Jordan–Schönflies extension for plane curves

Assume AC [A1]. By [F1] the complement of the Jordan curve $C$ has exactly two components, the bounded $D$ and the unbounded $E$, with $\mathrm{Fr}(D)=\mathrm{Fr}(E)=C$, and $\bar{D}=D\cup C$ is closed and bounded, hence compact by [L1]; write $D^{\mathrm{\prime }}$ for the bounded component of ${\mathbb{R}}^2\setminus C^{\mathrm{\prime }}$. Put $Q=[-1,1{]}^2$ and $S=\mathrm{\partial }Q$. Reduction: it suffices to prove that for every Jordan curve $C$ and every homeomorphism $u:C\to S$ there is a homeomorphism $\bar{u}:\bar{D}\to Q$ with $\bar{u}{\mathrm{\mid }}_C=u$. Indeed, a Jordan curve is by definition the image of an embedding $\gamma$ of the unit circle [L2], so ${\gamma }^{-1}:C\to S^1$ is a homeomorphism [L2]; the map $g(x)=x\mathrm{/}\max (\mathrm{\mid }{x}_1\mathrm{\mid },\mathrm{\mid }{x}_2\mathrm{\mid })$ carries $S^1$ homeomorphically onto $S$, because its two components are quotients of continuous functions by the nonvanishing continuous function $x\mapsto \max (\mathrm{\mid }{x}_1\mathrm{\mid },\mathrm{\mid }{x}_2\mathrm{\mid })$ [L2], it maps $S^1$ onto $S$, and its inverse is the radial normalisation $y\mapsto y\mathrm{/}\parallel y{\parallel }_2$ restricted to $S$, continuous by [L2]; hence $\phi =g\circ {\gamma }^{-1}:C\to S$ is a homeomorphism. Given $h:C\to C^{\mathrm{\prime }}$, apply the square statement to $u=\psi \circ h:C\to S$ and to $\psi :C^{\mathrm{\prime }}\to S$, where $\psi$ is any homeomorphism onto $S$, obtaining homeomorphisms $\mathrm{\Phi }:\bar{D}\to Q$ and $\mathrm{\Psi }:\overline{D^{\mathrm{\prime }}}\to Q$; then ${\mathrm{\Psi }}^{-1}\circ \mathrm{\Phi }$ is a homeomorphism $\bar{D}\to \overline{D^{\mathrm{\prime }}}$ agreeing with $h$ on $C$. The second clause follows from the same reduction with $C^{\mathrm{\prime }}=S^1$: every closed bounded Jordan region $\bar{D}$ is then homeomorphic to the closed unit disk ${\bar{B}}_2(0,1)$.

For $q\in D$ the continuous function $y\mapsto \mathrm{\mid }q-y\mathrm{\mid }$ on the nonempty compact set $C$ attains a minimum at some $a(q)\in C$ [L1]; then $r:=\mathrm{\mid }q-a(q)\mathrm{\mid }>0$ because $q\mathrm{\notin }C$, and $B(q,r)\cap C=\mathrm{\varnothing }$; the convex ball $B(q,r)$ is connected and contains $q\in D$, so it lies in the single component of ${\mathbb{R}}^2\setminus C$ containing $q$, namely $D$. Call $p\in C$ strongly accessible if $B(q,r)\subseteq D$ for some $q\in D$ with $\mathrm{\mid }q-p\mathrm{\mid }=r$; every nearest point $a(q)$ is strongly accessible, and if $p=a(q)$, $v=(q-p)\mathrm{/}r$ and $w$ is a unit vector with $⟨v,w⟩>1\mathrm{/}2$, then for every $0<t<r$ one has $\mathrm{\mid }p+tw-q{\mathrm{\mid }}^2=r^2+t^2-2tr⟨v,w⟩<r^2+t(t-r)<r^2$, so the whole open cone of short segments $p+tw$ lies in $B(q,r)\subseteq D$ and meets $C$ only at $p$. Let $A:=\{a(q):q\in {\mathbb{Q}}^2\cap D\}$, a countable set. $A$ is dense in $C$: given $p\in C$ and $\epsilon >0$, choose $y\in D$ with $\mathrm{\mid }y-p\mathrm{\mid }<\epsilon \mathrm{/}4$ (possible since $p\in \mathrm{Fr}(D)$ and $D$ is open) and then $q\in {\mathbb{Q}}^2$ with $\mathrm{\mid }q-y\mathrm{\mid }<\epsilon \mathrm{/}4$ by density of the rationals in each coordinate [L5]; then $\mathrm{\mid }p-q\mathrm{\mid }<\epsilon \mathrm{/}2$ and $\mathrm{\mid }p-a(q)\mathrm{\mid }\le \mathrm{\mid }p-q\mathrm{\mid }+\mathrm{\mid }q-a(q)\mathrm{\mid }\le 2\mathrm{\mid }p-q\mathrm{\mid }<\epsilon$, because $a(q)$ is nearest to $q$. Finally, for distinct $a,b\in A$ the two cone segments at $a$ and $b$ have interior points in the region $D$, which is polygonally connected [L3], so joining those interior points by a polygonal arc and extracting a simple subpath [L4] produces a simple polygonal arc $P$ with endpoints $a,b$, relative interior in $D$ and $P\cap C=\{a,b\}$.

Crosscut splitting. Let $J$ be a Jordan curve with bounded region $V$, and let $P$ be a simple polygonal arc with distinct endpoints $p,q\in J$ and relative interior in $V$; let $A_1,A_2$ be the two closed arcs of $J$ from $p$ to $q$ and put $J_i=A_i\cup P$. Then ${\mathbb{R}}^2\setminus (J\cup P)$ has exactly three regions, with frontiers $J,J_1,J_2$; in particular $V\setminus P$ has exactly two components, with frontiers $J_1$ and $J_2$. Proof. Each $J_i$ is a Jordan curve: pasting parametrizations of $A_i$ and $P$ at their common endpoints gives a continuous bijection from the circle onto $J_i$, which is a homeomorphism because the circle is compact and $J_i$ is a Hausdorff subspace of the plane, and $J_i$ is compact and hence closed [L1, L4]. Let $W$ be the region of ${\mathbb{R}}^2\setminus J$ other than $V$, and let $Z_i$ be the region of ${\mathbb{R}}^2\setminus J_i$ containing $W$, whose companion region $Y_i$ has frontier $J_i$ by [F1]. Every point of $J\setminus J_i$ has a ball missing $J_i$ and meeting $W$, hence lies in $Z_i$; so $Y_i$ misses $J$, and being connected it lies in a single region of ${\mathbb{R}}^2\setminus J$, necessarily $V$ because it is not $W$. For each $i$, the frontier of $Y_i$ contains the open arc $A_i\setminus \{p,q\}$. At a point of that arc away from $P$, a small ball misses $J_{3-i}$ and meets $W$, hence meets $Z_{3-i}$. Since the ball also meets $Y_i$, and $Y_i$ avoids $J_{3-i}$ and is connected, $Y_i\subseteq Z_{3-i}$. Thus $Y_1\cap Y_2=\mathrm{\varnothing }$. At a point $x$ interior to $P$ take a small ball $B\subseteq V$ meeting $J\cup P$ only in the local subarc of $P$. If $x$ lies in a straight segment, this is a diameter; if $x$ is a polygonal bend, an explicit piecewise-linear local homeomorphism straightens its two incident rays to a diameter. The two resulting local sectors of $B\setminus P$ are connected and avoid $J\cup P$, and for each $i$ the two local sectors lie one in $Z_i$ and one in $Y_i$, since $x$ lies in the frontier of both $Z_i$ and $Y_i$; no local sector lies in both $Y_1$ and $Y_2$, so one local sector lies in $Y_1$ and the other in $Y_2$. Every component $N$ of $V\setminus P$ meets $Y_1\cup Y_2$: otherwise its closure in $V$ would be disjoint from $P\setminus \{p,q\}$, making $N$ closed in $V$ and, being a component of the open set $V\setminus P$, also open in ${\mathbb{R}}^2$ hence in $V$ [L3], so $N=V$, contradicting that the nonempty set $P\setminus \{p,q\}$ lies in $V$ and misses $N$; choose $x\in {\mathrm{cl}}_V(N)\cap (P\setminus \{p,q\})$ and take a fresh local ball $B$ centered at $x$ as above; then $N$ meets $B\setminus P\subseteq Y_1\cup Y_2$. Finally each $Y_i$ is closed in $V\setminus P$, since its frontier is $J_i$, so a connected $N$ meeting $Y_1\cup Y_2$ lies in $Y_1$ or in $Y_2$; hence $V\setminus P=Y_1\bigsqcup Y_2$ is exactly the two components, while $W$ is a region of ${\mathbb{R}}^2\setminus (J\cup P)$ with frontier $J$, so there are exactly three regions with the stated frontiers.

Initial matched pair. Fix $C$ and $u$. Choose $a,b\in A$ in the relative interiors of two arcs of $C$ cut out by the four points $u^{-1}$ of the corners of $Q$, such that $u(a)$ and $u(b)$ lie in the relative interiors of two opposite sides of $S$; this is possible because $A$ is dense in $C$ (step 1.2) and only finitely many points are forbidden. Subdivide $C$ at those six points and $S$ at the four corners together with $u(a),u(b)$, matching the subdivisions by $u$. By step 1.2 there is a polygonal crosscut $P$ of $D$ from $a$ to $b$; let $P^{\mathrm{\prime }}$ be the closed straight segment from $u(a)$ to $u(b)$, whose relative interior $P_{\circ }^{\mathrm{\prime }}$ lies in $Q^{\circ }$ because the endpoints are distinct boundary points of the convex square $Q$ lying on opposite sides. By step 1.3 applied to $(C,P)$ and to $(S,P^{\mathrm{\prime }})$, the closed domains $\bar{D}$ and $Q$ are each divided into exactly two closed cells whose common part is $P$, respectively $P^{\mathrm{\prime }}$; writing $A_1,A_2$ for the two boundary paths from $a$ to $b$ on $C$ and $R_1,R_2$ for the two components of $D\setminus P$, one has $\overline{R_i}\cap C=A_i$, and for the corresponding sides $R_i^{\mathrm{\prime }}$ of $Q\setminus P^{\mathrm{\prime }}$ one has $\overline{R_i^{\mathrm{\prime }}}\cap S=u(A_i)$. A cellulation of $\bar{D}$, respectively of $Q$, is a finite partition of the closed domain into cells --- vertices (single points), edges (open arcs) and faces (open connected sets) --- obtained from this initial decomposition by finitely many applications of the constructors (i) edge subdivision: insert a point in the relative interior of an edge, replacing the edge by the two edges and the new vertex; and (ii) face split: choose a face $R$ and two distinct vertices on its boundary, insert a polygonal arc with endpoints at those vertices and relative interior in $R$, and replace $R$ by that interior and the two components the arc cuts off. Cells contained in $C$, respectively $S$, are outer and the remaining cells inner; every inner edge is a simple polygonal arc with relative interior in $D$, respectively $Q^{\circ }$; the carrier of a point is the cell containing it; $\sigma ⪯\tau$ means $\sigma \subseteq \bar{\tau }$; and the closed star $\mathrm{St}(\sigma )$ is the union of the closures of the cells $\tau$ with $\sigma ⪯\tau$. A matched pair is a cellulation of $\bar{D}$, a cellulation of $Q$, and a homeomorphism $g$ of the two $1$-skeletons carrying every vertex and edge to its counterpart and restricting to $u$ on $C$; a matched cellulation is a matched pair whose two skeletons are $2$-connected and whose inner parts $\mathrm{\mid }G\mathrm{\mid }\setminus C$ and $\mathrm{\mid }{G}^{\mathrm{\prime }}\mathrm{\mid }\setminus S$ are connected. To apply a constructor to a matched pair is to apply it simultaneously on both sides: a subdivision inserts corresponding interior points via $g$, hence via $u$ on outer edges, and a split inserts corresponding crosscuts between corresponding vertices and extends $g$ over them. The initial pair is a matched cellulation: its skeleton is the six-vertex boundary cycle together with the inner edge $P$, a theta graph with branch vertices $a,b$ and three internally disjoint arcs, hence $2$-connected with connected inner part, and $g$ is $u$ on $C$ and any homeomorphism $P\to P^{\mathrm{\prime }}$ fixing endpoints.

Cell geometry. For every cellulation built from the initial decomposition of step 2.1 by the constructors: (a) the open cells partition the closed domain and every closed cell is the disjoint union of the open cells it contains; (b) if an open cell meets the closure of a cell then it is contained in that cell; (c) every face is the interior of a Jordan curve assembled from closed cells --- its boundary cycle --- every boundary cycle contains an inner edge, and every outer edge lies on exactly one boundary cycle; (d) distinct faces are incomparable under $⪯$. Proof by induction along the constructor sequence. At the initial stage the closure identities $\overline{R_i}=R_i\cup P\cup A_i$ supplied by step 1.3 give (a)--(c): the cells below $R_i$ are the cells of the path $A_i$ and of the crosscut $P$, the cells below an edge are its two endpoints, the two boundary paths meet only in $a$ and $b$, every boundary cycle contains the inner edge $P$ or $P^{\mathrm{\prime }}$, and each outer edge lies on exactly one boundary path. An edge subdivision replaces $\bar{e}$ by $\overline{e_1}\cup \overline{e_2}$, which meet in the new vertex; the only old closed cells that can contain a new cell are $\bar{e}$ and the cells $\bar{\tau }$ with $e⪯\tau$, by the old frontier property, so (a) and (b) persist, no face changes, and a boundary cycle that contained $e$ as an inner edge retains an inner subedge. A face split applies step 1.3 inside the Jordan region bounded by the face's boundary cycle: if the inserted crosscut has endpoints $x,y$ splitting that cycle into the two closed paths $B_1,B_2$, then the closure identities $\overline{V_i}=V_i\cup P_0\cup B_i$ give (a)--(c) for the new cells, each of the two new boundary cycles contains the inner crosscut $P_0$, and every outer edge of the split face lies on exactly one of $B_1,B_2$. For (d) suppose $R⪯T$ for distinct faces; then $R$ is disjoint from $T$ by (a), so $R\subseteq \bar{T}\setminus T$, and every point of the Jordan curve $\bar{T}\setminus T$ is a limit of points of the nonempty open set $T$, so $R$ would meet $T$, a contradiction.

Order, parents, stars and correspondence. (a) $⪯$ is a partial order. (b) Every cell produced by a constructor lies in the closure of a unique $⪯$-least old cell, its parent: for a subdivision the subdivided edge, for its three new cells; for a split the split face, for the crosscut's interior and the two new faces; each unchanged cell is its own parent. (c) A closed cell meets $C$, respectively $S$, if and only if it has an outer subcell. (d) In a matched pair the cell bijection $\sigma \mapsto {\sigma }^{\mathrm{\prime }}$ preserves $⪯$, outerness and parents, so that $\bar{\sigma }$ meets $C$ exactly when $\overline{{\sigma }^{\mathrm{\prime }}}$ meets $S$, and so that $\mathrm{St}(\sigma )\cap \mathrm{St}(\tau )\mathrm{\ne }\mathrm{\varnothing }$ exactly when $\mathrm{St}({\sigma }^{\mathrm{\prime }})\cap \mathrm{St}({\tau }^{\mathrm{\prime }})\mathrm{\ne }\mathrm{\varnothing }$. (e) Every cell is a subcell of a face, so $\mathrm{St}(\sigma )$ is the union of the closures of the faces above $\sigma$; if each of those closed faces has diameter $<\eta$, then $\mathrm{diam}\mathrm{St}(\sigma )<2\eta$. (f) Parents preserve $⪯$, so $\mathrm{St}({\sigma }^{+})\subseteq \mathrm{St}(\sigma )$ whenever $\sigma$ is the parent of ${\sigma }^{+}$, and the parent of the carrier of a point is its carrier at the previous stage: the closed stars of a fixed point shrink weakly under refinement. Proof. (a) Reflexivity and transitivity are immediate, and antisymmetry follows as in step 3.1(d): if $\sigma \mathrm{\ne }\tau$ with $\sigma ⪯\tau ⪯\sigma$ then each lies in the closure of the other minus itself, which is impossible for a vertex, because that set is empty; for an edge, because only vertices lie in its closure minus itself, and those are excluded by the vertex case; and for two distinct faces, by step 3.1(d). (b) As in step 3.1, every old closed cell other than the subdivided edge, respectively the split face, is a union of other open cells and hence disjoint from the open cell containing the new cells, so the parent relations are exactly those listed. (c) Inner vertices and edges lie in $D$, respectively $Q^{\circ }$, and faces are disjoint from $C$ and $S$, so a closed cell meeting $C$, respectively $S$, contains a point whose carrier is outer, and the frontier property makes that carrier a subcell; the converse is immediate. (d) Each constructor creates its new cells with the same subcell relations and the same outer status on both sides, and the parent rule of (b) is combinatorial, so induction along the shared constructor sequence proves the preservation; for the star criterion, a point of $\mathrm{St}(\sigma )\cap \mathrm{St}(\tau )$ lies in $\bar{\alpha }\cap \bar{\beta }$ for cells $\alpha ⪰\sigma$ and $\beta ⪰\tau$, and its carrier is by (b) a common subcell of $\alpha$ and $\beta$, so the corresponding target cells have a common subcell and the target stars meet, and symmetrically. (e) Every cell is a subcell of a face: initially every cell lies in $\overline{R_1}$ or $\overline{R_2}$; a subdivision's new cells lie in the closure of the subdivided edge, which is below a face by the induction hypothesis; and a split's new cells are the crosscut, below both new faces, and the two new faces. If $\tau ⪰\sigma$ and $\tau ⪯R$ with $R$ a face, then $\bar{\tau }\subseteq \bar{R}$, so the star is the union of the closures of the faces above $\sigma$; every such closed face contains $\sigma$, so choosing $x\in \sigma$ shows that any two points of the star are within $2\eta$ of each other through $x$. (f) In each constructor all new cells share one parent and every surviving cell is its own parent; a relation between two new cells holds on both sides, relations with surviving cells are inherited, and if an old $\sigma$ lies below a new $\tau$ then $\bar{\tau }\subseteq \bar{\rho }$ for the parent $\rho$, so $\sigma ⪯\rho$, while a new cell below a surviving old cell is excluded by (b). For the carriers, if $x$ has carrier $\sigma$ before and ${\sigma }^{+}$ after a constructor, then ${\sigma }^{+}\subseteq \bar{\sigma }$ by (b), so $\mathrm{par}({\sigma }^{+})⪯\sigma$, while $x\in \overline{\mathrm{par}({\sigma }^{+})}$ forces $\sigma ⪯\mathrm{par}({\sigma }^{+})$; antisymmetry gives equality.

Local sectors and face access. Let $G$ be the skeleton of one of our cellulations, and let $x\in \mathrm{\mid }G\mathrm{\mid }$ be a point not on $C$ when $G$ is the source skeleton, and an arbitrary point of $\mathrm{\mid }G\mathrm{\mid }$ when $G$ is the target skeleton. A small closed disk $B$ about $x$ meets $G$ in finitely many straight radial segments with pairwise distinct directions: this is clear for the finitely many polygonal arcs making up $G$ that pass through $x$, while the arcs not through $x$ are avoided by a small disk, and in the source case $B$ can moreover be chosen meeting $C$ in the empty set because $C$ is compact and disjoint from a neighbourhood of $x$ [L1]. Hence $B\setminus G$ is a union of open convex sectors, each connected and disjoint from $G$, and each sector lies in a single face. If $F$ is a face whose closure contains $x$, then some sector meets $F$, and a short segment from $x$ into that sector meets $G$ only at $x$ and otherwise lies in $F$; in the target case the sector is connected, meets $Q^{\circ }$, which is a union of faces, and misses $S$, hence lies in $Q^{\circ }$ and in $F$. Consequently every face is accessible from each of its boundary points by a straight polygonal segment, at every boundary point off $C$ on the source side and at every boundary point including points of $S$ on the target side.

Finite transfer. Let $(G_0,G_0^{\mathrm{\prime }})$ be a matched cellulation with skeleton homeomorphism $g$. (a) If $H\supseteq G_0$ is a finite $2$-connected plane graph with outer cycle $C$, all new edges polygonal with relative interiors in $D$, and $\mathrm{\mid }H\mathrm{\mid }\setminus C$ connected, then the refinement $G_0\to H$ can be reproduced on the target: there is a matched cellulation extending the pair whose source skeleton is $H$. (b) Symmetrically from the target side, provided every new inner edge meeting $S$ has exactly one endpoint on $S$, at a fresh point of $u(A)$ that becomes a boundary vertex with exactly one incident new inner edge. Proof. First pass to a common subdivision so that the old skeleton is a subgraph of the new one, transferring each new point on an old edge by the edge parametrization, which is $u$ on outer edges; in (b) every fresh anchor is inserted into the outer cycle first. Repeat [F4] proof step 1.2 from the whole current $2$-connected subgraph $G_0$: an unused component with only one attachment would make that attachment a cut vertex of $H$, so a shortest path through it between two distinct attachments is an ear; an unused edge whose endpoints already lie in $G_0$ is a one-edge ear. Each addition consumes an unused edge, hence a finite rooted-ear decomposition starts at $G_0$. We transfer those ears one at a time; an ear is inserted by a single face split preceded by at most two subdivisions, to make its endpoints vertices, and followed by one subdivision for each internal vertex, so every intermediate stage satisfies steps 3.1 and 4.1 whether or not its inner part is momentarily disconnected. The relative interior of the next ear $Q_0$ is connected and disjoint from the current skeleton, hence lies in a single current face $F$ by step 3.1(a); the current source and target graphs satisfy the hybrid hypothesis of [F3], since one cycle is drawn as the Jordan curve $C$, respectively the polygon $S$, and every other edge is a simple polygonal arc with relative interior in the bounded component, so [F3] certifies that the boundary of $F$ is a graph cycle, a Jordan curve containing the ear's endpoints $v,w$; let $F^{\ast }$ be the corresponding face on the other side. In case (a) the points $v^{\ast },w^{\ast }$ corresponding to $v,w$ lie on the skeleton off $S$ or on $S$ as images of boundary vertices already present, and are accessible from $F^{\ast }$ by step 5.1. In case (b) an endpoint of the reproduced crosscut is either an inner skeleton point, accessible by step 5.1, or a fresh anchor $a\in A$, where the open cone of short straight segments from step 1.2 can be shrunk, using that the rest of the skeleton is at positive distance from $a$ [L1], until its short segments avoid the rest of the skeleton and lie in the unique current source face incident with $a$: that face is $F^{\ast }$, because $a$ was inserted as a boundary vertex incident with exactly one face, every outer edge lying on exactly one boundary cycle by step 3.1(c), and $a$ is an endpoint of no later splitting crosscut, since all later new inner edges meeting $C$ are spokes at fresh anchors. Now join $v^{\ast },w^{\ast }$ by a polygonal crosscut of $F^{\ast }$, which exists because $F^{\ast }$ is a region and hence polygonally connected [L3], and use step 1.3 to split both faces by the two corresponding crosscuts, extending $g$ over $Q_0$ by a homeomorphism fixing the endpoints. After the last ear the prescribed-side skeleton is the given graph, and the reproduced skeleton is $2$-connected, because it grew from the $2$-connected old skeleton by subdivisions and by adding paths with distinct endpoints, which preserves $2$-connectivity; it also has connected inner part, which is homeomorphic to $\mathrm{\mid }H\mathrm{\mid }\setminus C$ under $g$. Hence the final pair is a matched cellulation.

The refinement scheme. Fix an enumeration of ${\mathbb{Q}}^2$ in which every point occurs infinitely often and let $B$ be the set of its members lying in $D$, again enumerated with every point occurring infinitely often; put ${\epsilon }_n=2^{-n}$, $r(p)={\mathrm{dist}}_{\mathrm{\infty }}(p,C)$, and at stage $n$ choose $s_n=s_n(b_n)$ generically in $(r(b_n)-\min (r(b_n)\mathrm{/}2,{\epsilon }_n),r(b_n)-\min (r(b_n)\mathrm{/}4,{\epsilon }_n\mathrm{/}2))$ and let $W_n(b_n)$ be the closed coordinate square of that radius about $b_n$; then $W_n(b_n)\subseteq D$ and $r(b_n)-s_n<{\epsilon }_n$. Claim: starting from any matched cellulation $(G_0,G_0^{\mathrm{\prime }})$ there is a sequence $(G_n,G_n^{\mathrm{\prime }})$ of matched cellulations such that $\mathrm{diam}{\mathrm{St}}_{G_n}(x)\to 0$ for every $x\in D$ and $\mathrm{diam}{\mathrm{St}}_{G_n^{\mathrm{\prime }}}({\sigma }_n^{\mathrm{\prime }}(x))\to 0$ uniformly in $x$, where ${\sigma }_n(x)$ is the carrier of $x$ and ${\sigma }_n^{\mathrm{\prime }}(x)$ its corresponding target cell; moreover every anchor used keeps an incident inner edge at all later stages, the set $A_{\mathrm{\infty }}$ of anchors used is dense in $C$, and $u(A_{\mathrm{\infty }})$ is dense in $S$. Stage $n$ starts from $(G_{n-1},G_{n-1}^{\mathrm{\prime }})$ and has four parts. (i) The old matched source skeleton H contains the outer Jordan cycle C, is 2-connected, and H\C is connected (Step 2.1/6.1 invariant). Every bounded face of H has an inner edge on its boundary (Step 3.1(c)). Put W=W_n(b_n). The generic radius excludes the finitely many horizontal and vertical levels supporting old polygonal edge subsegments; the radius bounds just stated put W compactly inside D and ensure every fixed source point lies in int W for infinitely many repeats of a sufficiently close b_n. Subdivide W into a rectangular grid of sufficiently small squares, choosing interior mesh line offsets generically as well. No grid edge, including ∂W, then shares a nondegenerate segment with any of the finitely many old polygonal inner edges; polygonal intersections are finite, and the grid misses the wild outer C because W⊂D. The grid 1-skeleton K is 2-connected. Proof: start with one square's 4-cycle; order remaining squares by an edge-adjacency spanning tree. Each new square shares at least one full edge with the preceding square union. Therefore the intersection of its boundary 4-cycle with the old 1-skeleton contains an edge. Each component of the complement of that intersection in the 4-cycle is an open path whose two endpoints are distinct old vertices; the shared full edge rules out the same-endpoint loop case. Its closure is a rooted ear, including the case of one new edge. Subdivide at any extra old corner contacts before adding the ear paths. This covers one, two adjacent, two opposite, three, or all four shared edges, and preserves 2-connectivity. Finite subdivisions at old-grid crossings preserve it. If K meets H in at least two distinct points, the union H∪K is 2-connected: after deleting any vertex, both surviving graphs are connected and at least one other common point remains. All contacts are inside D, so their inner parts connect. If K meets H in exactly one point, K minus that point lies in one bounded Jordan face F of H; if K∩H is empty, all of K lies in one bounded face F. In either case choose a free point y on an inner edge of ∂F. Choose a free open grid-edge point p of K\H inside F, away from its finite intersections with H, and choose the free old-inner-edge point y away from all vertices. Endpoint collars enter F at both points. Polygonal connectivity of F gives a path between the collars; perturb its finitely many segments generically so it misses every vertex of K and crosses its edges transversely, and erase loops. The initial subpath from y to its first K contact ends in a free open grid edge, lies otherwise in F\K, and is the desired simple connector. This justifies the free-edge endpoint rather than assuming an arbitrary first-contact path has one. In the one-contact case, choose p and y distinct from that contact; the extra connector makes the union 2-connected by the deletion test and joins the inner parts. In the zero-contact case, thicken a compact interior segment of the connector to a narrow ribbon within the same component of F\K; its two long sides are disjoint connectors with distinct K/H endpoints. Deleting any vertex leaves one surviving attachment, so the union is 2-connected, and the inner parts connect. Endpoint collars exist because all edges here are polygonal and the disjoint nonincident finite pieces have positive distance; face openness and polygonal connectedness give the connector. The free inner edge exists by Step 3.1(c), so no connector is allowed to terminate solely on C. Before invoking the rooted-ear transfer, split all but one of every finite family of parallel edges in the overlaid graph at a new free interior point. The resulting abstract graph is simple; these degree-two subdivisions preserve the deletion test for 2-connectivity, preserve connectedness after removing C, and leave every grid line geometrically present. No loop edge is created by the transverse overlay. Treat the extra subdivision points as vertices of H_n in the transfer, and choose them off the fresh-boundary-spoke edges so Step 6.1(b) is unchanged. The resulting H_n contains every grid line and ∂W. Thus every source face incident to a point in int W lies within one small grid rectangle (its complement cannot cross any grid line or ∂W), giving the desired source-star diameter estimate. The graph is 2-connected and its inner part connected, so it meets Step 6.1(a)'s exact transfer hypotheses. (ii) Transfer the source refinement to the target by step 6.1(a). (iii) Let S be the target polygonal boundary. Choose at least two distinct fresh anchors a_1,a_2 (the ε_n/4 boundary mesh in fact gives many more). Choose λ close to 1 and the inner grid-line levels generically so neither the core boundary ∂(λQ) nor its internal mesh shares a segment with an old polygonal inner edge. A radial spoke at anchor a can share a segment with an old edge only when an old edge lies on the same ray from the square centre; the finite old edge set excludes only finitely many such anchor points, so the dense allowable anchor set still supplies the required ε_n/4 mesh. Take the fine core grid K inside λQ and one radial spoke from each a_j to its distinct point λa_j on the core boundary, subdividing at all finite meetings with the old target inner skeleton H'. The union S∪K∪spokes is 2-connected: S and K are cycles-containing 2-connected graphs joined by at least two internally disjoint spokes with distinct endpoints; deleting a vertex leaves a surviving spoke and both surviving graph pieces connected. Its union with old H' is 2-connected because they contain the full outer S. But 2-connectivity alone does not imply the required (new graph)\S is connected: if K/spokes miss H' off S, their inner parts remain separate. In that case the connected new inner core with its spoke interiors is disjoint from H'\S, so it lies entirely in one component F of the complement of H'; because it contains the core inside S, F is a bounded old face. Its frontier has a free inner H' edge by Step 3.1(c). Choose free open edge points y on this old inner edge and p on a new inner core edge inside F. Endpoint collars enter F at both. Join them by a polygonal path in F, perturb finitely many segments to avoid all graph vertices and be transverse to new edges, then retain the initial subpath from y to its first new-edge contact. That contact is a free open edge point; loop erasure gives a simple connector in one face of the full overlaid arrangement. Thus its interior misses every existing edge, it joins the old and new inner parts, and it cannot harm 2-connectivity. If K/spokes meet H' at an inner point, no extra connector is needed. All new inner edges meeting S are still exactly the single spokes at the fresh anchors, meeting Step 6.1(b)'s boundary condition. Before the Step 6.1 transfer, subdivide all but one edge in each parallel-endpoint family of the finite overlay at a fresh inner point, making the abstract graph simple without changing its geometric trace, 2-connectivity, inner connectivity or fresh S-anchor spokes. For the target-star estimate, a spoke-bounded collar sector has diameter at most the diameter of its outer S-arc plus twice its radial thickness. Choose the anchor spacing and λ with strict total margin below ε_n; the core cells have diameter below the remaining margin. Overlay subdivisions only shrink these cells, so every resulting target face-star has diameter < ε_n. (iv) Transfer back by step 6.1(b), obtaining $(G_n,G_n^{\mathrm{\prime }})$. The alternation shrinks both sides. On the target, all new faces lie in core mesh cells or spoke-bounded collar sectors of diameter below ${\epsilon }_n$ by the strict anchor-spacing and radial-thickness bound just proved; later overlays only subdivide them, so $\mathrm{diam}{\mathrm{St}}_{G_n^{\mathrm{\prime }}}({\sigma }_n^{\mathrm{\prime }}(x))<2{\epsilon }_n$ uniformly by step 4.1(e),(f). On the source, fix $x\in D$ and $d=r(x)>0$. Pick $b\in B$ with $\mathrm{\mid }b-x{\mathrm{\mid }}_{\mathrm{\infty }}<d\mathrm{/}8$ and use its infinitely many repetitions at indices $n$ with ${\epsilon }_n<d\mathrm{/}8$. Then $x\in \mathrm{int}{W}_n(b_n)$; every source face incident with $x$ has a parent face in the pre-pullback grid overlay whose closure contains $x$. That parent face cannot cross a grid line or $\mathrm{\partial }{W}_n$, so it lies in one rectangle of diameter below ${\epsilon }_n$. Hence $\mathrm{diam}{\mathrm{St}}_{G_n}(x)<2{\epsilon }_n$, and later stars shrink by step 4.1(f). The stage-$n$ anchors cut $S$ into arcs of diameter below ${\epsilon }_n\mathrm{/}4$, so their union is dense in $S$, and its inverse image is dense in $C$ because $u$ is a homeomorphism. Choices are finite or from the fixed enumerations.

The interior map. For $x\in D$ put $T_n(x)={\mathrm{St}}_{G_n^{\mathrm{\prime }}}({\sigma }_n^{\mathrm{\prime }}(x))$. These sets are nonempty and compact, and by steps 4.1(f) and 7.1 they are nested with $\mathrm{diam}{T}_n(x)\to 0$; hence their intersection is a single point: it is nonempty because a decreasing sequence of nonempty compact sets has nonempty intersection, since otherwise the open complements would cover the compact set $T_1(x)$, finitely many of them would cover it, and for the largest index $N$ occurring one would have $T_N(x)=\mathrm{\varnothing }$, a contradiction; and it contains at most one point because two points of the intersection would be at distance at most $\mathrm{diam}{T}_n(x)$ for every $n$. Define $F(x)$ by $\{F(x)\}={\bigcap }_n{T}_n(x)$. Then $F$ is a homeomorphism $D\to Q^{\circ }$, and $F=g_n$ on $G_n\cap D$ for every $n$. Agreement: if $x\in G_N\cap D$ and $n\ge N$, the skeleton map carries the carrier of $x$ onto the corresponding cell, so $g_N(x)=g_n(x)\in T_n(x)$ and $F(x)=g_N(x)$. Continuity: given $x$ and $\eta$, choose $n$ with $\mathrm{diam}{T}_n(x)<\eta$; in the stage-$n$ star neighbourhood of $x$, the complement of the union of the closed cells not containing $x$, shrunk into $D$, every point $z$ has carrier above the carrier of $x$, so by step 4.1(d) $T_n(z)\subseteq T_n(x)$ and $\mathrm{\mid }F(z)-F(x)\mathrm{\mid }<\eta$. Interior image: for $n$ with the source star of $x$ smaller than $\mathrm{dist}(x,C)$, the corresponding target star misses $S$ by step 4.1(d), so $F(x)\in Q^{\circ }$. Injectivity: if $F(x)=F(y)$ then the target stars of $x$ and $y$ meet at every stage, hence so do the source stars by step 4.1(d), and a common point gives $\mathrm{\mid }x-y\mathrm{\mid }\le \mathrm{diam}{\mathrm{St}}_{G_n}(x)+\mathrm{diam}{\mathrm{St}}_{G_n}(y)\to 0$. Surjectivity: the skeleton points off $S$ are dense in $Q^{\circ }$, since at every stage every face has diameter less than ${\epsilon }_n$ and its boundary cycle contains an inner edge by step 3.1(c); given $y\in Q^{\circ }$, take skeleton points $y_k\to y$ off $S$, corresponding at a late common stage to points $x_k\in D$; for $n$ with the target star of $y$ off $S$, the finitely many source closed cells corresponding to the supercells of the carrier of $y$ form a compact set $K\subseteq D$, and for large $k$ the points $y_k$ lie in the star neighbourhood of $y$, so $x_k\in K$; by [L1] a subsequence converges to some $x\in K\subseteq D$, and by continuity and agreement $F(x)=y$. Cells map to cells: $F=g_n$ on inner vertices and edges; for a face $R$ one has $F(R)\subseteq \overline{R^{\mathrm{\prime }}}$ by step 4.1(e), and $F(R)\subseteq R^{\mathrm{\prime }}$ because a value on the target skeleton would have a source preimage in $D$ at which $F$ agrees with $g_n$, contradicting injectivity; conversely a preimage in $D$ of a point of $R^{\mathrm{\prime }}$ has a face carrier $R_0$ with $F(R_0)$ meeting $R^{\mathrm{\prime }}$, so $R_0=R$ and $F(R)=R^{\mathrm{\prime }}$. Inverse continuity: given $y=F(x)$ and $\eta$, choose $n$ with the source star of $x$ smaller than $\eta$ and take the target star neighbourhood $V$ of $y$; for $z\in V\cap Q^{\circ }$ the target carrier of $y$ lies below the carrier of $z$, so by step 4.1(d) the source carrier of $x$ lies below that of $F^{-1}(z)$, whence $F^{-1}(z)\in {\mathrm{St}}_{G_n}(x)$ and $\mathrm{\mid }{F}^{-1}(z)-x\mathrm{\mid }<\eta$.

Skeleton crosscuts and corresponding sides. For distinct $a,b\in A_{\mathrm{\infty }}$ occurring in a common stage there is a crosscut $P$ of $D$ from $a$ to $b$ inside that stage's skeleton whose image $P^{\mathrm{\prime }}=g(P)$ is a crosscut of $Q^{\circ }$ from $u(a)$ to $u(b)$; and if $U_1,U_2$ are the two components of $D\setminus P$ labelled by $\overline{U_i}\cap C=A_i$ and $U_1^{\mathrm{\prime }},U_2^{\mathrm{\prime }}$ the two components of $Q^{\circ }\setminus P^{\mathrm{\prime }}$ labelled by $\overline{U_i^{\mathrm{\prime }}}\cap S=u(A_i)$, then $F(U_i)=U_i^{\mathrm{\prime }}$. Proof. Put $X=\mathrm{\mid }G\mathrm{\mid }\setminus C$, a connected set. If some inner edge has both endpoints on $C$, then its relative interior is open and closed in $X$ and hence equals $X$, and its endpoints are exactly $a$ and $b$, since each anchor is incident with an inner edge at every later stage; take that edge as $P$. Otherwise let $\mathrm{\Lambda }$ be the subgraph consisting of the vertices and edges entirely in $D$; attaching to each component of $\mathrm{\Lambda }$ the half-open relative interiors of the edges with exactly one endpoint on $C$ partitions $X$ into clopen pieces, one per component of $\mathrm{\Lambda }$, so $\mathrm{\Lambda }$ meets every component of $X$ and is therefore connected, and it is nonempty because the retained inner edge at the boundary anchor $a$ has an interior endpoint that is a vertex of $\mathrm{\Lambda }$; the union of $\mathrm{\Lambda }$ with the two edges at $a$ and $b$ is connected and meets $C$ exactly in $\{a,b\}$, and a simple graph path from $a$ to $b$ inside it is a polygonal crosscut. The image under the skeleton homeomorphism is simple, meets $S$ exactly in $u(a),u(b)$ and is polygonal in $Q^{\circ }$. For the sides, $F$ is a homeomorphism agreeing with $g$ on $P\cap D$, so it maps the components of $D\setminus P$ onto the components of $Q^{\circ }\setminus P^{\mathrm{\prime }}$; to identify the labels, choose $c\in A_{\mathrm{\infty }}$ in the relative interior of the boundary arc $A_i$ at a stage containing $P,c$ and an inner edge $e$ at $c$; a short initial subarc $J$ of $e$ with $c$ removed avoids $\overline{U_{3-i}}$, whose intersection with $C$ is $A_{3-i}$, and avoids $P$, so $J\subseteq U_i$; its image $g(J)$, short enough, lies in $U_i^{\mathrm{\prime }}$, and $F=g$ on $J$, so the labels match.

Boundary continuity and the square extension. Extend $F$ to $\bar{u}:\bar{D}\to Q$ by $\bar{u}=u$ on $C$. Then $\bar{u}$ is a homeomorphism, so the square statement of step 1.1 holds. Proof. $\bar{u}$ is a bijection because $F:D\to Q^{\circ }$ and $u:C\to S$ are bijections; it is continuous on $D$ and on $C$, and we verify sequential continuity at each $p\in C$, which suffices by [L2]. Suppose $x_k\to p$ with $x_k\in D$ but $\bar{u}(x_k)\to ̸u(p)$; passing to a subsequence, which is legitimate by the Bolzano–Weierstrass property of the bounded set $Q$ [L1], we may suppose $\bar{u}(x_k)\to q\mathrm{\ne }u(p)$; then $q\in S$, because a limit $q\in Q^{\circ }$ would give $x_k=F^{-1}(\bar{u}(x_k))\to F^{-1}(q)\in D$ by continuity of the inverse homeomorphism $F^{-1}$ (step 8.1), contradicting $x_k\to p\mathrm{\notin }D$. Let $r=u^{-1}(q)\mathrm{\ne }p$. Since $A_{\mathrm{\infty }}$ is dense in $C$ (step 7.1), choose $a,b\in A_{\mathrm{\infty }}$, distinct from $p$ and $r$, that separate $p$ from $r$ on $C$, and take them in a common finite stage; step 9.1 gives crosscuts $P,P^{\mathrm{\prime }}$. Let $U_1$ be the side of $D\setminus P$ whose closure meets $C$ in the closed arc $A_1$ containing $p$ in its relative interior, so that $r$ lies in the other arc $A_2$. The crosscut $P$ meets $C$ only in $a,b$ and $\overline{U_2}\cap C=A_2$, so both $P$ and $\overline{U_2}$ have positive distance from $p$; hence all $x_k$ sufficiently close to $p$ lie in $U_1$, and their images lie in $U_1^{\mathrm{\prime }}$ by step 9.1, so the limit $q$ lies in $\overline{U_1^{\mathrm{\prime }}}$, whose intersection with $S$ is $u(A_1)$; but $q=u(r)$ with $r$ in the relative interior of $A_2$, a contradiction. Thus $\bar{u}$ is a continuous bijection from the compact space $\bar{D}$ onto the Hausdorff space $Q$, hence a homeomorphism [L1].

The exterior. First note the pointed form of the square statement: for Jordan curves $C,C^{\mathrm{\prime }}$ and interior points $z\in D$, $z^{\mathrm{\prime }}\in D^{\mathrm{\prime }}$, every homeomorphism $f:C\to C^{\mathrm{\prime }}$ has an extension to a homeomorphism $\bar{D}\to \overline{D^{\mathrm{\prime }}}$ carrying $z$ to $z^{\mathrm{\prime }}$. Indeed, choose homeomorphisms $\mathrm{\Phi }:\bar{D}\to Q$ and $\mathrm{\Psi }:\overline{D^{\mathrm{\prime }}}\to Q$ by steps 1.1 and 10.1, with $\mathrm{\Phi }{\mathrm{\mid }}_C=\psi \circ f$ and $\mathrm{\Psi }{\mathrm{\mid }}_{C^{\mathrm{\prime }}}=\psi$, where $\psi :C^{\mathrm{\prime }}\to S$ is a boundary homeomorphism. Put $a=\mathrm{\Phi }(z)$ and $b=\mathrm{\Psi }(z^{\mathrm{\prime }})$, both interior points of $Q$. The four triangles spanned by $a$ and the sides of $Q$ triangulate $Q$. Mapping each by the vertex-affine map to the corresponding triangle spanned by $b$, fixing the side vertices and sending $a$ to $b$, gives a boundary-fixing homeomorphism $M:Q\to Q$: maps agree on shared segments and paste by [L2], and reversing $a,b$ builds its inverse. The pointed extension is ${\mathrm{\Psi }}^{-1}\circ M\circ \mathrm{\Phi }:\bar{D}\to \overline{D^{\mathrm{\prime }}}$, which extends $f$ and sends $z$ to $z^{\mathrm{\prime }}$. Next, for $a\in D$ the inversion $I_a(x)=a+(x-a)\mathrm{/}\mathrm{\mid }x-a{\mathrm{\mid }}^2$ is an involutive homeomorphism of ${\mathbb{R}}^2\setminus \{a\}$, and $C^a:=I_a(C)$ is a Jordan curve; the set $U:=I_a(E)\cup \{a\}$, where $E$ is the unbounded component of ${\mathbb{R}}^2\setminus C$, is open --- away from $a$ because $I_a$ is a homeomorphism there, and at $a$ because the complement of a large disk maps into a punctured disk about $a$ --- connected, since $a$ lies in the closure of the connected image of $E$, and bounded, because $E$ avoids a disk about $a$ of radius $\mathrm{dist}(a,C)$; away from $a$ its frontier is $C^a$ and $a$ is an interior point, so $U$ is open and closed in ${\mathbb{R}}^2\setminus C^a$ and hence a component of that complement [L3]; being bounded it is the bounded component $V^a$ of ${\mathbb{R}}^2\setminus C^a$ by [F1], so $I_a$ restricts to a homeomorphism $C\cup E\to (C^a\cup V^a)\setminus \{a\}$. Now let $h:C\to C^{\mathrm{\prime }}$ be a homeomorphism and fix $a\in D$, $b\in D^{\mathrm{\prime }}$; set $f^{\ast }=I_b\circ h\circ I_a:C^a\to (C^{\mathrm{\prime }}{)}^b$ and extend $f^{\ast }$ with ${\mathrm{\Phi }}^{\ast }(a)=b$ to a homeomorphism ${\mathrm{\Phi }}^{\ast }:\overline{V^a}\to \overline{V^{\mathrm{\prime }b}}$, which the pointed form supplies because $a\in V^a$ and $b\in V^{\mathrm{\prime }b}$ are interior points; then $F_{\mathrm{e}\mathrm{x}\mathrm{t}}:=I_b\circ {\mathrm{\Phi }}^{\ast }\circ I_a$ is a homeomorphism $C\cup E\to C^{\mathrm{\prime }}\cup E^{\mathrm{\prime }}$ extending $h$ on $C$, since the inversions restrict to homeomorphisms of those closed exteriors and remove the points $a,b$ correspondingly.

Conclusion. The interior extension $F_{\mathrm{i}\mathrm{n}\mathrm{t}}:={\mathrm{\Psi }}^{-1}\circ \mathrm{\Phi }$ of step 1.1 is a homeomorphism $C\cup D\to C^{\mathrm{\prime }}\cup D^{\mathrm{\prime }}$ extending $h$, and $F_{\mathrm{e}\mathrm{x}\mathrm{t}}$ of step 11.1 is a homeomorphism $C\cup E\to C^{\mathrm{\prime }}\cup E^{\mathrm{\prime }}$ extending $h$; the two agree on $C$, and their closed domains cover ${\mathbb{R}}^2$, so pasting them gives a continuous bijection $H:{\mathbb{R}}^2\to {\mathbb{R}}^2$ extending $h$; pasting the two inverse homeomorphisms, which agree on $C^{\mathrm{\prime }}$, gives continuity of $H^{-1}$ as well, so $H$ is a homeomorphism [L2]. Since $H$ restricts to a homeomorphism of ${\mathbb{R}}^2\setminus C$ onto ${\mathbb{R}}^2\setminus C^{\mathrm{\prime }}$, it carries complementary components to complementary components; and it carries bounded sets to bounded sets, because the closure of a bounded set is compact [L1], a continuous image of a compact set is compact and hence closed and bounded [L1]; therefore $H(D)=D^{\mathrm{\prime }}$ and $H(E)=E^{\mathrm{\prime }}$, the bounded and unbounded components corresponding. Finally, taking $C^{\mathrm{\prime }}=S^1$ and any homeomorphism $h:C\to S^1$ exhibits the closed bounded Jordan region $\bar{D}$ as homeomorphic to the closed unit disk ${\bar{B}}_2(0,1)$. AC is used only as declared in [A1].