A finite triangulated surface has a one-polygon schema

Write $V,E_1,T$ for the sets of vertices, $1$-simplices and $2$-simplices of $K$; all are finite. By hypothesis 1 each $e\in E_1$ lies in exactly two triangles, so it determines a two-element subset $\{\sigma ,{\sigma }^{\mathrm{\prime }}\}\subseteq T$, and distinct $e,f\in E_1$ determine distinct subsets: two distinct simplices of $K$ intersect in the common face $\sigma \cap {\sigma }^{\mathrm{\prime }}$, so they share at most one $1$-simplex, and a pair of triangles therefore has at most one common edge. Hence $D$, with vertex set $T$ and one edge per $e\in E_1$ joining the two triangles containing $e$, is a finite simple graph, the dual graph, and $e\mapsto \{\sigma ,{\sigma }^{\mathrm{\prime }}\}$ is injective. Hypothesis 1 also shows that every $1$-simplex lies in a triangle, and hypothesis 2 shows that every vertex of $K$ lies in a $1$-simplex, because $\mathrm{lk}(v)$ is a cycle and a cycle has at least three vertices, each of them a $1$-simplex of $K$ containing $v$.

Fix a vertex $v$ and define $\mathrm{\Lambda }(v)$: its vertices are the triangles containing $v$, and two of them are adjacent when they share a $1$-simplex that contains $v$. Under $\{v,x,y\}\mapsto \{x,y\}$ the edges of $\mathrm{\Lambda }(v)$ correspond to pairs of edges of $\mathrm{lk}(v)$ sharing a vertex, so $\mathrm{\Lambda }(v)$ is the line graph of $\mathrm{lk}(v)$; by hypothesis 2 the graph $\mathrm{lk}(v)$ is a cycle, hence connected. A connected graph with at least one edge has connected line graph: if $\{x_0,x_1\}$ and $\{z_0,z_1\}$ are edges and $x_0=u_0,u_1,\dots ,u_n=z_0$ is a walk, then the edges $\{u_{i-1},u_i\}$ form a chain in which consecutive terms share the vertex $u_i$, so $\{x_0,x_1\}$ is joined to $\{z_0,z_1\}$. Therefore $\mathrm{\Lambda }(v)$ is connected.

Local structure of a simple polygonal region. Let $R\subseteq {\mathbb{R}}^2$ be a simple polygonal region with boundary chain $v_0{v}_1\cdots v_{n-1}$, so that $\mathrm{\partial }R$ is the polygon $v_0{v}_1\cdots v_{n-1}$, and abbreviate $s=[v_0,v_1]$, $s^{\circ }=(v_0,v_1)$. By [L6] the complement of $\mathrm{\partial }R$ has exactly two regions, a bounded one $B$ and an unbounded one $C$, with $\mathrm{Fr}(B)=\mathrm{Fr}(C)=\mathrm{\partial }R$. The interior $\mathrm{int}R$ is nonempty, connected and contained in ${\mathbb{R}}^2\setminus \mathrm{\partial }R$, hence contained in $B$ or in $C$; say $\mathrm{int}R\subseteq G$ with $G\in \{B,C\}$. Since $R$ is closed and $G\cap \mathrm{\partial }R=\mathrm{\varnothing }$, the set $G\cap R=G\cap \mathrm{int}R=G\cap \overline{\mathrm{int}R}$ is open in $G$ and nonempty, while $G\setminus R$ is open in $G$; as $G$ is connected, $G\setminus R=\mathrm{\varnothing }$, that is, $G\subseteq R$. If $G=C$ the unbounded region would be contained in the compact, hence bounded, set $R$, which is impossible; hence $G=B$, so $B\subseteq R$, and from $R=\mathrm{int}R\cup \mathrm{\partial }R\subseteq B\cup \mathrm{\partial }R=\bar{B}$ we get $R=\bar{B}$ and $\mathrm{\partial }R=\mathrm{Fr}(B)$. Now fix $x\in s^{\circ }$. The set $\mathrm{\partial }R\setminus s^{\circ }$ is compact, being a closed subset of the compact polygon $\mathrm{\partial }R$, and does not contain $x$, so $\rho :=\mathrm{dist}(x,\mathrm{\partial }R\setminus s^{\circ })\mathrm{/}2$ is positive by [L7]; also $\rho \le \mathrm{dist}(x,v_0)$ and $\rho \le \mathrm{dist}(x,v_1)$, because $v_0,v_1\in \mathrm{\partial }R\setminus s^{\circ }$. Hence $N_{\rho }(x)\cap \mathrm{\partial }R=s\cap N_{\rho }(x)=:\sigma$ is a chord of the disk $N_{\rho }(x)$ with both endpoints on its boundary circle. The disk minus this chord has exactly two components, and each lies in $B$ or in $C$, being connected and disjoint from $\mathrm{\partial }R$. They cannot both lie in $B$: since $\mathrm{Fr}(C)=\mathrm{\partial }R$, the point $x$ is a limit point of $C$, so $N_{\rho }(x)\cap C\mathrm{\ne }\mathrm{\varnothing }$, and $C\cap N_{\rho }(x)$ meets only components of $N_{\rho }(x)\setminus \mathrm{\partial }R$ that lie in $C$. Interchanging $B$ and $C$ in this argument, using $\mathrm{Fr}(B)=\mathrm{\partial }R$, shows that they cannot both lie in $C$ either. Hence one component lies in $B$ and the other in $C$, and $R\cap N_{\rho }(x)=(B\cap N_{\rho }(x))\cup (\mathrm{\partial }R\cap N_{\rho }(x))$ is exactly the closed half-disk cut out by $\sigma$ on the side of $B$. At a vertex, say $v_0$, the same reasoning with a disk $N_{\delta }(v_0)$ of radius $\delta :=\mathrm{dist}(v_0,\mathrm{\partial }R\setminus ([v_{n-1},v_0]\cup [v_0,v_1]))>0$ shows that its complement is divided into exactly two sectors by the two edge germs, one sector lying in $B$ and the other in $C$; since $B\subseteq R$, the sector in $B$ is the one filled by $R$, so $R\cap N_{\delta }(v_0)$ is the closed sector bounded by the two edge germs, and the other sector lies in the unbounded region $C$.

The dual graph $D$ is connected. Let $A\subseteq T$ be the vertex set of a connected component of $D$ and put $X_A={\bigcup }_{\sigma \in A}\mathrm{\mid }\sigma \mathrm{\mid }$, a nonempty subset of $\mathrm{\mid }K\mathrm{\mid }$. It is closed: for all simplices $\sigma ,\tau$ of $K$ one has $\mathrm{\mid }\sigma \mathrm{\mid }\cap \mathrm{\mid }\tau \mathrm{\mid }=\mathrm{\mid }\sigma \cap \tau \mathrm{\mid }$, since a point of $\mathrm{\mid }K\mathrm{\mid }$ lies in both closed simplices exactly when its support is contained in both and $\sigma \cap \tau$ is a simplex; hence each $\mathrm{\mid }\sigma \mathrm{\mid }$ is closed in $\mathrm{\mid }K\mathrm{\mid }$, its trace on $\mathrm{\mid }\rho \mathrm{\mid }$ being the closed face $\mathrm{\mid }\sigma \cap \rho \mathrm{\mid }$, and a finite union of closed sets is closed. It is open: let $x\in X_A$ have support $\tau$, so that $\tau \subseteq \sigma$ for some $\sigma \in A$. If $\dim \tau =2$ then $\tau =\sigma \in A$, and $\mathrm{st}(\tau ):=\{y\in \mathrm{\mid }K\mathrm{\mid }:\mathrm{supp}(y)\supseteq \tau \}$ equals $\{y\in \mathrm{\mid }K\mathrm{\mid }:\mathrm{supp}(y)=\tau \}=\mathrm{\mid }\tau \mathrm{\mid }\setminus \bigcup \{\mathrm{\mid }\rho \mathrm{\mid }:\rho ⊊\tau \}$, which is open in $\mathrm{\mid }K\mathrm{\mid }$ because its trace on $\mathrm{\mid }\tau \mathrm{\mid }$ is $\mathrm{\mid }\tau \mathrm{\mid }$ minus the union of the finitely many closed faces $\mathrm{\mid }\rho \mathrm{\mid }$ with $\rho ⊊\tau$, and its trace on a simplex not containing $\tau$ is empty. If $\dim \tau =1$ then $\mathrm{st}(\tau )$ is open in $\mathrm{\mid }K\mathrm{\mid }$, because its trace on $\mathrm{\mid }\rho \mathrm{\mid }$ is $\{y\in \mathrm{\mid }\rho \mathrm{\mid }:y_w>0\text{ for }w\in \tau \}$, an intersection of finitely many subsets that are open in $\mathrm{\mid }\rho \mathrm{\mid }$, and its trace on simplexes not containing $\tau$ is empty; and $\mathrm{st}(\tau )\subseteq X_A$, because a point of $\mathrm{st}(\tau )$ lies in a simplex containing $\tau$, hence in $\tau$ itself or in one of the exactly two triangles containing $\tau$, and the triangle $\sigma \in A$ with $x\in \mathrm{\mid }\sigma \mathrm{\mid }$ is one of those two, so the other triangle containing $\tau$ is dual-adjacent to $\sigma$ and also lies in $A$. If $\tau =\{v\}$ is a vertex then some triangle $\sigma$ containing $v$ lies in $A$, and by step 1.2 every triangle containing $v$ is joined to $\sigma$ in the connected graph $\mathrm{\Lambda }(v)$, hence also lies in $A$; every simplex containing $v$ is contained in a triangle containing $v$, by hypothesis 1 and step 1.1, so $\mathrm{st}(\{v\})=\{y\in \mathrm{\mid }K\mathrm{\mid }:\mathrm{supp}(y)\ni v\}\subseteq X_A$, and this set is open by the previous case. Thus $X_A$ is clopen and nonempty in the connected space $\mathrm{\mid }K\mathrm{\mid }$, which is connected because $h$ is a homeomorphism onto the connected space $S$; hence $X_A=\mathrm{\mid }K\mathrm{\mid }$, and then $A=T$, because a triangle $\tau \in T\setminus A$ would give a point of $\mathrm{\mid }\tau \mathrm{\mid }\subseteq X_A$ whose support is not contained in any simplex of $A$. Therefore $D$ is connected.

Attachment of a thin triangle. In the situation of step 1.3 let $L$ be the line through $s$ and let $m$ be the midpoint of $s$. For each $y\in s^{\circ }$, step 1.3 says that $R\cap N_{\rho (y)}(y)$, where $\rho (y):=\mathrm{dist}(y,\mathrm{\partial }R\setminus s^{\circ })\mathrm{/}2>0$, is the relative closed half-disk on one side of $L$. This side is locally constant along $s^{\circ }$: when $z\in s^{\circ }$ is sufficiently close to $y$, a smaller disk about $z$ lies in both $N_{\rho (y)}(y)$ and $N_{\rho (z)}(z)$, and its occupied half-disk must be the same in both descriptions. Since $s^{\circ }$ is connected, one fixed side is occupied everywhere. Let $n$ be a unit normal to $L$ pointing into the other, exterior side. Write ${\delta }_0,{\delta }_1>0$ for the two vertex radii in step 1.3, and choose $r>0$ so that $r<\mathrm{\mid }s\mathrm{\mid }\mathrm{/}4$ and $2r<\min \{{\delta }_0,{\delta }_1\}$. On the nonempty compact middle subsegment $M:=\{y\in s:\mathrm{dist}(y,\{v_0,v_1\})\ge r\}$, the continuous positive function $\rho$ has a positive minimum $c$. Choose $\epsilon >0$ with $\epsilon <\min \{c,r\}$ and small enough that, for $q_{\epsilon }:=m+\epsilon n$, the narrow wedge at each $v_i$ between the ray toward $v_{1-i}$ and the ray toward $q_{\epsilon }$, with the original edge ray removed, lies inside the exterior open sector of step 1.3. Such a choice exists because those two rays approach each other from the exterior side as $\epsilon \downarrow 0$, while the other incident boundary edge makes a nonzero angle with $s$. Put ${\mathrm{\Delta }}_{\epsilon }:=\mathrm{conv}\{v_0,v_1,q_{\epsilon }\}$ and $R_{\epsilon }:=R\cup {\mathrm{\Delta }}_{\epsilon }$. Then ${\mathrm{\Delta }}_{\epsilon }\setminus s$ misses $R$. Indeed, for $y\in {\mathrm{\Delta }}_{\epsilon }\setminus s$, let $y_0$ be its nearest point of $s$; then $\mathrm{\mid }y-y_0\mathrm{\mid }\le \epsilon$ and $y$ is on the exterior side of $L$. If $y_0\in M$, then $\rho (y_0)\ge c>\epsilon$, so $y\in N_{\rho (y_0)}(y_0)$ on the unoccupied side of its local half-disk, whence $y\mathrm{\notin }R$. Otherwise $\mathrm{\mid }{y}_0-v_i\mathrm{\mid }<r$ for some endpoint $v_i$, so $\mathrm{\mid }y-v_i\mathrm{\mid }<r+\epsilon <2r<{\delta }_i$. Writing $y=\lambda v_0+\mu v_1+\nu q_{\epsilon }$ with $\lambda ,\mu ,\nu \ge 0$, $\lambda +\mu +\nu =1$ and $\nu >0$, we see that $y-v_i$ lies in the chosen exterior wedge at $v_i$; step 1.3 again gives $y\mathrm{\notin }R$. Thus ${\mathrm{\Delta }}_{\epsilon }\setminus s\subseteq {\mathbb{R}}^2\setminus R$. Consequently the open segments of the two new sides $[v_1,q_{\epsilon }]$ and $[q_{\epsilon },v_0]$ meet $\mathrm{\partial }R\subseteq R$ only at $v_0$ and $v_1$, and the chain obtained from that of $R$ by replacing $[v_0,v_1]$ by the two new segments is an irredundant simple closed polygonal chain: its vertices are the old ones with $q_{\epsilon }$ replacing the old edge, consecutive edges meet exactly in the common endpoints, nonconsecutive edges are disjoint, and no three consecutive vertices are collinear for small $\epsilon$, since the directions from $v_i$ to $q_{\epsilon }$ tend to the direction of $s$ as $\epsilon \to 0$ and consecutive edges of the given chain are not collinear. This chain is the boundary of $R_{\epsilon }$, whose interior is $\mathrm{int}R\cup s^{\circ }\cup \mathrm{int}{\mathrm{\Delta }}_{\epsilon }$, a connected set, and whose closure of interior is $R_{\epsilon }$. Hence $R_{\epsilon }$ is a simple polygonal region with that chain as boundary, and $R\subseteq R_{\epsilon }$.

Choose a subset $S_0\subseteq E(D)$ maximal by inclusion among the acyclic subsets of the finite edge set $E(D)$; the empty set is acyclic, so such a subset exists. If $(T,S_0)$ were disconnected then some edge of $D$ would join two of its components, and adding that edge keeps acyclicity, contradicting maximality; hence $(T,S_0)$ is a spanning tree of $D$ and $\mathrm{\mid }{S}_0\mathrm{\mid }=\mathrm{\mid }T\mathrm{\mid }-1$. We need an ordering ${\sigma }_1,\dots ,{\sigma }_m$ of $T$ in which every term after the first has exactly one neighbour among the earlier terms: induct on $m$, using that a finite tree with $m\ge 2$ vertices has a vertex of degree one, remove such a leaf, order the remaining tree by induction and append the leaf; the appended leaf has its unique neighbour among the earlier terms, and the earlier terms gain no earlier neighbour. Fix such an ordering, and for $i\ge 2$ let $e_i\in E_1$ be the $1$-simplex whose dual edge joins ${\sigma }_i$ to its unique earlier neighbour ${\sigma }_{p(i)}$; the $e_i$ are distinct because $e\mapsto \{\sigma ,{\sigma }^{\mathrm{\prime }}\}$ is injective. The sets $E_{\mathrm{t}\mathrm{r}}:=\{e_2,\dots ,e_m\}$ and $E_{\mathrm{f}\mathrm{r}}:=E_1\setminus E_{\mathrm{t}\mathrm{r}}$ are finite, $\mathrm{\mid }{E}_{\mathrm{t}\mathrm{r}}\mathrm{\mid }=m-1$, and $\mathrm{\mid }{E}_{\mathrm{f}\mathrm{r}}\mathrm{\mid }=\mathrm{\mid }{E}_1\mathrm{\mid }-m+1$.

Building the polygon. Place the triangle ${\mathrm{\Delta }}_{{\sigma }_1}$ as a nondegenerate triangle in the plane. After stage $i$ we maintain the invariant: $R_i:={\bigcup }_{j\le i}{\mathrm{\Delta }}_{{\sigma }_j}$ is a simple polygonal region; any two placed triangles meet in the empty set, a common corner or a full common side, and a full common side is exactly a glued pair $(p(l),l)$, $l\le i$; and the boundary chain of $R_i$ consists of the copies $(j,e)$ with $j\le i$ and $e\mathrm{\notin }\{e_2,\dots ,e_i\}$, each of them a full edge of the chain labelled by $e$, with its two endpoints labelled by the two vertices of $e$. For $i=1$ this holds, since the three sides of the single triangle form its boundary chain and none of them is among $e_2,\dots ,e_i$. Assume it for $i<m$. The tree edge $e_{i+1}$ is a $1$-simplex of ${\sigma }_{p(i+1)}$ whose other triangle is ${\sigma }_{i+1}$, so the copy $(p(i+1),e_{i+1})$ is a free side $A$ of $R_i$, a full edge of the boundary chain with labelled endpoints; place ${\mathrm{\Delta }}_{{\sigma }_{i+1}}$ so that its $e_{i+1}$-side coincides with $A$, with equal vertex labels, and its third corner is a point of the form $m_A+{\epsilon }_{i+1}{n}_A$ as in step 2.2; this is possible because the placement only prescribes the images of the three labelled corners. By step 2.2, $R_{i+1}=R_i\cup {\mathrm{\Delta }}_{{\sigma }_{i+1}}$ is a simple polygonal region and ${\mathrm{\Delta }}_{{\sigma }_{i+1}}\setminus A$ misses $R_i$, so ${\mathrm{\Delta }}_{{\sigma }_{i+1}}$ meets $R_i$ exactly in $A$; hence it meets each earlier triangle in the empty set, in a common corner, or in a full side, the last case occurring only for the triangle ${\mathrm{\Delta }}_{{\sigma }_{p(i+1)}}$ sharing $A$, because every other common set would lie in $A$ and therefore in the corners of the earlier chain. Its two remaining sides are new full edges of the new boundary chain, labelled by the two remaining $1$-simplices of ${\sigma }_{i+1}$; if $e$ is such a $1$-simplex then $e\mathrm{\ne }{e}_{i+1}$, and $e\mathrm{\ne }{e}_l$ for all $l\le i$ because otherwise ${\sigma }_{i+1}$ would be one of ${\sigma }_{p(l)},{\sigma }_l$ and would already be placed; so $e\mathrm{\notin }\{e_2,\dots ,e_{i+1}\}$. The chain update replaces the glued side $A$ by these two new sides, so the boundary chain of $R_{i+1}$ consists of exactly the copies $(j,e)$ with $j\le i+1$ and $e\mathrm{\notin }\{e_2,\dots ,e_{i+1}\}$, which is the invariant at stage $i+1$. This proves the invariant for all stages, so $P:=R_m$ is a simple polygonal region whose boundary chain has exactly two copies $(j,e)$ and $(j^{\mathrm{\prime }},e)$ of each $e\in E_{\mathrm{f}\mathrm{r}}$, namely the two copies in the two triangles containing $e$, and no other edge.

The finished polygon and its word. A $1$-simplex $e\in E_1$ lies in $E_{\mathrm{t}\mathrm{r}}$ exactly when its two copies were glued to one another at the attachment of the later of its two triangles, in which case both copies are interior edges of $P$ and no copy of $e$ remains on the boundary, and it lies in $E_{\mathrm{f}\mathrm{r}}$ otherwise, in which case no gluing was performed along it and, by step 4.1, both of its copies remain boundary edges of the chain of $P$. Declare a polygon with this chain, oriented as the boundary of $P$, and identify, for every $e\in E_{\mathrm{f}\mathrm{r}}$, the two boundary edges labelled $e$ by the affine homeomorphism matching corners with equal vertex labels. This is a partition of the sides of $P$ into pairs with affine homeomorphisms, so with the single polygon it constitutes a one-polygon polygonal schema with realization $Y$, the quotient of $P$ by the equivalence relation generated by these pairings; identifying the sides by equal labels amounts to identifying each of the two copies of $\mathrm{\mid }e\mathrm{\mid }$ by the map induced by the canonical identifications with $\mathrm{\mid }e\mathrm{\mid }$.

The canonical map. Each placed triangle ${\mathrm{\Delta }}_{{\sigma }_i}$ carries a homeomorphism onto $\mathrm{\mid }{\sigma }_i\mathrm{\mid }\subseteq \mathrm{\mid }K\mathrm{\mid }$ determined by the vertex labels of its corners, sending the corner labelled $w$ to the vertex $w$ of $\mathrm{\mid }{\sigma }_i\mathrm{\mid }$. These homeomorphisms agree on every pair of placed triangles meeting in a full side, because by step 4.1 such a side is a glued pair, glued by the identification of two copies of a $1$-simplex with equal labels, and on such a side both maps are the affine map onto $\mathrm{\mid }e\mathrm{\mid }$ matching the labels; they also agree on a common corner, which carries the same label in both triangles by step 4.1. Hence, since the placed triangles are finitely many closed subsets covering $P$ and the restriction to each of them is continuous, the maps paste to a continuous map $f:P\to \mathrm{\mid }K\mathrm{\mid }$ by [L2], and $f$ is surjective because every point of $\mathrm{\mid }K\mathrm{\mid }$ lies in some closed simplex $\mathrm{\mid }{\sigma }_i\mathrm{\mid }$ and the map on ${\mathrm{\Delta }}_{{\sigma }_i}$ is onto $\mathrm{\mid }{\sigma }_i\mathrm{\mid }$. Moreover $f$ respects the schema relation of step 5.1: two paired boundary edges carry the same label $e$ and are identified by the map matching equal vertex labels, while the maps of the two copies of $\mathrm{\mid }e\mathrm{\mid }$ are the affine maps onto $\mathrm{\mid }e\mathrm{\mid }$ determined by the same two vertex labels, so points of a paired pair of boundary edges are sent to points of $\mathrm{\mid }e\mathrm{\mid }$ with equal barycentric coordinates. Therefore $f$ induces a continuous surjection $\psi :Y\to \mathrm{\mid }K\mathrm{\mid }$ by the universal property of the quotient [L4].

$\psi$ is injective. Let $x,y\in P$ with $f(x)=f(y)=:z$, let $\tau =\mathrm{supp}(z)$, and note that $x$ lies in a placed triangle ${\mathrm{\Delta }}_{\sigma }$ with $\tau \subseteq \sigma$ and $y$ lies in a placed triangle ${\mathrm{\Delta }}_{{\sigma }^{\mathrm{\prime }}}$ with $\tau \subseteq {\sigma }^{\mathrm{\prime }}$. If $\dim \tau =2$ then $\tau$ is a triangle of $K$ and $\sigma ={\sigma }^{\mathrm{\prime }}=\tau$, and $x=y$ because the canonical map on ${\mathrm{\Delta }}_{\tau }$ is injective. If $\dim \tau =1$, say $\tau =e$, then the coordinates of $f(x)$ force $x$ to be an interior point of the $e$-side of ${\mathrm{\Delta }}_{\sigma }$ and $y$ to be an interior point of the $e$-side of ${\mathrm{\Delta }}_{{\sigma }^{\mathrm{\prime }}}$; if $\sigma ={\sigma }^{\mathrm{\prime }}$ then again $x=y$ by injectivity of the map on ${\mathrm{\Delta }}_{\sigma }$, while if $\sigma \mathrm{\ne }{\sigma }^{\mathrm{\prime }}$ then, since $e$ lies in exactly two triangles, $\{\sigma ,{\sigma }^{\mathrm{\prime }}\}$ are those two triangles, and their copies of $e$ are identified in $Y$ either by the gluing performed when the later triangle was attached, when the dual edge of $e$ is the tree edge $e_l$, or by the declared pairing of $e$ when $e\in E_{\mathrm{f}\mathrm{r}}$; in both cases $x$ and $y$ have the same barycentric coordinate on the copy of $\mathrm{\mid }e\mathrm{\mid }$, hence have the same image in $Y$. If $\dim \tau =0$, say $\tau =\{v\}$, then $x$ is the corner labelled $v$ of ${\mathrm{\Delta }}_{\sigma }$ and $y$ the corner labelled $v$ of ${\mathrm{\Delta }}_{{\sigma }^{\mathrm{\prime }}}$; by step 1.2 the triangles containing $v$ form a connected graph in which consecutive triangles share a $1$-simplex containing $v$, and for two such triangles the corners labelled $v$ are identified in $Y$ by the argument of the previous case applied to that shared $1$-simplex, whose two copies are identified in $Y$ and whose identification matches the corners with equal labels; hence the corners labelled $v$ of ${\mathrm{\Delta }}_{\sigma }$ and ${\mathrm{\Delta }}_{{\sigma }^{\mathrm{\prime }}}$ are identified and $x,y$ have the same image in $Y$. Therefore $\psi$ is injective.

Conclusion. The space $Y$ is a quotient of the compact space $P$, hence compact, and $\mathrm{\mid }K\mathrm{\mid }$ is Hausdorff by [L2]; a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism by the compact-to-Hausdorff assertion of [L3]. Hence $Y\cong \mathrm{\mid }K\mathrm{\mid }\cong S$, so the one-polygon schema of step 5.1 realizes a surface homeomorphic to $S$, and $Y$ is connected because $S$ is. Its vertex classes are the sets of corners labelled by a fixed vertex of $K$, its edge classes the pairs of equal-label boundary edges, and its one face is the interior of the polygon; every edge class therefore has exactly two incident face-sides. Fix a vertex $v$ of $K$. The triangles at $v$ occur in the cyclic order of the graph $\mathrm{\Lambda }(v)$ of step 1.2. The edges of the dual spanning tree that correspond to edges containing $v$ form a forest in this cycle, since they are a subset of a tree; each of its components is a path, possibly an isolated triangle. Gluing along these tree edges concatenates the corresponding consecutive triangle-corner arcs into one polygon-corner arc per path, while every remaining edge containing $v$ supplies paired boundary side germs that join the endpoints of these arcs in their original cyclic order. Thus the schema link at the vertex class of $v$ is obtained from the original cyclic triangle link by suppressing the intermediate vertices along those paths; it is a single topological cycle, though it need not be literally the graph $\mathrm{lk}(v)$. Its vertex star is consequently a disk, so the schema is a connected surface schema by [L5]. All selections above were made from finite explicit collections — the maximal acyclic set, the leaf-removal ordering and the finitely many placement parameters ${\epsilon }_{i+1}$ — so no choice principle was used.