ∼ is the cellular equivalence relation defined by u ∼ v if u a 1-cell in ∂t and v a 1-cell in ∂t ′ and f (u) = f (v), as in Theorem 5.1.Define C jk = D jk for (j, k) ≠ (i, j).We get {C ij } 1≤i<j≤4 is a new deflated tetrahedron with fewer cells, as all is still satisfied (T 123 is still a spiky triangle as explained in the proof of Theorem 5.1), and the induced map from f is a filling map, so we get a contradiction.
Hence t, t ′ must be in different cell complexes.Without loss of generality assume that t ⊆ D 12 , t ′ ⊆ D 13 .
We would like to "fold" t, t ′ and insert the folded copy to D 14 , as in Fig. 22.We divide into cases depending on t ∩ t ′ : Case 1.The intersection t ∩ t ′ is an interval (one or two 1-cells).C 12 to be a complex obtained from D 12 removing int(t) and the open interval of t ∩ t ′ .Similarly, define C 13 to be the complex obtained from D 12 removing int(t ′ ) and the open interval of t ∩ t ′ .Set C 14 = D 14 ∪ t and through o and that will give us contradiction to minimality of ({D ij } 1≤i<j≤4 , f ).
We have that C 12 and C 13 are deformation retracts of D 12 and D 13 respectively and D 14 is a deformation retract of C 14 , hence they are planar, finite, contractible ∆-complexes.Choose y i , y j , ω to be the same marked points as in D ij and get that {C ij } 1≤i<j≤4 are generalized triangles.t ′ )/ ∼ where ∼ identifies the boundary of t ∪ t ′ to an interval, so T ′ 123 is a spiky triangle.Lastly, T ′ 134 = (T 134 − t ′ ) ∪ t/ ∼ is a spiky triangle, as topologically we deleted a disc and glued another disc instead.

The induced map between y
Clearly, f ′ is a filling map.This gives a contradiction to minimality.Case 2. The intersection t ∩ t ′ is a disjoint union of a 1-cell and 0-cell v.
There are two further cases, either v ∈ y 1 ω, as in Fig. 23, or v ∈ y 2 ω, as in Fig. 24 (where in this case it happens to be that D 23 has a skinny part).
Case 2.1.v ∈ y 1 ω.We argue identically to case 1, with T, T ′ instead of t, t ′ , to show that ({C ij } 1≤i<j≤4 , f ′ ) is a deflated tetrahedron filling x 1 , x 2 , x 3 , x 4 through o.This is again a contradiction to minimality.
Case 2.2.v ∈ y 2 ω.Let A be the connected component of y 1 in T 123 − (t ∪ t ′ ).It is a generalized annulus with spikes and ω ∉ A, as in Theorem 5.1.Consider ∼ the cellular equivalence relation defined by u ∼ v if u a 1-cell in ∂t and v a 1-cell in ∂t ′ and f (u) = f (v), as in Theorem 5.1.It is a good gluing on A and so from Claim 3.4, A/ ∼ is a spiky triangle, in particular, it is simply connected.The map f induced from f maps the boundary of A homeomorphically onto △(x 1 , x 2 , x 3 ), hence ▲ (x 1 , x 2 , x 3 ) ⊆ f (T ) which is a contradiction to f being a filling map and satisfying f | −1 Cij ({o}) = {ω} and the fact that ω ∉ A while f (ω) = o ∈ ▲ (x 1 , x 2 , x 3 ).
Proof.f is injective on cells by assumption.Assume towards contradiction that there are t, t ′ ⊆ Z twin cells.As f | T ijk is injective, by Proposition 10.4 we have that t ⊆ D ij , t ′ ⊆ D kl for {i, j, k, l} = {1, 2, 3, 4}.From this, we have in X and this is a contradiction to the fact that the intersection is a 1-dimensional complex, by Theorem 7.1.
is not injective then it must be that there are two points on the boundary which are identified, that is, D 23 , D 13 have a skinny part, as in Fig. 25.There are two cases, either e 1 , e 2 , the two 1-cells adjacent to ω in the boundary of D 12 , are identified in T 123 , or not.In the latter case, we consider the pair of points in D 12 which are identified and are closest to ω and get that D 12 bounds a disc in T 123 .
Note q ij separates D ij for every 1 ≤ i < j ≤ 4. Consider C ij the connected component of y i in D ij − {q ij }.Their image in T 123 is simply connected by Van Kampen.This is a contradiction to o ∈ ▲ (x 1 , x 2 , x 3 ).

Future directions
Higher dimensions.It is natural to try and generalize this work to higher dimensions.To do this, one needs to generalize triangles and n-median spaces.We suggest the following inductive definition for the generalisation of triangles.
One can study intersections of ▲ n (x 1 , . . ., x n+1 ) for all x 1 , . . ., x n+1 in a set of n + 2 points.A median space with the median map is clearly a median algebra.A natural question to ask is if there is a meaningful definition of "2-median algebras", such that 2-median spaces will satisfy this definition, and characterize them.It is natural to try to define coarse 2-median spaces and study which spaces are coarse 2-median.In particular, are rank-2 symmetric spaces coarse 2median? of the connected component of ∂T in T − t.Let e 1 , e 2 be the two 1-cells on the boundary of t other than e oriented such that t(e 1 ) = o(e 2 ).Since e is an outermost loop, e 1 , e 2 are not loops, hence their concatenation forms a simple loop which is the inner boundary of A.
As f is cellular and not injective on t we have f (t) = f (∂t), which is a 0-cell or 1-cell in X, so f (e 1 ) = f (e 2 ).Let ∼ be the cellular equivalence relation defined by e 1 ∼ e −1 2 .The relation ∼ is a good gluing, as it identifies the two 1-cells on the inner boundary of A in the correct orientation, hence by Claim 3.4, A/ ∼ is a disc.
The induced ∆-complex structure on the disc A/ ∼ has fewer cells than the number of cells in T and the map induced from f is still a filling map.This is a contradiction to the minimality of (f, T ).T (1) contains no double 1-cells.Assume e 1 , e 2 are oriented 1-cells with o(e 1 ) = o(e 2 ), t(e 1 ) = t(e 2 ), then as X is a simplicial complex f (e 1 ) = f (e 2 ).The concatenation of e 1 and e 2 separates T into a generalized annulus A and a disc, as in Fig. 27.The inner boundary of A is the cycle e 1 , e −1 2 .Let ∼ be the cellular equivalence relation e 1 ∼ e 2 .As in the loop case, ∼ is a good gluing, as it identifies the two 1-cells on the inner boundary of A in the correct orientation, hence by Claim 3.4, A/ ∼ is a disc.
The induced ∆-complex structure on the disc A/ ∼ has less cells than the number of cells in T and the map induced from f is still a filling map.This is a contradiction to the minimality of (f, T ).If f | T is not a near-immersion then either there is a cell t such that f (t) = f (∂t) or there are 2-cells t, t ′ which share at least one 1-cell on their boundary and such that f (t) = f (t ′ ).We will show both cases cannot happen.
Case 1. Assume f (t) = f (∂t).In this case, there exists a 1-cell, e, in ∂t that is mapped to a point.As f | ∂T is injective, e must be in the interior of T .Let t ′ be the other 2-cell incident to e. Let e, e 1 , e 2 and e, d 1 , d 2 be the 1-cells in ∂t and ∂t ′ respectively, oriented as a cycle, as in Fig. 13.
Let A be the closure of the connected component of ∂T in T − ( t ∪ t′ ).The subcomplex A is the closure of the complement of a closed disc in T , hence it is a generalized annulus.By the claim above, Claim A.2, the inner boundary of A is a path consisting of four oriented 1-cells e 1 , e 2 , d −1 2 , d −1 1 in this order, see Fig. 28.
We have that f (e 1 ) is the inverse of f (e 2 ) and f (d 1 ) is the inverse of f (d 2 ).Let ∼ be the cellular equivalence relation defined by e 1 ∼ e −1 2 , d 1 ∼ d −1 2 .It is a good gluing, as f is injective on the boundary of T , from strongly noncollinearity of x, y, z.From Claim 3.4, A/ ∼ is a disc which is a finite ∆-complex.The map induced from f , denoted f ∶ A/ ∼→ X, is well defined, cellular and f | ∂A ∶ ∂A → △(x, y, z) is a homeomorphism.But A/ ∼ has fewer cells than T , this is a contradiction to the minimality of (f, T ).Case 2. Assume f (t) = f (t ′ ) for t, t ′ sharing a 1-cell e.Let e, e 1 , e 2 and e, d 1 , d 2 be the 1-cells in ∂t and ∂t ′ respectively, oriented as a cycle, as in Fig. 29.
Let A be the closure of the connected component of ∂T in T − ( t ∪ t′ ).As above, A is a generalized annulus and the inner boundary of A is a path consisting of four oriented 1-cells e 1 , e 2 , d −1 2 , d −1 1 in this order, as in Fig. 29.Because f (t) = f (t ′ ), we have that f (e 1 ) is the inverse of f (d 2 ) and f (e 2 ) is the inverse of f (d 1 ).
Let ∼ be the cellular equivalence relation defined by e 1 ∼ d −1 2 , d 1 ∼ e −1 2 .It is a good gluing, as before, as x, y, z are strongly non-collinear.By Claim 3.4, we have that A/ ∼ is homeomorphic to a disc.It is a finite ∆-complex with fewer cells than T , and the induced map f ∶ A/ ∼→ X is still cellular.This contradicts the minimality of (f, T ).Hence f is a near-immersion.By Theorem 9.2, we get that f is injective and this concludes the proof.

3 oFigure 1 :
Figure 1: Four points in the Euclidean plane and the triangles they span

y 1 y 2 y 3 y 4 ωFigure 2 :
Figure 2: Deflated tetrahedron Injectivity will follow using Theorem 1.4 and the two facts we prove in §7 about intersections of two and four full triangles.In §11 we discuss further directions and in §A we give an elementary proof of Stadler's result for CAT(0) polygonal complexes.

pFigure 4 :
Figure 4: The four cases, lk(p) on the left of the configuration

•
A generalized triangle is a generalized disc D with boundary cycle α ∶ C → D together with 3 vertices x, y, z ∈ C on its boundary.We denote xy D = α| xy where xy is the path in C connecting x and y and not passing through z.The images of x, y, z and the arcs xy, xz, yz under α are well-defined and by abuse of notation we regard the points x, y, z as the corresponding points in D. • Given a generalized disc D, points in D that are the image of two points in the boundary cycle of D are referred to as skinny parts of D.

Figure 5 :
Figure 5: Example of generalized disc

Figure 13 :
Figure 13: lk(p) for the case p ∈ X (1) − X (0) and four relevant longitudes we have by Corollary 2.6 that x1 x2 ⊆ C i and so C 3 ∩ C 4 ≠ ∅ and we are done.So assume q ∉ x1 x3 ∩ x2 x3 , x1 x4 ∩ x2 x4 .In this case, again by Corollary 2.6, each of C 3 , C 4 is a cycle or an arc depending on whether q ∈ △(x 1 , x 2 , x i ) or not.If C 3 is a cycle, then from Lemma 4.5 C 3 = △(ẑ 1 , ẑ2 , ẑ3 ) where ẑi is the point closest to xi on C 3 , that is it is the point furthest away from xi on xi xj ∩ xi xk where {i, j, k} = {1, 2, 3}.Similarly if C 4 is a cycle then C 4 = △( ŵ1 , ŵ2 , ŵ4 ) where ŵi is the point closest to xi on C 4 , that is it is the point furthest away from xi on xi xj ∩ xi xk where {i, j, k} = {1, 2, 4}.If C 3 is an arc then q ∈ x 1 x 3 or x 2 x 3 and C 3 = x1 ẑ3 ∪ x1 ẑ3 where ẑ3 is defined as above.Similarly if C 4 is an arc C 4 = x1 ŵ4 ∪ x1 ŵ4 In all cases C i ∩ x1 x2 is non-trivial and connected, hence an arc (perhaps degenerate, i.e. a point).Assume towards contradiction that C 3 , C 4 do not intersect on x1 x2 , that is C 3 ∩ x1 x2 , C 4 ∩ x1 x2 are two disjoint arcs.Without loss of generality C 3 ∩ x1 x2 is closer to x1 than C 4 ∩ x1 x2 .The endpoint of C 3 ∩ x1 x2 furthest from x1 is ẑ2 and the endpoint of C 4 ∩ x1 x2 furthest from x2 is ŵ1 , as in Fig. 14.By assumption x1 ẑ2 ∩ ŵ1 x2 = ∅.

Notation 9 . 5 .
Let S be a piecewise Euclidean pseudo-surface.For v ∈ int(S) we denote K v = 2π − ∑ α where the sums run over all angles around v in the 2-cells containing v. For v ∈ ∂S, let i be the number of connected component of lk(v); we denote k v = (2 − i)π − ∑ α where the sums run over all angles around v in the 2-cells containing v.

13 Figure 22 :
Figure 22: of t, t ′ is a bijection.Denote by T ′ ijk the gluing induced by the bijections on the boundary parts of C ij , C jk , C ik .T ′ ijk are spiky triangles for 1 ≤ i < j < k ≤ 4: indeed, T ′ 124 = T 124 by working out the definitions of C ij .We have T ′ 234 = T 234 as C ij = D ij for the relevant cell complexes.As in the proof of Theorem 5.1, T ′ 123 = T 123 − (int(t ∪

Figure 23 :
Figure 23: Case 2.1.We have that D 12 − t has two connected components.Denote by C 12 the closure of the connected component containing ω and by S the closure of the other connected component and let T = S ∪ t.Define C 13 , S ′ similarly to be the connected components of D 13 −t ′ containing ω and not containing ω respectively and let T ′ = S ′ ∪ t ′ .Set C 14 = D 14 ∪ T and C ij = D ij for 2 ≤ i < j.Let f ′ be the induced map.We argue identically to case 1, with T, T ′ instead of t, t ′ , to show that ({C ij } 1≤i<j≤4 , f ′ ) is a deflated tetrahedron filling x 1 , x 2 , x 3 , x 4 through o.This is again a contradiction to minimality.Case 2.2.v ∈ y 2 ω.

Figure 26 :
Figure 26: is a loop in T

T e 1 e 2 A
Figure 27: e 1 , e 2 are double 1-cells