# CAT(0) Polygonal Complexes are 2-Median

Shaked Bader  Department of Mathematics, University of Oxford, Oxford, UK  Nir
Lazarovich  Department of Mathematics, Technion—Israel Institute of
Technology, Haifa, Israel

###### Abstract

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.

##  1 Introduction

It is a well known fact that $\mathbb{R}$ 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 $x,y\in X$ we define the _interval_ between $x,y$ to be

$[x,y]=\left\\{z\;\mid\;d(x,z)+d(z,y)=d(x,y)\right\\}.$

  * •

The space $X$ is _median_ if for any $x,y,z\in X$, we have that
$[x,y]\cap[x,z]\cap[y,z]=\left\\{m(x,y,z)\right\\}$ is a unique point.

Examples of median spaces include: metric trees, $\ell^{1}$ products of median
spaces, $\ell^{1}(X)$ and $L^{1}(X,\mu)$, the vertices of a CAT(0) cube
complexes with the induced graph metric, see Roller [18]. In fact, by a result
of Chepoi [6] the class of 1-skeleta of CAT(0) cube complexes coincides with
the class of median graphs.

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
$\mathbb{E}^{2}$. Clearly, $\mathbb{E}^{2}$ 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
$x_{1},x_{2},x_{3},x_{4}\in\mathbb{E}^{2}$ and the four full triangles they
span $\left\\{\blacktriangle\left(x_{i},x_{j},x_{k}\right)\right\\}_{1\leq
i<j<k\leq 4}$. 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.

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

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
$\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}}$ for
$\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}$ for which that
intersection is non-empty.

We prove that a similar phenomenon, which we call _2-median_ , holds in
2-dimensional 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
$x,y,z\in X$. In Definition 4.1, we define the full triangle
$\blacktriangle\left(x,y,z\right)$ as the minimal subset of $X$ necessary in
order to fill the geodesic triangle $\triangle(x,y,z)$ by a disc. If
$\triangle(x,y,z)$ is a Jordan curve, the full triangle
$\blacktriangle\left(x,y,z\right)$ 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 $\blacktriangle\left(x,y,z\right)$ is homeomorphic
to a disc.

######  Definition 1.2.

The space $X$ is a _2-median space_ if for any $x_{1},x_{2},x_{3},x_{4}\in X$
either there exists $\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}$ such
that $\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}}$ is an interval, in which
case $\bigcap_{1\leq i<j<k\leq
4}{\blacktriangle\left(x_{i},x_{j},x_{k}\right)}=\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}}$,
or $\bigcap_{1\leq i<j<k\leq 4}{\blacktriangle\left(x_{i},x_{j},x_{k}\right)}$
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 3-dimensional Euclidean space $\mathbb{E}^{3}$ is empty;
it should be a non-positive curvature phenomenon – e.g. the intersection of
the triangles of the regular tetrahedron in the sphere $\mathbb{S}^{2}$ 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
$\blacktriangle\left(x,y,z\right)$ 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 §6 we show that the intersection of the four full triangles is non-empty by
showing that
$\blacktriangle\left(x_{1},x_{2},x_{3}\right)\subseteq\bigcup_{(i,j,k)\neq(1,2,3)}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$,
and using Sperner’s Lemma.

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:

######  Definition 1.3.

[See [14] Definition 2.13] A cellular map $Y\to X$ between two CW-complexes is
a _near-immersion_ if the restriction $Y-Y^{(0)}\to X$ is locally injective.

######  Theorem 1.4.

Let $X$ be a CAT(0) polygonal complex, and let $\Gamma$ be an embedded
polygonal curve with ordered vertices $x_{1},...,x_{n}$ such that
$\kappa(\Gamma)<4\pi$. Assume $f\mathrel{\mathop{\mathchar 58\relax}}\Delta\to
X$ is a near-immersion, where $\Delta$ is a finite $\Delta$-complex
homeomorphic to a disc and $f|_{\partial\Delta}\mathrel{\mathop{\mathchar
58\relax}}\partial\Delta\to\Gamma$ 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.

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.

### Acknowledgements.

The authors would like to thank Tobias Hartnick and Nima Hoda for fruitful
discussions at an early stage of this project, and Michah Sageev for numerous
helpful suggestions and corrections of this manuscript. The authors are
indebted to the anonymous referee for the careful reading of this manuscript
and the valuable suggested improvements. Both authors were partially supported
by the Israel Science Foundation (grant no. 1562/19).

##  2 CAT(0) polygonal complexes and their links

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].

######  Notation 2.1.

  * •

If $\xi,\zeta$ are points in $\operatorname{lk}(p)$, we denote by
$\angle_{p}(\xi,\zeta)$ the angle between $\xi$ and $\zeta$ in $X$ (as defined
in [4, Definition I.1.12]). This defines a metric $\angle_{p}$ on
$\operatorname{lk}(x)$. We denote the length of a path $\alpha$ in
$\operatorname{lk}(p)$ by $\measuredangle_{p}\alpha$, and by
$\measuredangle_{p}(\xi,\zeta)$ the induced length metric, i.e. the length of
the shortest path connecting them in $\operatorname{lk}(p)$. Note that
$\angle_{p}$ does not necessarily equal the metric $\measuredangle_{p}$.

We will abuse notation and for $x,y\in X$ and $\gamma$ a path in $X$ we denote
by $\angle_{p}(x,y),\measuredangle_{p}(\alpha),\measuredangle_{p}(x,y)$ the
angle, length and distance of the projections of $x,y$ and $\gamma$ to
$\operatorname{lk}(p)$.

In cases in which the point referred to is clear, we will sometimes omit the
subscript $p$,

  * •

Let $Y$ be a geodesic space, $x,y\in Y$. When there exists a unique geodesic
between $x,y$ (as is the case in CAT(0) spaces and when
$x,y\in\operatorname{lk}(q)$ and $\angle_{q}(x,y)<\pi$ for $q$ in a CAT(0)
space), we denote it by $\overline{xy}$.

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
$\operatorname{lk}(v)$ (i.e. the _girth_ of $\operatorname{lk}(v)$) has length
at least $2\pi$.

In the remainder of this section, we discuss the connection between the link
of point $p\in X$ and $X-p$.

Recall that there exists $\epsilon>0$ such that the sphere around $p$ of
radius $\epsilon$, $S$, can be identified with $\operatorname{lk}(p)$, and the
closest point projection, $\pi_{p}\mathrel{\mathop{\mathchar 58\relax}}X-p\to
S$, 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 $\pi_{p}(x)$
by $\hat{x}$.

######  Proposition 2.3.

Let $X$ be a CAT(0) polygonal complex, $p\in X$. Let $x,y\in X$, then
$\angle_{p}(x,y)=\min\left\\{\measuredangle_{p}(\hat{x},\hat{y}),\pi\right\\}$.

In particular, if $p\notin\overline{xy}$, then $\angle_{p}(x,y)<\pi$, so
$\angle_{p}(x,y)=\measuredangle_{p}(\hat{x},\hat{y})$ and there exists a
unique path in $\operatorname{lk}(p)$ between $\hat{x},\hat{y}$ of length
$\angle_{p}(x,y)$.

###### Proof.

By [4, Theorem I.7.16] there exists $r>0$ such that $S$, the sphere or radius
$r$ around $p$, is homeomorphic to $\operatorname{lk}(p)$ and the ball or
radius $r$ around $p$ is isometric to $C(\operatorname{lk}(p))$; where
$C(\operatorname{lk}(p))$ is the cone over $\operatorname{lk}(p)$ with the
metric defined such that for $u,v\in\operatorname{lk}(p)$ the angle between
$\overline{pu},\overline{pv}$ in $\operatorname{lk}(p)$ is
$\min\left\\{\measuredangle_{p}(x,y),\pi\right\\}$.

We can assume $x,y\in S$. As isometries preserve angles, we have that
$\angle_{p}(x,y)=\min\left\\{\measuredangle_{p}(x,y),\pi\right\\}$. Hence, if
$\angle_{p}(x,y)<\pi$, it must be that
$\measuredangle_{p}(x,y)=\angle_{p}(x,y)$. ∎

As the next example shows, in general if $X$ is a CAT(0) polygonal complex,
$p\in X$ and $q,r\in X$ such that $p\notin\overline{qr}$, it might be that
$\pi_{p}(\overline{qr})\neq\overline{\pi_{p}(q)\pi_{p}(r)}$. However, we will
see in 2.5 that $\pi_{p}(\overline{qr}),\overline{\pi_{p}(q)\pi_{p}(r)}$ are
homotopic relative to endpoints in $\operatorname{lk}(p)$.

######  Example 2.4.

Let $T$ be the metric graph with one vertex $o$ of degree three and three
1-cells of length 1 with vertices $a,b,c$. Consider $T\times[0,1]$ with its
CAT(0) metric, that is the $\ell^{2}$ metric, and the points
$p=(a,1),q=(b,0),r=(c,0)$, as in Fig. 3. Let $s$ be the intersection of
$\overline{qr}$ with $\left\\{o\right\\}\times[0,1]$, that is, $s=(o,0)$. Note
that
$\pi_{p}(s)\notin\overline{\pi_{p}(q)\pi_{p}(r)}=\left\\{\pi_{p}(q)\right\\}$.

Figure 3: Example in which
$\pi_{p}(\overline{qr})\neq\overline{\pi_{p}(q)\pi_{p}(r)}$

######  Claim 2.5.

Let $X$ be a CAT(0) polygonal complex and $p\in X$. Given $q,r\in X$ such that
$p\notin\overline{qr}$ we have that $\pi_{p}(\overline{qr})$ is homotopic
relative to endpoints in $\operatorname{lk}(p)$ to the geodesic between
$\pi_{p}(q),\pi_{p}(r)$. In particular, there exists a unique geodesic between
$\pi_{p}(q),\pi_{p}(r)$.

###### Proof.

Let $S$ be as in Proposition 2.3 and identify $\operatorname{lk}(p)$ with $S$.
Let $\alpha_{q}\mathrel{\mathop{\mathchar 58\relax}}[0,1]\to X$ be a
parametrisation of $\overline{q\hat{q}}$ where $\hat{q}\in
S=\operatorname{lk}(p)$, such that $\alpha_{q}(0)=q$ and
$\alpha_{q}(1)=\hat{q}$. Similarly consider $\alpha_{r}$. Let
$\gamma_{t}\mathrel{\mathop{\mathchar 58\relax}}[0,1]\to X$ be a
parametrisation of $\overline{\alpha_{q}(t)\alpha_{r}(t)}$. Let
$G\mathrel{\mathop{\mathchar 58\relax}}[0,1]\times[0,1]\to X-p$ be
$G(s,t)=\gamma_{t}(s)$.

We have that $G$ is continuous and well defined into $X-p$, so $\pi_{p}\circ
G$ gives a homotopy between $\pi_{p}\circ\gamma_{0}$ and
$\pi_{p}\circ\gamma_{1}$.

As in Proposition 2.3, the ball bounded by $S$ is isometric to the cone over
$S$. The cone over $\overline{\hat{q}\hat{r}}$ is convex and contains
$\gamma_{1}$, so $\pi_{p}\circ\gamma_{1}=\overline{\hat{q}\hat{r}}$. We get
that $\pi_{p}\circ G$ is a homotopy relative to
$\left\\{\hat{q},\hat{r}\right\\}$ between $\pi_{p}(\overline{qr})$ and
$\overline{\hat{q}\hat{r}}=\overline{\pi_{p}(q)\pi_{p}(r)}$.

∎

As an immediate corollary, we get the following.

######  Corollary 2.6.

Let $X$ be a CAT(0) polygonal complex, let $\Delta$ be a geodesic triangle
with vertices $x,y,z\in X$, and let $p\in X$.

  * •

If $p\in X-\Delta$, then $\pi_{p}(\Delta)$ is homotopic to the geodesic
triangle in $\operatorname{lk}(p)$ with vertices $\hat{x},\hat{y},\hat{z}$.

  * •

If $p\in\overline{xy}-(\overline{xz}\cup\overline{yz})$ then
$\pi_{p}(\Delta-p)$ is homotopic to
$\overline{\hat{x}\hat{z}}\cup\overline{\hat{y}\hat{z}}$.

  * •

If $p\in\overline{xy}\cap\overline{xz}-\overline{yz}$ then $\pi_{p}(\Delta-p)$
is homotopic to $\left\\{\hat{x}\right\\}\cup\overline{\hat{y}\hat{z}}$ if
$p\neq x$ and to $\overline{\hat{y}\hat{z}}$ if $p=x$.

  * •

If $p\in\overline{xy}\cap\overline{xz}\cap\overline{yz}$ then
$\pi_{p}(\Delta-p)=\left\\{\hat{u}\mid u\in\left\\{x,y,z\right\\}-p\right\\}$.

where all homotopies are relative to
$\left\\{\hat{x},\hat{y},\hat{z}\right\\}$ when they are defined. See Fig. 4
for an illustration of the four cases.

Figure 4: The four cases, $\operatorname{lk}(p)$ on the left of the
configuration

##  3 Some planar complexes

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.

######  Definition 3.1.

  * •

A _generalized disc_ $D$ is a contractible finite $\Delta$-complex (see [11]
Chapter 2 for definition of $\Delta$-complex) with a piecewise linear
embedding to $\mathbb{R}^{2}$, see Fig. 5. A _boundary cycle_ is a map
$\alpha\mathrel{\mathop{\mathchar 58\relax}}C\to D^{(1)}$ such that
$\alpha(C)$ 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 $\alpha$ then
$e_{i}^{-1},e_{i+1}$ 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 $\mathbb{R}^{2}$.

Any two parametrized boundary cycles of $D$ differ by precomposition by an
automorphism of $C$. By abuse of language, we call $\alpha$ _the_ boundary
cycle of $D$.

  * •

A _generalized triangle_ is a generalized disc $D$ with boundary cycle
$\alpha\mathrel{\mathop{\mathchar 58\relax}}C\to D$ together with 3 vertices
$x,y,z\in C$ on its boundary. We denote
$\overline{xy}^{D}=\alpha|_{\overline{xy}}$ where $\overline{xy}$ is the path
in $C$ connecting $x$ and $y$ and not passing through $z$. The images of
$x,y,z$ and the arcs $\overline{xy},\overline{xz},\overline{yz}$ under
$\alpha$ 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: Example of generalized disc

######  Definition 3.2.

A _generalized annulus_ is a finite $\Delta$-complex with a piecewise linear
embedding to $\mathbb{R}^{2}$ which is homotopy equivalent to $\mathbb{S}^{1}$
and has no separating points. For example see Fig. 6.

Given a generalized annulus $A$, it separates $\mathbb{R}^{2}$ 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 $A\cup D$ is a generalized disc and we define the _outer boundary_
of $A$ to be the boundary cycle of $A\cup D$.

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 $\alpha\mathrel{\mathop{\mathchar
58\relax}}C\to A^{(1)}$ be the inner boundary of $A$ where $C$ is a cycle
graph. The equivalence relation $\sim$ on $\alpha(C)$ is a _good gluing_ if
the relation $\sim$ identifies $\alpha(e_{i})$ with $\alpha(j_{i})$ where
$e_{1},\dots,e_{k},j_{k}^{-1},...,j_{1}^{-1}$ are the edges of $C$ in this
order, and at most one of
$\operatorname{int}(e_{i}),\operatorname{int}(j_{i})$ is on the outer boundary
of $A$. For example see Fig. 6.

Figure 6: generalized annulus with good gluing

######  Claim 3.4.

If $A$ is a generalized annulus and $\sim$ is a good gluing of $A$, then
$A/\sim$ is homeomorphic to a disc.

###### Proof.

Assume $A$ is a generalized annulus and $\sim$ is a good gluing of $A$. The
quotient $A/\sim$ 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 $\mathbb{S}^{1}$), 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 $A/\sim$.

The generalized annulus $A$ is homotopic to a circle and so $\chi(A)=0$. The
CW complex $A/\sim$ 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
$\chi(A/\sim)=\chi(A)+1=1$. Since the compact surface $A/\sim$ has one
boundary component and $\chi(A/\sim)=1$ it must be a disc. ∎

##  4 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.

######  Definition 4.1.

Let $x,y,z\in X$.

  1. 1.

Denote by $\triangle(x,y,z)=\overline{xy}\cup\overline{yz}\cup\overline{zx}$
the _geodesic triangle_ spanned by $x,y,z$.

  2. 2.

Define the _full triangle_ $\blacktriangle\left(x,y,z\right)$ to be the union
of $\triangle(x,y,z)$ and the set of points $p$ in $X-\triangle(x,y,z)$ for
which $\triangle(x,y,z)$ is not trivial in $\pi_{1}(X-p)$.

######  Example 4.2.

In Fig. 3 we see $\triangle(p,q,r)$ in green in the space $T\times[0,1]$ and
its corresponding full triangle in yellow. Note that
$\blacktriangle\left(p,q,r\right)$ 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
$\blacktriangle\left(x,y,w\right)$, but is in $\operatorname{conv}(x,y,w)$,
and it is in $\blacktriangle\left(x,z,w\right)=\operatorname{conv}(x,z,w)$ and
in
$\overline{zy}\subseteq\blacktriangle\left(x,y,z\right)\cap\blacktriangle\left(y,z,w\right)$.
So the intersection of the four convex hulls is not a single point.

Figure 7: Intersection of the convex hulls is not a single point. Top left:
$\blacktriangle\left(x,y,z\right)=\operatorname{conv}(x,y,z)$ and
$\blacktriangle\left(x,z,w\right)=\operatorname{conv}(x,z,w)$; bottom left:
$\blacktriangle\left(x,y,w\right),\blacktriangle\left(z,y,w\right)$; right:
$\operatorname{conv}(x,y,w)$

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.

######  Definition 4.3.

A _spiky triangle (resp. full spiky triangle)_ is the topological space
obtained from gluing a possibly degenerate interval to each of the vertices of
a possibly degenerate Euclidean triangle (resp. full Euclidean triangle), as
in Fig. 8.

Figure 8: A full spiky triangle

######  Lemma 4.4.

Let $x,y,z\in X$. Then the geodesic triangle $\triangle(x,y,z)$ is a spiky
triangle.

###### Proof.

The intersection of any two geodesics in a CAT(0) space is either empty or a
(possibly degenerate) segment. Consider $\overline{xy}\cap\overline{xz}$, this
is a (possibly degenerate) segment, as it is non-empty. Let $x^{\prime}$ be
the point furthest away from $x$ on that interval and define
$y^{\prime},z^{\prime}$ similarly. We get
$\triangle(x,y,z)=\triangle(x^{\prime},y^{\prime},z^{\prime})\cup\overline{xx^{\prime}}\cup\overline{yy^{\prime}}\cup\overline{zz^{\prime}}$,
as in Fig. 8.

It is easy to see that unless $\triangle(x,y,z)$ is a tripod,
$\overline{xx^{\prime}}\cap\overline{yz}=\emptyset,\overline{yy^{\prime}}\cap\overline{xz}=\emptyset,\overline{zz^{\prime}}\cap\overline{xy}=\emptyset$.
It follows that $\triangle(x,y,z)$ is a spiky triangle. ∎

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 $2\pi$ and let $x,y,z\in X$ be
such that $d(x,y),d(y,z),d(x,z)<\pi$, then the geodesic triangle
$\triangle(x,y,z)$ (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 $<\pi$ in $G$ is either empty or a (possibly
degenerate) segment by assumption on the girth”.

We want to understand how $\blacktriangle\left(x,y,z\right)$ looks locally.
Consider $x,y,z\in X$ and $q\in\blacktriangle\left(x,y,z\right)$. Let $B$ be a
small enough ball around $q$. We want to describe
$B\cap\blacktriangle\left(x,y,z\right)$.

We start by analyzing points $q$ in
$\blacktriangle\left(x,y,z\right)-\triangle(x,y,z)$. By Proposition 2.3,
$q\in\blacktriangle\left(x,y,z\right)-\triangle(x,y,z)$ if and only if
$c=\pi_{q}(\triangle(x,y,z))$ is a homotopically non-trivial cycle in
$\operatorname{lk}(q)$ or $q\in\triangle(x,y,z)$. We claim that for a small
enough ball $B$ around $q$ the points of
$B\cap\blacktriangle\left(x,y,z\right)$ 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 non-
backtracking 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 $q\in\blacktriangle\left(x,y,z\right)-\triangle(x,y,z)$ and let $B$ be a
small enough ball around $q$ such that $\partial B\cong\operatorname{lk}(q)$
and $B\cap\triangle(x,y,z)=\emptyset$. Let $C$ be the essential part of
$\pi_{q}(\triangle(x,y,z))$. Then for $p\in B$, $\hat{p}\in C$ if and only if
$p\in\blacktriangle\left(x,y,z\right)$.

###### Proof.

By Corollary 2.6, $\triangle(x,y,z)$ is homotopic to
$\triangle(\hat{x},\hat{y},\hat{z})\subseteq\operatorname{lk}(q)\cong\partial
B$ and
$\measuredangle(\hat{x},\hat{y}),\measuredangle(\hat{x},\hat{z}),\measuredangle(\hat{y},\hat{z})<\pi$
by assumption that $q\notin\triangle(x,y,z)$. By Lemma 4.5,
$\triangle(\hat{x},\hat{y},\hat{z})$ is a spiky triangle, and since
$q\in\blacktriangle\left(x,y,z\right)$, it is non-degenerate. The essential
part of $\triangle(\hat{x},\hat{y},\hat{z})$, $C$, is thus an embedded circle
in $\partial B$. In conclusion, we got a homotopy in $X-p$ of
$\triangle(x,y,z)\to C$.

Let $\hat{C}$ be the cone over $C$ with respect to $q$. It is a convex disc
and $\hat{p}\in C$ if and only if $p$ is in the cone. If $p\notin\hat{C}$ then
$C$ is filled with a disk $\hat{C}$ in $X-p$, and so
$p\notin\blacktriangle\left(x,y,z\right)$. If $p\in\hat{C}$ then $C$ projects
to a non-trivial loop in $\operatorname{lk}(p)$ (since $\hat{C}$ is convex).
It follows that $C$ is non trivial in $\pi_{1}(X-p)$, whence
$p\in\blacktriangle\left(x,y,z\right)$.

∎

Similarly one proves the following claim

######  Claim 4.8.

Let $q\in\triangle(x,y,z)$ and let $B$ be a small enough ball around $q$ such
that $\partial B\cong\operatorname{lk}(q)$ and $B$ does not intersect any side
of $\triangle(x,y,z)$ not containing $q$. Let $C$ be the essential part of
$\pi_{q}(\triangle(x,y,z)-q)$. Then for $p\in B$, $\hat{p}\in C$ if and only
if $p\in\blacktriangle\left(x,y,z\right)$.

###### Proof.

We argue similarly to the proof of 4.7, by analysing $\hat{C}$ (in the
notations of the proof of 4.7) for the different cases in Corollary 2.6. ∎

We remind that Corollary 2.6 characterizes the connected components of
$\pi_{q}(\triangle(x,y,z)-q)$.

In Theorem 5.1 we will show that full triangles in $X$ are homeomorphic to
full spiky triangles.

######  Definition 4.9.

We say $x_{1},x_{2},x_{3}\in X$ are _strongly non-collinear_ if they are
distinct and for any $\left\\{i,j,k\right\\}=\left\\{1,2,3\right\\}$ we have
$\overline{x_{i}x_{j}}\cap\overline{x_{i}x_{k}}=\left\\{x_{i}\right\\}$.

######  Definition 4.10.

The points $x_{1},x_{2},x_{3},x_{4}\in X$ form a _lanky quadrilateral_ if
$\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}}$ is a non-degenerate segment
for some $\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}$. E.g. see Fig. 9.
Note that this does not imply that the four points lie on a geodesic.

Figure 9: $x_{1},x_{2},x_{3},x_{4}$ form a lanky quadrilateral

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.

##  5 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 $x,y,z\in X$ and let $T$ be an (abstract) full spiky triangle with
$\partial T\cong\triangle(x,y,z)$. Then there exists an embedding
$f\mathrel{\mathop{\mathchar 58\relax}}T\hookrightarrow X$ such that
$f|_{\partial T}$ maps $\partial T$ homeomorphically onto $\triangle(x,y,z)$.

Any such map has image $\blacktriangle\left(x,y,z\right)$.

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 $\Gamma$ is a piecewise linear curve in a CAT(0) polygonal complex $X$
and assume there exists a $\Delta$-complex homeomorphic to $\mathbb{D}^{2}$
and a cellular and injective map $f\mathrel{\mathop{\mathchar 58\relax}}D\to
X$ such that $f(\partial D)=\Gamma$, then $f(D)$ equals the union of $\Gamma$
with the set of points $p\in X-\Gamma$ such that $\Gamma$ is not trivial in
$\pi_{1}(X-p)$.

In particular, for any $g\mathrel{\mathop{\mathchar
58\relax}}\mathbb{D}^{2}\to X$ cellular with respect to some $\Delta$-complex
structure on $\mathbb{D}^{2}$ and injective such that
$g(\partial\mathbb{D})=\Gamma$, we have $g(\mathbb{D}^{2})=f(D)$.

###### Proof.

Denote by $M$ the union of $\Gamma$ with the set of points $p\in X$ such that
$\Gamma$ is not trivial in $\pi_{1}(X-p)$. Clearly $M\subseteq f(D)$, as if
$p\notin f(D)$ then $f$ provides a null homotopy of $\Gamma$ in $X-p$ and thus
$p\notin M$.

For the other direction, let $p\in D$. There are projections, as in 2.5,
$\pi_{p}\mathrel{\mathop{\mathchar 58\relax}}\partial
D\to\operatorname{lk}(p)$ and $\pi_{f(p)}\mathrel{\mathop{\mathchar
58\relax}}X-f(p)\to\operatorname{lk}(f(p))$ and as $f$ is injective, we have
$f_{p}\circ\pi_{p}$ is homotopic equivalent to $\pi_{f(p)}\circ f$.

As $\pi_{p}(\partial D)$ is a cycle, so is $f_{p}(\pi_{p}(\partial D))$, hence
it is not null homotopic in $\operatorname{lk}(f(p))$ which is a graph. By the
equality above $\pi_{f(p)}\circ f(\partial D)=\pi_{f(p)}(\Gamma)$ is not
trivial, so $\Gamma$ is not trivial in $\pi_{1}(X-p)$. ∎

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
$\partial T$ homeomorphically onto $\triangle(x,y,z)$ and the triangle part of
$T$ is mapped injectively, then $f$ is injective.

###### Proof.

Let $x,y,z\in X$ and denote by $x^{\prime}\in X$ the point in which
$\overline{xy},\overline{xz}$ diverge and similarly denote
$y^{\prime},z^{\prime}$, as in Fig. 8. By Lemma 4.4, $\triangle(x,y,z)$ is
homeomorphic to a spiky triangle and
$\triangle(x^{\prime},y^{\prime},z^{\prime})$ is homeomorphic to a triangle.

By Theorem 78 in [20], there exists a full triangle $D$ and
$f\mathrel{\mathop{\mathchar 58\relax}}D\to X$ an embedding such that
$f|_{\partial D}\mathrel{\mathop{\mathchar 58\relax}}\partial
D\to\triangle(x^{\prime},y^{\prime},z^{\prime})$ is a homeomorphism.

By Lemma 5.2
$f(D)=\blacktriangle\left(x^{\prime},y^{\prime},z^{\prime}\right)$.

Glue to $D$ the three spikes, i.e. let $I_{1},I_{2},I_{3}$ be three intervals
homeomorphic to
$\overline{xx^{\prime}},\overline{yy^{\prime}},\overline{zz^{\prime}}$
respectively and let $a_{j}\in I_{j}$ be one of their endpoints. Consider
$T=D\sqcup I_{1}\sqcup I_{2}\sqcup I_{3}/\sim$ where $a_{1}\sim
f^{-1}(x^{\prime}),a_{2}\sim f^{-1}(y^{\prime}),a_{3}\sim f^{-1}(z^{\prime})$
and consider the map $\bar{f}\mathrel{\mathop{\mathchar 58\relax}}T\to X$,
induced from $f$ and the homeomorphisms of $I_{j}$ to the geodesic segments.

In order to show $\bar{f}$ is injective it is enough to show that for
$p\in\partial T-D$ we have
$\bar{f}(p)\notin\bar{f}(D)=\blacktriangle\left(x^{\prime},y^{\prime},z^{\prime}\right)$.
Indeed as $\bar{f}|_{\partial T}$ is a homeomoprhism, we have that
$\bar{f}(p)\notin\triangle(x^{\prime},y^{\prime},z^{\prime})$, hence, by 4.7,
$\bar{f}(p)\in\blacktriangle\left(x^{\prime},y^{\prime},z^{\prime}\right)$ if
and only if the concatenation of
$\overline{\hat{x^{\prime}}\hat{y^{\prime}}},\overline{\hat{y^{\prime}}\hat{z^{\prime}}},\overline{\hat{z^{\prime}}\hat{x^{\prime}}}$
in $\operatorname{lk}(p)$ is not null homotopic. But two out of the three
segments are constant in $\operatorname{lk}(p)$, 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
$\triangle(x,y,z)$. We are left to prove that any such map
$f\mathrel{\mathop{\mathchar 58\relax}}T\to X$ has image
$\blacktriangle\left(x,y,z\right)$.

Let $D$ be the triangle part of $T$ and let $x^{\prime},y^{\prime},z^{\prime}$
as above. We have that $f$ maps $\partial D$ homeomorphically onto
$\triangle(x,y,z)$, hence
$f(D)=\blacktriangle\left(x^{\prime},y^{\prime},z^{\prime}\right)$ by Lemma
5.2. We get $f(T)=f(\partial T)\cup
f(D)=\triangle(x,y,z)\cup\blacktriangle\left(x^{\prime},y^{\prime},z^{\prime}\right)=\blacktriangle\left(x,y,z\right)$.
∎

######  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].

##  6 Existence of intersection

For $x_{1},x_{2},x_{3},x_{4}\in X$ let $W_{1},W_{2},W_{3},W_{4}$ be the four
full triangles they define, where $W_{i}$, $1\leq i\leq 4$, 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 $\bigcap_{i=1}^{4}W_{i}\neq\emptyset$. 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,
$f_{i}\mathrel{\mathop{\mathchar 58\relax}}F_{i}\to W_{i}$ 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

$S\cong\bigcup_{i=1}^{4}F_{i}/\sim\text{ where }x\sim y\iff x\in\partial
F_{i},y\in\partial F_{j},f_{i}(x)=f_{j}(y).$

######  Claim 6.1.

If $p\in W_{i}$ then there exists $j\neq i$ such that $p\in W_{j}$.

###### Proof.

Without loss of generality $p\in W_{4}$. Assume towards contradiction that
$p\notin\bigcup_{i=1}^{3}W_{i}$. By assumption, $f(S-F_{4})\subseteq X-p$.
Since $S-F_{4}$ is contractible, $f|_{\partial
F_{4}}\mathrel{\mathop{\mathchar 58\relax}}\partial
F_{4}\to\triangle(x_{1},x_{2},x_{3})$ is null homotopic in $X-p$, in
contradiction to the definition of $W_{4}$. ∎

######  Proposition 6.2.

$\bigcap_{1\leq i\leq 4}W_{i}\neq\emptyset$.

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 $\partial
F_{4}$ we define it as in Fig. 10, i.e. for a 0-cell $v\in
f^{-1}(\overline{x_{2}x_{3}})-f^{-1}(x_{3})$ we let its colour be 1, the
colours of 0-cells in $f^{-1}(\overline{x_{1}x_{3}})-f^{-1}(x_{1})$ and
$f^{-1}(\overline{x_{1}x_{2}})-f^{-1}(x_{2})$ are 2 and 3 respectively. For
$v$ a 0-cell in the interior of $F_{4}$ we define its colour to be
$\min\left\\{j\mid f\left(v\right)\in W_{j}\right\\}$ 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 $p^{\prime}_{n}$ be the barycenter of $T_{n}$,
by compactness (up to passing to a subsequence) $p^{\prime}_{n}$ converges to
$p^{\prime}\in F_{4}$. We now claim that $p=f\left(p^{\prime}\right)$ is in
$\bigcap_{1\leq i\leq 4}W_{i}$.

Note that for $j\in\left\\{1,2,3\right\\}$ the sequence of vertices of $T_{n}$
coloured $j$ also converges to $p^{\prime}$, as the bound on the diameter of
2-cells goes to 0, hence $p^{\prime}$ is arbitrarily close to $W_{j}$ for all
$j\in\left\\{1,2,3\right\\}$, hence $p\in W_{j}$, and of course $p\in W_{4}$,
as $p^{\prime}\in F_{4}$, so we are done. ∎

##  7 Discreteness of intersection

In this section we will prove the following theorem.

######  Theorem 7.1.

Let $x_{1},x_{2},x_{3},x_{4}\in X$, denote $W_{1},W_{2},W_{3},W_{4}$ the four
full triangles they define, then

  * •

If $x_{1},x_{2},x_{3},x_{4}$ do not form a lanky quadrilateral, then
$\bigcap_{i=1}^{4}W_{i}$ is a finite set of points.

  * •

If $x_{1},x_{2},x_{3},x_{4}$ form a lanky quadrilateral, then
$\bigcap_{i=1}^{4}W_{i}$ is the union of a finite set and
$\bigcup_{\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}}(\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}})$,
in particular it is 1-dimensional.

Let $x_{1},x_{2},x_{3},x_{4}\in X$, we will refer to them as vertices and to
the geodesic segments between them as sides. By opposite sides we mean
$\overline{x_{i}x_{j}},\overline{x_{k}x_{l}}$ for
$\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}$. Denote
$W_{1},W_{2},W_{3},W_{4}$ the four full triangles in $X$. Assume throughout
this section that $p\in\bigcap_{i=1}^{4}W_{i}$.

We will study the link of $p$ and show that the case
$p\in\bigcap_{i=1}^{4}W_{i}$ is very restricted. Indeed, denoting
$\hat{x}_{1},\hat{x}_{2},\hat{x}_{3},\hat{x}_{4}$ the image of
$x_{1},x_{2},x_{3},x_{4}$ in $\operatorname{lk}(p)$, we have from 4.7 and 4.8
that either two are antipodal, that is $p$ is on a side; or
$\triangle(\hat{x}_{i},\hat{x}_{j},\hat{x}_{k})\subseteq\operatorname{lk}(p)$
are well defined for all $i,j,k\in\left\\{1,2,3,4\right\\}$, distinct and, by
Corollary 2.6, not trivial in homotopy.

######  Claim 7.2.

If $p\in\bigcap_{i=1}^{4}W_{i}-X^{(1)}$, then $p$ is in the intersection of
two opposite sides.

###### Proof.

The link of $p$ is a circle of length $2\pi$, as in Fig. 11. Assume by
contradiction that $p$ is not on a side containing $x_{i}$, that is, the
antipodal point to $\hat{x}_{i}$ is not $\hat{x}_{j}$ for all
$j\in\left\\{1,2,3,4\right\\}$. The points $\hat{x}_{i}$ and its antipodal
separate $\operatorname{lk}(p)$ into two arcs. By the pigeonhole principle,
one of the arcs contains at least two out of
$\left\\{\hat{x}_{j}\right\\}_{j\neq i}$, say $\hat{x}_{k},\hat{x}_{l}$, so
$\triangle(\hat{x}_{i},\hat{x}_{k},\hat{x}_{l})$ is null homotopic in
$\operatorname{lk}(p)$ which is a contradiction to
$p\in\blacktriangle\left(x_{i},x_{k},x_{l}\right)$ and Corollary 2.6, which
states that $\triangle(\hat{x}_{i},\hat{x}_{k},\hat{x}_{l})$ is homotopic to
$\pi_{p}(\triangle(x_{i},x_{k},x_{l}))$.

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, $p\in\bigcap_{j=2,3,4}\overline{x_{1}x_{j}}$ and in particular
$\hat{x}_{2}=\hat{x}_{3}=\hat{x}_{4}$. That cannot be, as then
$\triangle(\hat{x}_{2},\hat{x}_{3},\hat{x}_{4})$ is null homotopic in
$\operatorname{lk}(p)$ which is again a contradiction to Corollary 2.6.

Figure 11: $\operatorname{lk}(p)$ for $p\notin X^{(1)}$

∎

######  Claim 7.3.

If $p\in\bigcap_{i=1}^{4}W_{i}$ and $p\in X^{(1)}-X^{(0)}$ 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 $\operatorname{lk}(p)$ 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 $\operatorname{lk}(p)$ as longitudes in
order to differentiate from the 1-cells in $X$. We may consider only the
longitudes on which $\hat{x}_{1},\hat{x}_{2},\hat{x}_{3},\hat{x}_{4}$ 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 $\pi$ and if
$p\notin\overline{x_{i}x_{j}}$ then by Proposition 2.3
$\measuredangle_{p}(\hat{x}_{i},\hat{x}_{j})<\pi$. Hence if
$p\notin\triangle(x_{i},x_{j},x_{k})$, then
$\triangle(\hat{x}_{i},\hat{x}_{j},\hat{x}_{k})$ is of length $<3\pi$, so the
non-backtracking representative of
$\triangle(\hat{x}_{i},\hat{x}_{j},\hat{x}_{k})$ passes through each of the
poles at most once.

Case 1. $\hat{x}_{1},\hat{x}_{2},\hat{x}_{3},\hat{x}_{4}$ lie on three
longitudes. without loss of generality $\hat{x}_{1},\hat{x}_{2}$ lie on the
same longitudes and $\hat{x}_{1}$ is closer to $S$ than $\hat{x}_{2}$, as in
Fig. 12.

Figure 12: $\operatorname{lk}(p)$ for the case $p\in X^{(1)}-X^{(0)}$ and
three relevant longitudes

Assume towards contradiction that $p$ is not on any side, so the triangles
$\left\\{\triangle(\hat{x}_{i},\hat{x}_{j},\hat{x}_{k})\right\\}_{1\leq
i<j<k\leq 4}$ are well defined in $\operatorname{lk}(p)$. They are not null-
homotopic, by assumption on $p$.

We have that $S\in\overline{\hat{x}_{1}\hat{x}_{3}}$, as if not,
$N\in\overline{\hat{x}_{1}\hat{x}_{3}}$. Hence
$N\in\overline{\hat{x}_{2}\hat{x}_{3}}$, as a subset of a geodesic is a
geodesic. It follows that $\triangle(\hat{x}_{1},\hat{x}_{2},\hat{x}_{3})$ is
null homotopic which is a contradiction to
$p\in\blacktriangle\left(x_{1},x_{2},x_{3}\right)$.

Similar reasoning replacing 3 by 4 gives
$S\in\overline{\hat{x}_{1}\hat{x}_{4}}$. Similar reasoning replacing 1 by 2
and $S$ by $N$ gives
$N\in\overline{\hat{x}_{2}\hat{x}_{3}},\overline{\hat{x}_{2}\hat{x}_{4}}$.

We arrive to a contradiction, as the geodesic between
$\hat{x}_{3},\hat{x}_{4}$ does not pass through one of the poles, say $S$,
hence the triangle $\triangle(\hat{x}_{2},\hat{x}_{3},\hat{x}_{4})$ is null
homotopic - does not pass through $S$.

Case 2. $\hat{x}_{1},\hat{x}_{2},\hat{x}_{3},\hat{x}_{4}$ lie on four
longitudes. It cannot be that in $\operatorname{lk}(p)$ 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
$\hat{x}_{1}$ pass through $S$. We get that all other geodesics pass through
$N$ because $\triangle(\hat{x}_{1},\hat{x}_{i},\hat{x}_{j})$ is not null
homotopic for $2\leq i<j\leq 4$. But then
$\triangle(\hat{x}_{2},\hat{x}_{3},\hat{x}_{4})$ is null homotopic, as all the
geodesic segments in it pass through $N$, so do not pass through $S$. This is
a contradiction to $p\in\blacktriangle\left(x_{2},x_{3},x_{4}\right)$.

Assume without loss of generality that $\hat{x}_{1}$ is the closest to $S$ and
$\hat{x}_{4}$ is the closest to $N$ (need not be unique), as in Fig. 13. If
the geodesic between $\hat{x}_{1},\hat{x}_{4}$ passes through $S$
(respectively $N$) then all geodesics emanating from $\hat{x}_{1}$
(respectively $\hat{x}_{4}$) pass through $S$ (respectively $N$). So all
geodesics emanating from $\hat{x}_{1}$ or $\hat{x}_{4}$ pass through the same
pole which is a contradiction by the previous paragraph.

Figure 13: $\operatorname{lk}(p)$ for the case $p\in X^{(1)}-X^{(0)}$ and four
relevant longitudes

∎

###### Proof of Theorem 7.1.

By 7.2 and 7.3, every point of $\bigcap_{i=1}^{4}W_{i}$ 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 $X$,
$X^{(0)}\cap\bigcup_{i=1}^{4}W_{i}$ 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 $\bigcap_{i=1}^{4}W_{i}$ is finite. If they
do, we have that $\bigcap_{i=1}^{4}W_{i}$ 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 $x_{1},x_{2},x_{3},x_{4}\in X$, $x_{1}\neq x_{2}$ and
$q\in\blacktriangle\left(x_{1},x_{2},x_{3}\right)\cap\blacktriangle\left(x_{1},x_{2},x_{4}\right)$,
then $q$ is not an isolated point in the intersection.

###### Proof.

Let $\hat{x}_{1},\hat{x}_{2},\hat{x}_{3},\hat{x}_{4}$ be the images of
$x_{1},x_{2},x_{3},x_{4}$ in $\operatorname{lk}(q)$.

Note that if $q\in\overline{x_{1}x_{2}}$ we are done, as
$\overline{x_{1}x_{2}}\subseteq\blacktriangle\left(x_{1},x_{2},x_{3}\right)\cap\blacktriangle\left(x_{1},x_{2},x_{4}\right)$
so $q$ is not an isolated point in the intersection. Assume this is not the
case.

By 4.7 and 4.8 it is enough to show that $C_{3},C_{4}$, the essential parts of
$\pi_{q}(\triangle(x_{1},x_{2},x_{3})-q),\pi_{q}(\triangle(x_{1},x_{2},x_{4})-q)$
in $\operatorname{lk}(q)$ respectively, intersect. We will actually show that
they intersect on $\overline{\hat{x}_{1}\hat{x}_{2}}$.

Clearly $C_{3},C_{4}$ intersect $\overline{\hat{x}_{1}\hat{x}_{2}}$. If for
some $i\in\left\\{3,4\right\\}$,
$q\in\overline{x_{1}x_{i}}\cap\overline{x_{2}x_{i}}$ we have by Corollary 2.6
that $\overline{\hat{x}_{1}\hat{x}_{2}}\subseteq C_{i}$ and so $C_{3}\cap
C_{4}\neq\emptyset$ and we are done. So assume
$q\notin\overline{\hat{x}_{1}\hat{x}_{3}}\cap\overline{\hat{x}_{2}\hat{x}_{3}},\overline{\hat{x}_{1}\hat{x}_{4}}\cap\overline{\hat{x}_{2}\hat{x}_{4}}$.
In this case, again by Corollary 2.6, each of $C_{3},C_{4}$ is a cycle or an
arc depending on whether $q\in\triangle(x_{1},x_{2},x_{i})$ or not.

If $C_{3}$ is a cycle, then from Lemma 4.5
$C_{3}=\triangle(\hat{z}_{1},\hat{z}_{2},\hat{z}_{3})$ where $\hat{z}_{i}$ is
the point closest to $\hat{x}_{i}$ on $C_{3}$, that is it is the point
furthest away from $\hat{x}_{i}$ on
$\overline{\hat{x}_{i}\hat{x}_{j}}\cap\overline{\hat{x}_{i}\hat{x}_{k}}$ where
$\left\\{i,j,k\right\\}=\left\\{1,2,3\right\\}$.

Similarly if $C_{4}$ is a cycle then
$C_{4}=\triangle(\hat{w}_{1},\hat{w}_{2},\hat{w}_{4})$ where $\hat{w}_{i}$ is
the point closest to $\hat{x}_{i}$ on $C_{4}$, that is it is the point
furthest away from $\hat{x}_{i}$ on
$\overline{\hat{x}_{i}\hat{x}_{j}}\cap\overline{\hat{x}_{i}\hat{x}_{k}}$ where
$\left\\{i,j,k\right\\}=\left\\{1,2,4\right\\}$.

If $C_{3}$ is an arc then $q\in\overline{x_{1}x_{3}}$ or
$\overline{x_{2}x_{3}}$ and
$C_{3}=\overline{\hat{x}_{1}\hat{z}_{3}}\cup\overline{\hat{x}_{1}\hat{z}_{3}}$
where $\hat{z}_{3}$ is defined as above. Similarly if $C_{4}$ is an arc
$C_{4}=\overline{\hat{x}_{1}\hat{w}_{4}}\cup\overline{\hat{x}_{1}\hat{w}_{4}}$

In all cases $C_{i}\cap\overline{\hat{x_{1}}\hat{x_{2}}}$ 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
$\overline{\hat{x}_{1}\hat{x}_{2}}$, that is
$C_{3}\cap\overline{\hat{x}_{1}\hat{x}_{2}},C_{4}\cap\overline{\hat{x}_{1}\hat{x}_{2}}$
are two disjoint arcs. Without loss of generality
$C_{3}\cap\overline{\hat{x}_{1}\hat{x}_{2}}$ is closer to $\hat{x}_{1}$ than
$C_{4}\cap\overline{\hat{x}_{1}\hat{x}_{2}}$. The endpoint of
$C_{3}\cap\overline{\hat{x}_{1}\hat{x}_{2}}$ furthest from $\hat{x}_{1}$ is
$\hat{z}_{2}$ and the endpoint of $C_{4}\cap\overline{\hat{x}_{1}\hat{x}_{2}}$
furthest from $\hat{x}_{2}$ is $\hat{w}_{1}$, as in Fig. 14. By assumption
$\overline{\hat{x}_{1}\hat{z}_{2}}\cap\overline{\hat{w}_{1}\hat{x}_{2}}=\emptyset$.

Figure 14:
$\triangle(\hat{x}_{1},\hat{x}_{2},\hat{x}_{3}),\triangle(\hat{x}_{1},\hat{x}_{2},\hat{x}_{4})$
in $\operatorname{lk}(q)$

Let $\alpha=d(\hat{w}_{1},\hat{x_{2}})$ and
$\beta=d(\hat{z}_{2},\hat{x_{1}})$.

Note that:

  * •

$d(\hat{w}_{1},\hat{x}_{4})\geq\pi-\alpha$.

This is because:

    * –

If $q\notin\overline{x_{2}x_{4}}$ then the concatenation of
$\overline{\hat{x_{2}}\hat{w}_{1}},\overline{\hat{w}_{1}\hat{x_{4}}},\overline{\hat{x_{4}}\hat{x_{2}}}$
includes an essential cycle, so is of length $\geq 2\pi$, but
$\overline{\hat{x}_{2}\hat{x}_{4}}$ is of length $<\pi$ by Proposition 2.3, so
$\alpha+d(\hat{w}_{1},\hat{x}_{4})\geq 2\pi-d(\hat{x}_{2},\hat{x}_{4})>\pi$
and this gives us what we want.

    * –

If $q\in\overline{x_{2}x_{4}}$ we can choose a path $\hat{\pi}$ of length
$\pi$ between $\hat{x}_{4},\hat{x}_{2}$ and the concatenation of
$\overline{\hat{x}_{2}\hat{w}_{1}},\overline{\hat{w}_{1}\hat{x}_{4}},\hat{\pi}$
either contains a cycle and then, as in the previous case, we get what we want
or is fully backtracking and then $\alpha+d(\hat{w}_{1},\hat{x}_{4})\geq\pi$.

  * •

$d(\hat{x}_{1},\hat{w}_{1})+d(\hat{w}_{1},\hat{x}_{4})<\pi$.

This is because the concatenation of
$\overline{\hat{x}_{1}\hat{w}_{1}},\overline{\hat{w}_{1}\hat{x}_{4}}$ is
exactly $\overline{\hat{x}_{1}\hat{x}_{4}}$, by choice of $\hat{w}_{1}$, so we
are done by Proposition 2.3.

From the second point we have $\beta=d(\hat{x}_{1},\hat{z}_{2})\leq
d(\hat{x}_{1},\hat{w}_{1})<\pi-d(\hat{w}_{1},\hat{x}_{4})$, and then from the
first point $\beta<\alpha$. But of course, this argument was symmetric so we
also have $\alpha<\beta$ which is a contradiction. ∎

##  8 Proof of Theorem for lanky quadrilateral case

The statement we will prove in this section is the second part of A:

######  Theorem 8.1.

Assume $x_{1},x_{2},x_{3},x_{4}\in X$ such that
$\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$ is a segment, then
$\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)=\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$

The situation described in the theorem corresponds to triangles whose union is
as in Fig. 15.

Figure 15: $\bigcup_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$ when $x_{1},x_{2},x_{3},x_{4}$
form a lanky quadrilateral

######  Lemma 8.2.

Given $x_{1},x_{2},x_{3},x_{4}\in X$, if
$\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}\neq\emptyset$ then

$\overline{x_{1}x_{3}}\cap\overline{x_{2}x_{4}},\overline{x_{1}x_{4}}\cap\overline{x_{2}x_{3}}\subseteq\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}},$

hence
$\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}}=\overline{x_{i^{\prime}}x_{j^{\prime}}}\cap\overline{x_{k^{\prime}}x_{l^{\prime}}}$
for
$\left\\{i,j,k,l\right\\}=\left\\{i^{\prime},j^{\prime},k^{\prime},l^{\prime}\right\\}=\left\\{1,2,3,4\right\\}$,
when both intersections are non-empty.

###### Proof.

We will prove
$\overline{x_{1}x_{3}}\cap\overline{x_{2}x_{4}}\subseteq\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$,
the proof of the other inclusion is the same up to renaming.

By assumption there exists
$p\in\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$. Let
$q\in\overline{x_{1}x_{3}}\cap\overline{x_{2}x_{4}}$.

Things are now as in Fig. 16, where lines of the same colour are geodesics and
the letters in gray are the lengths of each part of the geodesic.

Figure 16: The geodesic segments between $x_{i},x_{j}$

By the triangle inequality, we have the following

$\displaystyle l_{1}+l_{2}$ $\displaystyle\leq m_{1}+k_{2}$ $\displaystyle
j_{1}+j_{2}$ $\displaystyle\leq m_{2}+k_{1}$ $\displaystyle k_{1}+k_{2}$
$\displaystyle\leq j_{1}+l_{2}$ $\displaystyle m_{1}+m_{2}$ $\displaystyle\leq
l_{1}+j_{2}.$

Summing up these inequalities we get
$l_{1}+l_{2}+k_{1}+k_{2}+m_{1}+m_{2}+j_{1}+j_{2}\leq
l_{1}+l_{2}+k_{1}+k_{2}+m_{1}+m_{2}+j_{1}+j_{2}$, so all inequalities are
equalities.

Geodesics are unique in a CAT(0) space so we get

$\displaystyle\overline{x_{1}q}\cup\overline{qx_{2}}$
$\displaystyle=\overline{x_{1}x_{2}}$
$\displaystyle\overline{x_{4}q}\cup\overline{qx_{3}}$
$\displaystyle=\overline{x_{4}x_{3}}$

In particular $q\in\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$. ∎

######  Lemma 8.3.

In the conditions of Theorem 8.1,
$\blacktriangle\left(x_{1},x_{3},x_{4}\right)\cap\blacktriangle\left(x_{2},x_{3},x_{4}\right)$
is equal to $\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$.

###### Proof.

It suffices to show
$\blacktriangle\left(x_{1},x_{3},x_{4}\right)\cap\blacktriangle\left(x_{2},x_{3},x_{4}\right)\subseteq\bigcap_{1\leq
i<j<k\leq 4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$.

Denote by $y_{1},y_{2}$ the two endpoints of
$\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$ such that $y_{1}$ is closer
to $x_{1}$ than to $x_{2}$ and. Without loss of generality, $y_{1}$ is also
closer to $x_{3}$ than to $x_{4}$, as in Fig. 17 .

Figure 17: $\overline{x_{1}x_{2}}\cup\overline{x_{3}x_{4}}$

Note that
$\overline{x_{4}x_{1}}=\overline{x_{4}y_{1}}\cup\overline{y_{1}x_{1}}$ because
$\overline{x_{4}y_{1}},\overline{y_{2}x_{1}}$ are both geodesics that
intersect on an interval. Similarly
$\overline{x_{2}x_{3}}=\overline{x_{2}y_{1}}\cup\overline{y_{1}x_{3}}$. Hence
for $i\in\left\\{2,4\right\\}$ we have
$\blacktriangle\left(x_{1},x_{i},x_{3}\right)=\overline{x_{i}y_{1}}\cup\blacktriangle\left(y_{1},x_{1},x_{3}\right)$
and similarly for $j\in\left\\{1,3\right\\}$ we have
$\blacktriangle\left(x_{2},x_{j},x_{4}\right)=\overline{x_{j}y_{2}}\cup\blacktriangle\left(y_{2},x_{2},x_{4}\right)$.
So

$\blacktriangle\left(x_{1},x_{3},x_{4}\right)\cap\blacktriangle\left(x_{2},x_{3},x_{4}\right)=(\overline{x_{4}y_{1}}\cup\blacktriangle\left(y_{1},x_{1},x_{3}\right))\cap(\overline{x_{3}y_{2}}\cup\blacktriangle\left(y_{2},x_{2},x_{4}\right)).$

It is enough to show
$\blacktriangle\left(y_{1},x_{1},x_{3}\right))\cap\blacktriangle\left(y_{2},x_{2},x_{4}\right))\subseteq\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$.
If $p\in\blacktriangle\left(x_{1},y_{1},x_{3}\right)$, then
$\pi_{p}(x_{2})=\pi_{p}(x_{4})$, so
$p\notin\blacktriangle\left(y_{2},x_{1},x_{4}\right)-\triangle(y_{2},x_{1},x_{4})$.
Similarly interchanging the roles of the triangles. So

$\displaystyle\blacktriangle\left(y_{1},x_{1},x_{3}\right))\cap\blacktriangle\left(y_{2},x_{2},x_{4}\right))$
$\displaystyle=\triangle(y_{1},x_{1},x_{3})\cap\triangle(y_{2},x_{2},x_{4})$
$\displaystyle\subseteq(\overline{x_{1}x_{3}}\cap\overline{x_{2}x_{4}})\cup(\overline{x_{1}x_{4}}\cap\overline{x_{2}x_{3}}),$

which is in $\subseteq\overline{x_{1}x_{2}}\cap\overline{x_{3}x_{4}}$ by Lemma
8.2. ∎

Now for the proof of the theorem.

###### Proof of Theorem 8.1.

Denote
$T_{1}=\blacktriangle\left(x_{1},x_{3},x_{4}\right),T_{2}=\blacktriangle\left(x_{2},x_{3},x_{4}\right)$.
By Lemma 8.3 $T_{1}\cap T_{2}=\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$, which by Theorem 7.1 is equal
to
$\bigcup_{\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}}(\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}})$
union a finite set of points. By Lemma 7.4 $T_{1}\cap T_{2}$ has no isolated
points, thus $T_{1}\cap
T_{2}=\bigcup_{\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}}(\overline{x_{i}x_{j}}\cap\overline{x_{k}x_{l}})$
and by Lemma 8.2, all none empty segments among these coincide.

∎

##  9 Near-immersions

Recall Definition 1.3. We rewrite the definition for 2-dimensional complexes.

######  Definition 9.1.

A cellular map $f\mathrel{\mathop{\mathchar 58\relax}}A\to B$ 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 $t,t^{\prime}$
sharing a 1-cell such that $f(t)=f(t^{\prime})$.

######  Theorem 9.2.

Let $X$ be a CAT(0) polygonal complex, and let $\Gamma$ be an embedded
polygonal curve with ordered vertices $x_{1},...,x_{n}$ such that
$\kappa(\Gamma)<4\pi$. Assume $f\mathrel{\mathop{\mathchar 58\relax}}\Delta\to
X$ is a near-immersion, where $\Delta$ is a finite $\Delta$-complex
homeomorphic to a disc and $f|_{\partial\Delta}\mathrel{\mathop{\mathchar
58\relax}}\partial\Delta\to\Gamma$ is a homeomorphism, then $f$ is injective.

In particular, $\Gamma=\triangle(x,y,z)$ satisfies the conclusion of the
theorem when $x,y,z\in X$ are strongly non-collinear, as in this case
$\kappa(\Gamma)<4\pi$.

We get the following corollary using Lemma 5.3.

######  Corollary 9.3.

Let $X$ be a CAT(0) polygonal complex, $x,y,z\in X$ and
$f\mathrel{\mathop{\mathchar 58\relax}}T\to X$ a near-immersion such that $T$
is a full (abstract) spiky triangle and $f|_{\partial
T}\mathrel{\mathop{\mathchar 58\relax}}\partial T\to\triangle(x,y,z)$ is a
homeomorphism. Then $f$ is injective and its image is
$\blacktriangle\left(x,y,z\right)$.

###### 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 $f(T)=\blacktriangle\left(x,y,z\right)$.
∎

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_ ,
$\partial S$, is the union of the 1-cells which are incident to only one
2-cell. Its _interior_ , $\operatorname{int}(S)$, is defined to be $S-\partial
S$.

######  Notation 9.5.

Let $S$ be a piecewise Euclidean pseudo-surface. For
$v\in\operatorname{int}(S)$ we denote $K_{v}=2\pi-\sum\alpha$ where the sums
run over all angles around $v$ in the 2-cells containing $v$. For
$v\in\partial S$, let $i$ be the number of connected component of
$\operatorname{lk}(v)$; we denote $k_{v}=(2-i)\pi-\sum\alpha$ where the sums
run over all angles around $v$ in the 2-cells containing $v$.

######  Theorem 9.6 ([14] Theorem 4.6).

Let $S$ be a piecewise Euclidean pseudo-surface. Then,

$\sum_{v\in\operatorname{int}(S)}K_{v}+\sum_{v\in\partial S}k_{v}=2\pi\chi(S)$

where $\chi(S)$ is the Euler characteristic of $S$.

Note that the sums are finite, as for $v\notin S^{(0)}$ we have that the
appropriate summand is 0.

In the rest of this section we will prove Theorem 9.2.

Assume $\Delta$ is a finite $\Delta$-complex homeomorphic to a disc and
$f\mathrel{\mathop{\mathchar 58\relax}}\Delta\to X$ is a near-immersion
mapping the boundary of $\Delta$ homeomorphically onto $\Gamma$. We endow
$\Delta$ with a piecewise Euclidean structure by endowing each 2-cell $\sigma$
of $\Delta$ with the Euclidean metric of its image in $X$ under $f$. Let
$p\in\Delta$, since $f$ is a near-immersion, each arc $\alpha$ in
$\operatorname{lk}(p)$ is sent to an arc $f_{p}(\alpha)$ in
$\operatorname{lk}(f(p))$ which has the same length. It follows that the girth
of $\operatorname{lk}(p)$ is at least that of $\operatorname{lk}(f(p))$, which
is at least $2\pi$ since $X$ is CAT(0). By the link condition, 2.2, $\Delta$
is CAT(0).

Denote abusively by $x_{i}\in\Delta$ the unique point in $\partial\Delta$ with
image $x_{i}\in X$.

We collect a few important facts about $\Delta$ in the following claims.

######  Claim 9.7.

We have,

  1. (1)

for all $v\in\operatorname{int}(\Delta)$, $K_{v}\leq 0$,

  2. (2)

for all $v\in\partial\Delta-\left\\{x_{1},...,x_{n}\right\\}$, $k_{v}\leq 0$,

###### Proof.

Proof of (1): Since $\Delta$ is CAT(0), the girth of $\operatorname{lk}(v)$ is
at least $2\pi$. If $v\in\operatorname{int}\Delta$ then $\operatorname{lk}(v)$
is a circle, and $K_{v}=2\pi-\operatorname{girth}(\operatorname{lk}(v))\leq
0$, as claimed.

Proof of (2): If
$v\in\operatorname{int}(\gamma)\cap\partial\Delta-\left\\{x_{1},...,x_{n}\right\\}$,
then $f(v)$ 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 $\angle_{v}(x_{i},x_{i+1})=\pi$ (or,
in the latter case, $\angle_{v}(x_{n},x_{1})=\pi$), hence $k_{v}\leq 0$. ∎

Let $\gamma$ be a geodesic in $\Delta$ connecting the points $p,q$. For
$v\in\operatorname{int}(\gamma)$ define

$T_{v}=\pi-\angle_{f(v)}(f\circ\gamma)$

where $\angle_{f(v)}(f\circ\gamma)$ is the angle that $f\circ\gamma$ makes at
$f(v)$ in $X$. By definition, $0\leq T_{v}\leq\pi$.

######  Claim 9.8.

Assume $\gamma$ is a geodesic in $\Delta$ such that $f\circ\gamma$ is a closed
loop in $X$ Then:

  1. (i)

$\sum_{v\in\operatorname{int}(\gamma)}T_{v}\geq\pi$.

  2. (ii)

If $v\in\operatorname{int}(\gamma)\cap\operatorname{int}(\Delta)$ then
$K_{v}\leq-2T_{v}$.

  3. (iii)

If $v\in\operatorname{int}(\gamma)\cap\left\\{x_{1},\dots,x_{n}\right\\}$ then
$k_{v}\leq-2T_{v}+\pi-\angle_{x_{i}}\Gamma$, where
$T_{v}=\pi-\angle_{f(v)}f\circ\gamma$.

###### 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
$v_{1},\dots,v_{n}$ then $\sum_{i=1}^{n}(\pi-\angle_{v_{i}}P)\geq 2\pi$.∎

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 $2\leq i\leq n-2$
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 $\pi$ ([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 $f\circ\gamma$. We
have $\sum_{v\in\operatorname{int}(\gamma)}T_{v}=\sum_{w\in
P}(\pi-\angle_{w}P)-(\pi-\angle_{f(\gamma(0))}P)\geq 2\pi-\pi=\pi$ where the
last inequality follows from Lemma 9.9. This proves (i).

To prove (ii), let
$v\in\operatorname{int}(\gamma)\cap\operatorname{int}(\Delta)$. If $T_{v}=0$
then by 9.7(1) $K_{v}\leq 0=T_{v}$, therefore we may assume that $T_{v}>0$,
i.e. the angle that the path $f\circ\gamma$ makes at $f(v)$ in $X$ is strictly
less than $\pi$.

Let $\gamma_{+}$ and $\gamma_{-}$ be the two points in $\operatorname{lk}(v)$
corresponding to the geodesic $\gamma$ passing through $v$. The points
$\gamma_{+},\gamma_{-}$ break the circle $\operatorname{lk}(v)$ into two arcs
$\alpha_{1},\alpha_{2}$, as in the left hand side of Fig. 18. As the girth of
$\operatorname{lk}(v)$ is greater than $2\pi$, it follows that
$\measuredangle\alpha_{1}+\measuredangle\alpha_{2}\geq 2\pi$. Since $f$ is a
near-immersion, the paths $f_{v}\circ\alpha_{i}$ are non-backtracking in
$\operatorname{lk}(f(v))$ and
$\measuredangle\alpha_{i}=\measuredangle(f_{v}\circ\alpha_{i})$.

From Proposition 2.3 there is a path $\beta$ in $\operatorname{lk}(f(v))$
whose length realizes the angle that $f\circ\gamma$ makes at $f(v)$. In
$\operatorname{lk}(f(v))$, all three paths
$\beta,f_{v}\circ\alpha_{1},f_{v}\circ\alpha_{2}$ are non-backtracking and
connect $f_{v}(\gamma_{+}),f_{v}(\gamma_{-})$, as in the right hand side of
Fig. 18.

Figure 18: The links of $v$ (on the left) and $f(v)$ (on the right)

For $i\in\left\\{1,2\right\\}$,
$\measuredangle\alpha_{i}\geq\angle(\gamma_{-},\gamma_{+})=\pi$, since
$\gamma$ is a geodesic in $\Delta$, hence in particular
$\measuredangle(f_{v}\circ\alpha_{i})=\measuredangle\alpha_{i}\geq\pi>\measuredangle\beta$,
so the concatenation of $f_{v}\circ\alpha_{i}$ and $\beta$ forms an essential
cycle in $\operatorname{lk}(f(v))$. Hence
$\measuredangle(f_{v}\circ\alpha_{i})+\measuredangle\beta\geq 2\pi$ which
implies $\measuredangle\alpha_{i}\geq 2\pi-\measuredangle\beta$.

In conclusion

$K_{v}=2\pi-(\measuredangle\alpha_{1}+\measuredangle\alpha_{2})\leq
2\pi-2\cdot(2\pi-\measuredangle\beta)=-2(\pi-\measuredangle\beta)=-2T_{v},$

as claimed.

To prove (iii), let
$v\in\operatorname{int}(\gamma)\cap\left\\{x_{1},\dots,x_{n}\right\\}$. We
must have $k_{v}\leq 0$, because otherwise, we could find a shortcut of
$\gamma$ not passing through $v$, in contradiction to $\gamma$ being a
geodesic. Hence if $T_{v}=0$, then $k_{v}\leq 0=T_{v}$, as claimed. Therefore,
we may assume that $T_{v}>0$, i.e. that the angle that the path $f\circ\gamma$
makes at $f(v)$ in $X$ is less than $\pi$.

As in the proof of 9.8 (iii), let $\gamma_{+},\gamma_{-}$ be the two points in
$\operatorname{lk}(v)$ corresponding to $\gamma$. They break the arc
$\operatorname{lk}(v)$ into three arcs $\alpha_{1},\alpha_{2},\alpha_{3}$
where $\alpha_{2}$ is the middle arc and $\alpha_{1},\alpha_{3}$ may be
degenerate, as in the left-hand side of Fig. 19. Since $f$ is a near-
immersion, the paths $f_{v}\circ\alpha_{i}$ are non-backtracking in
$\operatorname{lk}(f(v))$ and
$\measuredangle\alpha_{i}=\measuredangle(f_{v}\circ\alpha_{i})$.

From Proposition 2.3 there is a path $\beta$ in $\operatorname{lk}(f(v))$
whose length realizes the angle that the path $f\circ\gamma$ makes at $f(v)$.
We have that $\beta,f_{v}\circ\alpha_{2}$ connect
$f_{v}(\gamma_{+}),f_{v}(\gamma_{-})$ while
$f_{v}\circ\alpha_{1},f_{v}\circ\alpha_{3}$ connect to
$f_{v}(\gamma_{+}),f_{v}(\gamma_{-})$ respectively, as in the right hand side
of Fig. 19.

We have $\measuredangle\alpha_{2}\geq\angle(\gamma_{-},\gamma_{+})=\pi$ where
the last equality follows, since $\gamma$ is a geodesic, hence in particular
$\measuredangle(f_{v}\circ\alpha_{2})=\measuredangle\alpha_{2}\geq\pi>\measuredangle\beta$,
so the concatenation of $f_{v}\circ\alpha_{2}$ and $\beta$ forms an essential
cycle in $\operatorname{lk}(f(v))$. We get
$\measuredangle(f_{v}\circ\alpha_{2})+\measuredangle\beta\geq 2\pi$ which
implies $\measuredangle\alpha_{2}\geq 2\pi-\measuredangle\beta$.

Figure 19: The links of $v$ (on the left) and $f(v)$ (on the right)

So
$k_{v}=\pi-(\measuredangle\alpha_{1}+\measuredangle\alpha_{2}+\measuredangle\alpha_{3})\leq\pi-(2\pi-\measuredangle\beta)-(\measuredangle\alpha_{1}+\measuredangle\alpha_{3})=-T_{v}-(\measuredangle\alpha_{1}+\measuredangle\alpha_{3})=-2T_{v}+(\pi-(\measuredangle\alpha_{1}+\measuredangle\beta+\measuredangle\alpha_{3})\leq-2T_{v}+(\pi-\angle_{v}\Gamma)$

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.

###### Proof of Theorem 9.2.

Let us assume towards contradiction that $f$ is not injective. Therefore there
are $p\neq q\in\Delta$ such that $f(p)=f(q)$ and consider
$\gamma\mathrel{\mathop{\mathchar 58\relax}}\left[0,1\right]\to\Delta$ a
geodesic between them in $\Delta$. Then
$f\circ\gamma\mathrel{\mathop{\mathchar 58\relax}}\left[0,1\right]\to X$ is a
non-trivial, piecewise linear loop. We choose $p,q$ closest such points, which
ensures that $f\circ\gamma$ is a simple closed curve.

(9.1)
$\begin{split}2\pi&\overset{(1)}{=}\sum_{v\in\operatorname{int}(D)}K_{v}+\sum_{v\in\partial(D)}k_{v}\\\
&=\sum_{v\in\operatorname{int}(\gamma)\cap\operatorname{int}(D)}K_{v}+\sum_{v\in\operatorname{int}(\gamma)\cap\partial(D)}k_{v}+\sum_{v\in\operatorname{int}(D)-\operatorname{int}(\gamma)}K_{v}+\sum_{v\in\partial(D)-\operatorname{int}(\gamma)}k_{v}\\\
&\overset{(2)}{\leq}-2(\sum_{v\in\operatorname{int}(\gamma)\cap\operatorname{int}(D)}T_{v}+\sum_{v\in\operatorname{int}(\gamma)\cap\partial(D)}T_{v})+\sum_{x_{i}\in\operatorname{int}(\gamma)}(\pi-\angle_{x_{i}}\Gamma)\\\
&+\sum_{v\in\operatorname{int}(D)-\operatorname{int}(\gamma)}K_{v}+\sum_{v\in\partial(D)-\operatorname{int}(\gamma)}k_{v}\\\
&\overset{(3)}{\leq}-2(\sum_{v\in\operatorname{int}(\gamma)}T_{v})+\sum_{x_{i}\in\operatorname{int}(\gamma)}(\pi-\angle_{x_{i}}\Gamma)+\sum_{x_{i}\notin\operatorname{int}(\gamma)}k_{x_{i}}\\\
&\overset{(4)}{<}-2\pi+4\pi-\sum_{x_{i}\notin\operatorname{int}(\gamma)}(\pi-\angle_{x_{i}}\Gamma)+\sum_{x_{i}\notin\operatorname{int}(\gamma)}k_{x_{i}}\overset{(5)}{\leq}2\pi\end{split}$

Where (1) is Theorem 9.6, (2) from 9.8(ii),9.8(iii), (3) from the fact that
$K_{v}\leq 0$ and $k_{v}\leq 0$ for
$v\notin\left\\{x_{1},\dots,x_{n}\right\\}$, (4) from $\sum T_{v}\geq\pi$
(9.8(i)) and $\kappa(\Gamma)<4\pi$, and finally (5) follows from
$k_{x_{i}}+\pi-\angle_{x_{i}}\Gamma\leq 0$. This is a contradiction. ∎

##  10 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 $\overline{x_{i}x_{j}}$, because then the “faces” of the
deflated tetrahedron must be mapped to
$\blacktriangle\left(x_{i},x_{j},x_{k}\right)$, and they intersect in a unique
point.

Figure 20: Deflated tetrahedron

In our minds, we start with a finite $\Delta$-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,
$\left\\{\blacktriangle\left(y_{i},y_{j},\omega\right)\right\\}_{i\leq i<j\leq
4}$, 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) $\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4}$ with the marked
points on $D_{ij}$ denoted $y_{i},y_{j},\omega$, (denoting $D_{ji}=D_{ij}$
when $i<j$),

    * –

bijections between $\overline{y_{i}\omega}$ in $D_{ij}$ and
$\overline{y_{i}\omega}$ 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 $\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4}$ let $Z$
be the complex obtained by gluing $\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4}$
along their boundaries, that is, $Z=\coprod_{1\leq i<j\leq 4}D_{ij}/\sim$
where $\sim$ is generated by the given bijections.

  * •

$f\mathrel{\mathop{\mathchar 58\relax}}\coprod_{1\leq i<j\leq 4}D_{ij}\to X$
is a _filling map_ of $x_{1},x_{2},x_{3},x_{4}$ through $o$ where
$\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4}$ is a deflated tetrahedron,
$x_{1},x_{2},x_{3},x_{4},o\in X$, if

    * –

$f$ is a cellular map, up to a refinement of the cellular structure on $X$,
and injective on every cell,

    * –

$f(y_{i})=x_{i}$,

    * –

$f|_{D_{ij}}^{-1}(\left\\{o\right\\})=\left\\{\omega\right\\}$,

    * –

$f|_{\overline{y_{i}y_{j}}}$ from $\overline{y_{i}y_{j}}$ to the geodesic
segment $\overline{x_{i}x_{j}}$ in $X$ is a homeomorphism and descends to a
map from $Z$.

  * •

A pair $(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ 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 $f|_{D_{ij}}$ 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.

######  Proposition 10.2.

For any $x_{1},x_{2},x_{3},x_{4}\in X$ and $o\in\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$ there exists a deflated
tetrahedron filling $x_{1},x_{2},x_{3},x_{4}$ through $o$.

###### Proof.

By Theorem 5.1,
$\left\\{\blacktriangle\left(x_{i},x_{j},o\right)\right\\}_{1\leq i<j\leq 4}$
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 $\Delta$-complexes
$D_{ij}\cong\blacktriangle\left(x_{i},x_{j},o\right)$ and cellular, injective
maps $f_{ij}\mathrel{\mathop{\mathchar 58\relax}}D_{ij}\to X$.

Define $f\mathrel{\mathop{\mathchar 58\relax}}\coprod_{1\leq i<j\leq
4}D_{ij}\to X$ to be the function induced from $\left\\{f_{ij}\right\\}_{1\leq
i<j\leq 4}$. There is a refinement of the triangulation of $X$ and of $D_{ij}$
such that $f$ is cellular.

The spiky triangles $D_{ij}$ are in particular generalized discs with three
marked points

$y_{i}\mathrel{\mathop{\mathchar
58\relax}}=f_{ij}^{-1}(x_{i}),y_{j}\mathrel{\mathop{\mathchar
58\relax}}=f_{ij}^{-1}(x_{j}),\omega\mathrel{\mathop{\mathchar
58\relax}}=f_{ij}^{-1}(o).$

We claim $(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ is a deflated
tetrahedron filling $x_{1},x_{2},x_{3},x_{4}$ through $o$. To establish that
$\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4}$ is a deflated tetrahedron we need
to show that $\left\\{T_{ijk}\right\\}_{1\leq i<j<k\leq 4}$ are spiky
triangles, the other parts of the definition are clear from the construction.
Define $a_{ij}$ to be the point in $D_{ij}$ in which $\overline{\omega
y_{i}},\overline{\omega y_{j}}$ diverge, as in Fig. 21.

Figure 21: $\blacktriangle\left(y_{i},y_{j},o\right)$

We will argue without loss of generality for $T_{123}$. If
$a_{12}=a_{13}=a_{23}=\omega$ then clearly $\left\\{T_{ijk}\right\\}_{1\leq
i<j<k\leq 4}$ are spiky triangles, as we are gluing three triangles with no
spike at $\omega$.

Suppose instead $a_{12}\neq\omega$.

Denote by $\hat{x}_{i}$ the projection of $x_{i}$ to $\operatorname{lk}(o)$.
As $a_{12}\neq\omega$ we have that $\hat{x}_{1}=\hat{x}_{2}$, hence
$\triangle(\hat{x}_{1},\hat{x}_{2},\hat{x}_{3})=\overline{\hat{x}_{1}\hat{x}_{3}}=\overline{\hat{x}_{2}\hat{x}_{3}}$.
As $o\in\blacktriangle\left(x_{1},x_{2},x_{3}\right)$ we have by Corollary 2.6
that $o\in\overline{x_{1}x_{3}},\overline{x_{2}x_{3}}$. In this case
$\blacktriangle\left(x_{1},x_{3},o\right)=\overline{x_{1}x_{3}}$ and
$\blacktriangle\left(x_{2},x_{3},o\right)=\overline{x_{2}x_{3}}$, so the
gluing of $T_{123}\cong\blacktriangle\left(x_{2},x_{3},o\right)$ which is of
course a spiky triangle.

We have that $f$ is a filling map of $x_{1},x_{2},x_{3},x_{4}$ through $o$. It
is piecewise linear and injective on every 2-cell, as it is injective on
$D_{ij}$, also from that follows that
$f_{ij}^{-1}(\left\\{(o)\right\\})=\left\\{\omega\right\\}$ and by
construction $f$ is a homeomorphism of the relevant boundary part to
$\overline{x_{i}x_{j}}$. ∎

######  Definition 10.3.

A deflated tetrahedron $(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ filling
$x_{1},x_{2},x_{3},x_{4}$ through some $o\in\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$ with a minimal number of cells
will be referred to as a minimal tetrahedron filling
$x_{1},x_{2},x_{3},x_{4}$.

######  Proposition 10.4.

Consider $(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ a minimal tetrahedron
filling $x_{1},x_{2},x_{3},x_{4}$, then for all $1\leq i<j<k\leq 4$ the
induced map $f\mathrel{\mathop{\mathchar 58\relax}}T_{ijk}\to X$ is near-
immersion and hence from Corollary 9.3 it is injective. In particular
$\left\\{T_{ijk}\right\\}_{1\leq I<j<k\leq 4}$ 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
$t,t^{\prime}$ that share a 1-cell and such that $f(t)=f(t^{\prime})$.

If $t,t^{\prime}\subseteq D_{ij}$ for some $i,j$ we will commence as in the
proof of Theorem 5.1, that is, define $C_{ij}$ the connected component of
$\omega$ in $D_{ij}-\operatorname{int}(t\cup t^{\prime})/\sim$, where $\sim$
is the cellular equivalence relation defined by $u\sim v$ if $u$ a 1-cell in
$\partial t$ and $v$ a 1-cell in $\partial t^{\prime}$ and $f(u)=f(v)$, as in
Theorem 5.1. Define $C_{jk}=D_{jk}$ for $(j,k)\neq(i,j)$. We get
$\left\\{C_{ij}\right\\}_{1\leq i<j\leq 4}$ 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 $t,t^{\prime}$ must be in different cell complexes. Without loss of
generality assume that $t\subseteq D_{12},t^{\prime}\subseteq D_{13}$.

We would like to “fold” $t,t^{\prime}$ and insert the folded copy to $D_{14}$,
as in Fig. 22.

Figure 22: Folding of $t,t^{\prime}$

We divide into cases depending on $t\cap t^{\prime}$:

Case 1. The intersection $t\cap t^{\prime}$ is an interval (one or two
1-cells).

Define $C_{12}$ to be a complex obtained from $D_{12}$ removing
$\operatorname{int}(t)$ and the open interval of $t\cap t^{\prime}$.
Similarly, define $C_{13}$ to be the complex obtained from $D_{12}$ removing
$\operatorname{int}(t^{\prime})$ and the open interval of $t\cap t^{\prime}$.
Set $C_{14}=D_{14}\cup t$ and $C_{ij}=D_{ij}$ for $2\leq i<j$. For notation
purposes $C_{ij}=C_{ji}$.

Denote by $f^{\prime}\mathrel{\mathop{\mathchar 58\relax}}\coprod_{1\leq
i<j\leq 4}C_{ij}\to X$ the induced map from $f\mathrel{\mathop{\mathchar
58\relax}}\coprod_{1\leq i<j\leq 4}D_{ij}\to X$. We claim that
$(\left\\{C_{ij}\right\\}_{1\leq i<j\leq 4},f)$ is a deflated tetrahedron
filling $x_{1},x_{2},x_{3},x_{4}$ through $o$ and that will give us
contradiction to minimality of $(\left\\{D_{ij}\right\\}_{1\leq i<j\leq
4},f)$.

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 $\Delta$-complexes. Choose
$y_{i},y_{j},\omega$ to be the same marked points as in $D_{ij}$ and get that
$\left\\{C_{ij}\right\\}_{1\leq i<j\leq 4}$ are generalized triangles.

The induced map between $\overline{y_{i}o}^{C_{ij}}$ to
$\overline{y_{i}o}^{C_{ik}}$ is a bijection.

Denote by $T^{\prime}_{ijk}$ the gluing induced by the bijections on the
boundary parts of $C_{ij},C_{jk},C_{ik}$. $T^{\prime}_{ijk}$ are spiky
triangles for $1\leq i<j<k\leq 4$: indeed, $T^{\prime}_{124}=T_{124}$ by
working out the definitions of $C_{ij}$. We have $T^{\prime}_{234}=T_{234}$ as
$C_{ij}=D_{ij}$ for the relevant cell complexes. As in the proof of Theorem
5.1, $T^{\prime}_{123}=T_{123}-(\operatorname{int}(t\cup t^{\prime})/\sim$
where $\sim$ identifies the boundary of $t\cup t^{\prime}$ to an interval, so
$T^{\prime}_{123}$ is a spiky triangle. Lastly,
$T^{\prime}_{134}=(T_{134}-t^{\prime})\cup t/\sim$ is a spiky triangle, as
topologically we deleted a disc and glued another disc instead.

Clearly, $f^{\prime}$ is a filling map. This gives a contradiction to
minimality.

Case 2. The intersection $t\cap t^{\prime}$ is a disjoint union of a 1-cell
and 0-cell $v$.

There are two further cases, either $v\in\overline{y_{1}\omega}$, as in Fig.
23, or $v\in\overline{y_{2}\omega}$, as in Fig. 24 (where in this case it
happens to be that $D_{23}$ has a skinny part).

Case 2.1. $v\in\overline{y_{1}\omega}$.

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 $\omega$ and by $S$ the closure
of the other connected component and let $T=S\cup t$. Define
$C_{13},S^{\prime}$ similarly to be the connected components of
$D_{13}-t^{\prime}$ containing $\omega$ and not containing $\omega$
respectively and let $T^{\prime}=S^{\prime}\cup t^{\prime}$. Set
$C_{14}=D_{14}\cup T$ and $C_{ij}=D_{ij}$ for $2\leq i<j$. Let $f^{\prime}$ be
the induced map.

We argue identically to case 1, with $T,T^{\prime}$ instead of $t,t^{\prime}$,
to show that $(\left\\{C_{ij}\right\\}_{1\leq i<j\leq 4},f^{\prime})$ 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. $v\in\overline{y_{2}\omega}$.

Figure 24: Case 2.2.

Let $A$ be the connected component of $y_{1}$ in $T_{123}-(t\cup t^{\prime})$.
It is a generalized annulus with spikes and $\omega\notin A$, as in Theorem
5.1. Consider $\sim$ the cellular equivalence relation defined by $u\sim v$ if
$u$ a 1-cell in $\partial t$ and $v$ a 1-cell in $\partial t^{\prime}$ and
$f(u)=f(v)$, as in Theorem 5.1. It is a good gluing on $A$ and so from 3.4,
$A/\sim$ is a spiky triangle, in particular, it is simply connected. The map
$\bar{f}$ induced from $f$ maps the boundary of $A$ homeomorphically onto
$\triangle(x_{1},x_{2},x_{3})$, hence
$\blacktriangle\left(x_{1},x_{2},x_{3}\right)\subseteq\bar{f}(T)$ which is a
contradiction to $f$ being a filling map and satisfying
$f|_{C_{ij}}^{-1}(\left\\{o\right\\})=\left\\{\omega\right\\}$ and the fact
that $\omega\notin A$ while
$f(\omega)=o\in\blacktriangle\left(x_{1},x_{2},x_{3}\right)$. ∎

######  Corollary 10.5.

For $(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ a minimal tetrahedron
filling $x_{1},x_{2},x_{3},$ $x_{4}$, the map $f\mathrel{\mathop{\mathchar
58\relax}}Z\to X$ is a near-immersion.

###### Proof.

$f$ is injective on cells by assumption. Assume towards contradiction that
there are $t,t^{\prime}\subseteq Z$ twin cells. As $f|_{T_{ijk}}$ is
injective, by Proposition 10.4 we have that $t\subseteq
D_{ij},t^{\prime}\subseteq D_{kl}$ for
$\left\\{i,j,k,l\right\\}=\left\\{1,2,3,4\right\\}$. From this, we have
$f(t)=f(t^{\prime})\subseteq\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$, but $f(t)$ is a 2-cell in $X$
and this is a contradiction to the fact that the intersection is a
1-dimensional complex, by Theorem 7.1. ∎

######  Corollary 10.6.

Assume $x_{1},x_{2},x_{3},x_{4}\in X$ do not form a lanky quadrilateral. If
$(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ is a minimal tetrahedron
filling $x_{1},x_{2},x_{3},x_{4}$, then each
$f|_{D_{ij}}\mathrel{\mathop{\mathchar 58\relax}}D_{ij}\to X$ is injective.

###### Proof.

Without loss of generality $(i,j)=(1,2)$. As $f|_{T_{123}}$ is injective we
have that if $f|_{D_{12}}$ 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.

Figure 25: non injectivity of $D_{12}$

There are two cases, either $e_{1},e_{2}$, the two 1-cells adjacent to
$\omega$ 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 $\omega$ and get that $D_{12}$ bounds a disc in
$T_{123}$.

Case 1. $e_{1},e_{2}$ are identified in $T_{123}$.

Assume
$e_{1}\subseteq\overline{y_{1}\omega}^{12},e_{2}\subseteq\overline{y_{2}\omega}^{12}$.
Note that $e_{1}\subseteq T_{124},T_{123},T_{134}$ and $e_{2}\subseteq
T_{234}$ so $f(e_{1})=f(e_{2})\subseteq\bigcap_{1\leq i<j<k\leq
4}f(T_{ijk})=\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$. By Theorem 7.1, we have that
$f(e_{1})$ is discrete, which is a contradiction to $f|_{e_{1}}$ is injective.

Case 2. $D_{12}$ bounds a disc in $T_{123}$.

The disc $D_{12}$ bounds in $T_{123}$ is a subset of $D_{13}\cup D_{23}$. As
$f|_{T_{124}}$ is injective we must have that under gluing the boundaries the
same points are identified in $T_{124}$ which gives us a disc in $D_{14}\cup
D_{24}\subseteq T_{124}$. In $Z$ these two discs glue to a sphere. By
Corollary 10.5, $f$ restricted to that sphere is a near-immersion which is a
contradiction, as we can pull back the CAT(0) structure from $X$ to the
sphere. ∎

######  Proposition 10.7.

Assume $x_{1},x_{2},x_{3},x_{4}\in X$ do not form a lanky quadrilateral and
$(\left\\{D_{ij}\right\\}_{1\leq i<j\leq 4},f)$ a minimal deflated tetrahedron
filling $x_{1},x_{2},x_{3},x_{4}$, then $f\mathrel{\mathop{\mathchar
58\relax}}Z\to X$ is injective.

###### Proof.

By Corollary 10.5 and Corollary 10.6, $f|_{Z}$ is a near-immersion and
$f|_{D_{ij}}$ is injective.

Assume towards contradiction $p,q\in Z$ distinct and $f(p)=f(q)$. Without loss
of generality $p\in D_{12}$. If $q\in\bigcup_{(i,j)\neq(3,4)}D_{ij}$ there
exists $i<j<k$ such that $p,q\in T_{ijk}$ are distinct. That is a
contradiction by Proposition 10.4 that states that $f|_{T_{ijk}}$ is
injective.

So it must be that $q\in\operatorname{int}(D_{34})\cup\overline{y_{3}y_{4}}$
and similarly $p\in\operatorname{int}(D_{12})\cup\overline{y_{1}y_{2}}$. We
get
$f(p)=f(q)\in\blacktriangle\left(x_{1},x_{2},x_{4}\right)\cap\blacktriangle\left(x_{1},x_{3},x_{4}\right)$,
so by Lemma 7.4 we have that the connected component of $f(p)$ in
$\blacktriangle\left(x_{1},x_{2},x_{4}\right)\cap\blacktriangle\left(x_{1},x_{3},x_{4}\right)$
is more than a point, i.e. there is a 1-cell or 2-cell $s\subseteq D_{12}$
containing $p$ such that
$f(s)\subseteq\blacktriangle\left(x_{1},x_{3},x_{4}\right)$. But from the
injectivity of $f|_{T_{123}},f|_{T_{124}}$ we get that $f(s)\subseteq
f(D_{34})$, hence $f(s)\subseteq\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$. This is a contradiction to
Theorem 7.1 that states that the intersection is discrete, as $f|_{s}$ is
injective. Hence $f\mathrel{\mathop{\mathchar 58\relax}}Z\to X$ is injective.
∎

Now we can complete the proof of A in the case where
$x_{1},x_{2},x_{3},x_{4}\in X$ do not form a lanky quadrilateral,:

######  Theorem.

Assume $x_{1},x_{2},x_{3},x_{4}\in X$ do not form a lanky quadrilateral, then
the intersection of $\blacktriangle\left(x_{i},x_{j},x_{k}\right)$ for ${1\leq
i<j<k\leq 4}$ is a single point.

###### Proof.

We will show that in a minimal deflated tetrahedron filling
$x_{1},x_{2},x_{3},x_{4}$, we have $\bigcap_{1\leq i<j<k\leq
4}T_{ijk}=\left\\{\omega\right\\}$. As $f\mathrel{\mathop{\mathchar
58\relax}}Z\to X$ is injective,
$f(T_{ijk})=\blacktriangle\left(x_{i},x_{j},x_{k}\right)$, which will show
that $\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)=\left\\{f(\omega)\right\\}=\left\\{o\right\\}$.

Assume $q\in\bigcap_{1\leq i<j<k\leq 4}T_{ijk}$. Without loss of generality
$q\in D_{12}$. We get $q\in D_{12}\cap T_{134}=\overline{y_{1}\omega},q\in
D_{12}\cap T_{234}=\overline{y_{2}\omega}$. It suffices to show that $\omega$
is the only point that satisfies this, i.e. that $D_{ij}$ are actually discs
with two spikes where $y_{i},y_{j}$ are the endpoints of each spike.

Assume towards contradiction $q\neq\omega$.

Note
$f(q)\in\blacktriangle\left(x_{1},x_{3},x_{4}\right)\cap\blacktriangle\left(x_{2},x_{3},x_{4}\right)$.
By Lemma 7.4, there is a 1-cell $e$ in $Z$ containing $q$ such that
$f(e)\subseteq\blacktriangle\left(x_{1},x_{3},x_{4}\right)\cap\blacktriangle\left(x_{2},x_{3},x_{4}\right)$.

From injectivity of $f$ on $Z$, the 1-cell $e$ must be in $D_{34}$ or in
$\overline{y_{1}\omega}\cap\overline{y_{2}\omega}$.

Case 1. $e$ in $D_{34}$.

In this case $q\in D_{34}$, that is $q\in\overline{y_{i}\omega}$ for $i=3$ or
$4$. Without loss of generality $q\in\overline{y_{3}\omega}$. For
$i,j\in\left\\{1,2,3\right\\}$ different $f|_{D_{ij}}$ is injective by
Corollary 10.6 hence there exists a unique point $q_{ij}\in D_{ij}$ such that
its image in $T_{123}$ is $q$.

Note $q_{ij}$ separates $D_{ij}$ for every $1\leq i<j\leq 4$. Consider
$C_{ij}$ the connected component of $y_{i}$ in
$D_{ij}-\left\\{q_{ij}\right\\}$. Their image in $T_{123}$ is simply connected
by Van Kampen. This is a contradiction to
$o\in\blacktriangle\left(x_{1},x_{2},x_{3}\right)$.

Case 2. $e$ in $\overline{y_{1}\omega}\cap\overline{y_{2}\omega}$.

By choice of $e$,
$f(e)\subseteq\blacktriangle\left(x_{1},x_{3},x_{4}\right)\cap\blacktriangle\left(x_{2},x_{3},x_{4}\right)$
and by the assumption in this case,
$f(e)\subseteq\blacktriangle\left(x_{1},x_{2},x_{3}\right)\cap\blacktriangle\left(x_{1},x_{2},x_{4}\right)$,
so $f(e)\subseteq\bigcap_{1\leq i<j<k\leq
4}\blacktriangle\left(x_{i},x_{j},x_{k}\right)$. This is a contradiction to
the fact that the intersection is discrete, stated in Theorem 7.1 and the fact
that $f|_{e}$ is injective.

∎

##  11 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.

######  Definition.

Let
$\blacktriangle^{2}(x_{1},x_{2},x_{3})=\blacktriangle\left(x_{1},x_{2},x_{3}\right)$.
Let $x_{1},\dots,x_{n+1}\in X$ where $X$ is a CAT(0) complex and define
$\blacktriangle^{n}(x_{1},\dots,x_{n+1})$ to be the minimal ball filling
$\bigcup_{i=1}^{n+1}\blacktriangle^{n-1}(x_{1},\dots,\hat{x}_{i},\dots,x_{n+1})$.

One can study intersections of $\blacktriangle^{n}(x_{1},\dots,x_{n+1})$ for
all $x_{1},\dots,x_{n+1}$ in a set of $n+2$ points.

### Median algebras.

We recall that for a set $M$ and a map $\mu\mathrel{\mathop{\mathchar
58\relax}}M^{3}\to M$ the pair $(M,\mu)$ is a _median algebra_ if for any
$a,b,c,d,e\in M$ we have $\mu(a,b,c)=\mu(b,c,a)=\mu(b,a,c)$, $\mu(a,a,b)=a$
and $\mu(a,b,\mu(c,d,e))=\mu(\mu(a,b,c),\mu(a,b,d),e)$. 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.

### Coarse median.

A geodesic metric space $(x,d)$ is _coarse median_ if

  * •

there is a map $\mu\mathrel{\mathop{\mathchar 58\relax}}X^{3}\to X$ and
$k,l>0$ such that for any $a,b,c,a^{\prime},b^{\prime},c^{\prime}\in X$ we
have
$d(\mu(a,b,c),\mu(a^{\prime},b^{\prime},c^{\prime})<k(d(a,a^{\prime})+d(b,b^{\prime})+d(c,c^{\prime}))+l$,

  * •

there is a map $h\mathrel{\mathop{\mathchar 58\relax}}\mathbb{N}\to[0,\infty)$
such that if $A\subseteq X$ such that $\mathinner{\\!\left\lvert
A\right\rvert}\leq p$ then there exists a finite median algebra $(M,m)$ and
maps $\pi\mathrel{\mathop{\mathchar 58\relax}}A\to
M,\lambda\mathrel{\mathop{\mathchar 58\relax}}M\to A$ such that for all $a\in
A$ $d(a,\lambda(\pi(a)))<h(p)$ and for all $a,b,c\in M$
$d(\mu(\lambda(a),\lambda(b),\lambda(c)),\lambda(m(a,b,c)))<h(p)$.

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
2-median?

##  Appendix A Appendix – Fary Milnor for CAT(0) polygonal complexes

In this appendix we will recover a special case of a result of Stadler [20],
proving a Fary-Milnor type theorem for CAT(0) polygonal complexes. Moreover,
in this case, the image of the embedded disc is unique.

Recall that for a polygonal curve $\Gamma$ we define its _total curvature_ to
be

$\kappa(\Gamma)=\sum_{i=1}^{k}(\pi-\angle_{x_{i}}\Gamma),$

where $x_{1},\ldots,x_{k}$ are the vertices of $\Gamma$.

######  Theorem A.1 (Fary-Milnor for CAT(0) polygonal complexes).

Let $X$ be a CAT(0) polygonal complex, and let $\Gamma$ be an embedded
polygonal curve. If $\kappa(\Gamma)<4\pi$ then $\Gamma$ bounds an embedded
disc in $X$.

The image of the disc is unique, by Lemma 5.2.

We note that $\kappa(\triangle(x,y,z))<4\pi$.

Theorem A.1 follows from Theorem 9.2, if we show that there exists a near-
immersion $f\mathrel{\mathop{\mathchar 58\relax}}\mathbb{D}^{2}\to X$ filling
$\Gamma$. This is similar to Lemma 3.3 in [17].

###### Proof.

By subdividing the cells of $X$, we may assume $X$ is a triangle complex, i.e.
it has a simplicial structure.

As $X$ is simply connected, there exists a map $f\mathrel{\mathop{\mathchar
58\relax}}\mathbb{D}^{2}\to X$ such that $f|_{\partial\mathbb{D}^{2}}$ maps
${\partial\mathbb{D}^{2}}$ homeomorphically onto $\Gamma$. The homeomorphism
$f|_{\partial\mathbb{D}^{2}}\mathrel{\mathop{\mathchar
58\relax}}\partial\mathbb{D}^{2}\to\Gamma$ can be made cellular by subdividing
$\partial\mathbb{D}^{2}$ and $X$. By the simplicial approximation theorem (see
e.g. [21]), there exists $T$, a triangulation of $\mathbb{D}^{2}$, and a map
$f^{\prime}\mathrel{\mathop{\mathchar 58\relax}}T\to X$ that is cellular with
respect to some simplicial structure of $X$ and that maps $\partial T$
homeomorphically onto $\Gamma$.

Fix the simplicial structure on $X$ from above and consider the collection of
all pairs $(h,D)$ where $D$ is a disc with some finite $\Delta$-complex
structure and $h\mathrel{\mathop{\mathchar 58\relax}}D\to X$ is a cellular
map, such that $h|_{\partial D}\mathrel{\mathop{\mathchar 58\relax}}\partial
D\to\Gamma$ is a homeomorphism. Consider $(f,T)$ in this collection such that
the number of cells in $T$ is minimal among all pairs in the collection. We
claim $f$ is a near-immersion.

To simplify future arguments we show first that $T$ has a simple structure -
all 1-cells in $T^{(1)}$ are intervals (not loops) and there are no double
1-cells.

######  Claim A.2.

The 1-skeleton of $T$ is a simple graph.

###### Proof.

$T^{(1)}$ contains no loops. As $X$ is a simplicial complex, if $e$ is a
1-cell in $T^{(1)}$ then $f(e)$ is either a point or a segment in $X^{(1)}$,
so if $e$ is a loop in $T$ then $f(e)$ is a point. Such $e$ must be in the
interior of $T$, as the boundary of $T$ is mapped injectively.

Consider $e$ an outermost such loop in $T$. We have that $e$ separates $T$
into a generalized annulus and a disc, as in Fig. 26, and it is on the
boundary of two 2-cells $t,t^{\prime}$, where $t$ is in the generalized
annulus. Consider the closure $A$ of the connected component of $\partial T$
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 $f(t)=f(\partial t)$,
which is a 0-cell or 1-cell in $X$, so $f(e_{1})=f(e_{2})$. Let $\sim$ be the
cellular equivalence relation defined by $e_{1}\sim e_{2}^{-1}$. The relation
$\sim$ is a good gluing, as it identifies the two 1-cells on the inner
boundary of $A$ in the correct orientation, hence by 3.4, $A/\sim$ is a disc.

The induced $\Delta$-complex structure on the disc $A/\sim$ 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 $(f,T)$.

Figure 26: $e$ is a loop in $T$

$T^{(1)}$ 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 $e_{1},e_{2}^{-1}$. Let $\sim$ be
the cellular equivalence relation $e_{1}\sim e_{2}$. As in the loop case,
$\sim$ is a good gluing, as it identifies the two 1-cells on the inner
boundary of $A$ in the correct orientation, hence by 3.4, $A/\sim$ is a disc.

The induced $\Delta$-complex structure on the disc $A/\sim$ 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 $(f,T)$.

Figure 27: $e_{1},e_{2}$ are double 1-cells

∎

If $f|_{T}$ is not a near-immersion then either there is a cell $t$ such that
$f(t)=f(\partial t)$ or there are 2-cells $t,t^{\prime}$ which share at least
one 1-cell on their boundary and such that $f(t)=f(t^{\prime})$. We will show
both cases cannot happen.

Case 1. Assume $f(t)=f(\partial t)$. In this case, there exists a 1-cell, $e$,
in $\partial t$ that is mapped to a point. As $f|_{\partial T}$ is injective,
$e$ must be in the interior of $T$. Let $t^{\prime}$ 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
$\partial t$ and $\partial t^{\prime}$ respectively, oriented as a cycle, as
in Fig. 13.

Let $A$ be the closure of the connected component of $\partial T$ in
$T-(\bar{t}\cup\bar{t^{\prime}})$. 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, A.2, the inner boundary of $A$ is a path consisting of four
oriented 1-cells $e_{1},e_{2},d_{2}^{-1},d_{1}^{-1}$ 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 $\sim$ be the cellular equivalence relation defined
by $e_{1}\sim e_{2}^{-1},d_{1}\sim d_{2}^{-1}$. It is a good gluing, as $f$ is
injective on the boundary of $T$, from strongly non-collinearity of $x,y,z$.
From 3.4, $A/\sim$ is a disc which is a finite $\Delta$-complex. The map
induced from $f$, denoted $\bar{f}\mathrel{\mathop{\mathchar
58\relax}}A/\sim\to X$, is well defined, cellular and $\bar{f}|_{\partial
A}\mathrel{\mathop{\mathchar 58\relax}}\partial A\to\triangle(x,y,z)$ is a
homeomorphism. But $A/\sim$ has fewer cells than $T$, this is a contradiction
to the minimality of $(f,T)$.

Figure 28: The case $f(t)=f(\partial t)$

Case 2. Assume $f(t)=f(t^{\prime})$ for $t,t^{\prime}$ sharing a 1-cell $e$.
Let $e,e_{1},e_{2}$ and $e,d_{1},d_{2}$ be the 1-cells in $\partial t$ and
$\partial t^{\prime}$ respectively, oriented as a cycle, as in Fig. 29.

Let $A$ be the closure of the connected component of $\partial T$ in
$T-(\bar{t}\cup\bar{t^{\prime}})$. As above, $A$ is a generalized annulus and
the inner boundary of $A$ is a path consisting of four oriented 1-cells
$e_{1},e_{2},d_{2}^{-1},d_{1}^{-1}$ in this order, as in Fig. 29. Because
$f(t)=f(t^{\prime})$, 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 $\sim$ be the cellular equivalence relation defined by $e_{1}\sim
d_{2}^{-1},d_{1}\sim e_{2}^{-1}$. It is a good gluing, as before, as $x,y,z$
are strongly non-collinear. By 3.4, we have that $A/\sim$ is homeomorphic to a
disc. It is a finite $\Delta$-complex with fewer cells than $T$, and the
induced map $\bar{f}\mathrel{\mathop{\mathchar 58\relax}}A/\sim\to X$ is still
cellular. This contradicts the minimality of $(f,T)$.

Figure 29: The case $f(t)=f(t^{\prime})$

Hence $f$ is a near-immersion.

By Theorem 9.2, we get that $f$ is injective and this concludes the proof.

∎

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