Finite surface normal forms, Jordan disks, and torsion control

(Compact subsurfaces.) For a nonempty compact $K$ choose finitely many chart disks with relatively compact larger chart disks. A $C^{\mathrm{\infty }}$ Euclidean bump supported in a larger disk, pulled back by its $C^2$ chart and extended by zero, is a $C^2$ function on $S$. Finitely many such bumps, positive on the smaller disks covering $K$, have a compactly supported sum $f$ with $f>0$ on $K$. Choose $0<a<{\min }_{K}f$ outside the critical values of $f$ on its support. This is possible by F2 applied in finitely many charts covering that compact support, since a finite union of null subsets of the line cannot contain an interval. At every point of $f^{-1}(a)$, F2 makes $f-a$ a $C^2$ coordinate. Hence $N_0=\{f\ge a\}$ is a compact $C^2$ subsurface and $K\subset \mathrm{Int}{N}_0$. A finite cover of $N_0$ by connected disk or half-disk neighborhoods shows it has finitely many components; their union gives the required $N$. If $K$ lies in one component of $S$, first connect the finitely many covering disk centers by finitely many paths in that component, and enlarge $K$ by those compact paths and the finitely many closed coordinate-core disks. The resulting compact connected set is inside $\{f>a\}$ after repeating the bump construction, so it lies in one component of $N_0$; take that component. For disconnected $S$, compactness meets only finitely many open components and the construction is performed in each of them. For $K=\mathrm{\varnothing }$, $N=\mathrm{\varnothing }$ suffices. For every compact subsurface used here, finitely many open chart domains cover it. Their union $W\subset S$ is an open Hausdorff boundaryless $C^2$ surface with a countable basis: take the union of the finitely many countable chart bases. Regard the subsurface as a subsurface of $W$ and apply F1 there. This supplies the finite cellulation, including any prescribed finite regular $C^2$ graph once $N$ contains it in its interior, without assuming that all of $S$ is second-countable.

(A boundary surface has a finite graph spine.) Let $N$ be connected and $\mathrm{\partial }N\mathrm{\ne }\mathrm{\varnothing }$, with the triangulation from F1. Its triangle dual graph, with an additional exterior vertex joined to triangles along their boundary edges, is connected: the cyclic or interval vertex links join all triangles incident to a vertex, and connectedness then joins all triangles. Choose a finite spanning tree rooted at the exterior vertex. Remove triangles in order from the root outward, always removing a triangle together with the edge connecting it to its already removed parent. That edge is free at its removal: its only other incident triangle has already been removed, or it was a boundary edge. A triangle with a free side strongly deformation retracts onto its other two sides; in coordinates on a planar reference triangle this is the elementary linear edge-collapse retraction. Performing these finitely many collapses leaves a connected finite embedded graph $B$, and $N$ deformation retracts onto $B$. More is true geometrically: $N$ is homeomorphic to a thickening of $B$. To check it, take small vertex disks and edge rectangles in the triangle charts. Reversing one free-side collapse attaches the missing triangular bulge along the two retained side strips. The union of those two strips and the bulge is a disk with the same two attaching arcs; parametrize its boundary arcs in the same order and extend their circle homeomorphism radially across the reference disk. This replaces the bulged neighborhood by the unbulged one relative to its attaching arcs. Induction reverses every collapse and identifies a neighborhood of the final graph with all of $N$, carrying boundary to boundary. Thus the thickening is made of finitely many vertex disks and edge bands with the orientation-induced cyclic orders. This argument proves the needed thickening assertion, rather than inferring it from a deformation retraction alone.

(Closed oriented polygon normal forms.) For a connected closed $N$, F1 verifies every edge and vertex hypothesis of F3, giving a one-face polygon schema. Opposite face-side orientations pair at every edge because $N$ is oriented. Reduce its vertex graph to one vertex by contracting a finite embedded spanning tree, unless the terminal sphere digon is reached first. Each edge contraction preserves the surface: a finite disk neighborhood of an edge with distinct endpoints is straightened to a segment $J$ strictly inside a convex disk; for $x\mathrm{\notin }J$ let $p(x)$ be its nearest point on $J$ and $b(x)$ the boundary point on the ray from $p(x)$ through $x$. The map $x\mapsto z_0+\frac{\mathrm{\mid }x-p(x)\mathrm{\mid }}{\mathrm{\mid }b(x)-p(x)\mathrm{\mid }}(b(x)-z_0)$, sending $J$ to its midpoint $z_0$, induces a homeomorphism from the disk modulo $J$ to the disk, fixing the outer boundary. Its inverse follows the normal ray indexed by the radial boundary point. Extend it by the identity outside that neighborhood. In the one-face polygon, collapsing the two occurrences of a tree side separately on the boundary still leaves a disk when other sides remain: a nondecreasing circle parameter constant on those intervals and strictly increasing elsewhere extends by $(r,t)\mapsto (r,(1-r)t+rq(t))$, which is strictly increasing for $r<1$. Thus a genuine polygon remains; when only $a{a}^{-1}$ remains, retain that actual sphere digon. For the remaining one-vertex opposite-pair word, there is no adjacent inverse pair, since its intermediate corner would be a separate vertex. An unprocessed pair must interlace another: otherwise $aX{a}^{-1}Y$ separates the corners of $X$ and $Y$ into distinct classes. F4 extracts the interlaced pair as a commutator block and leaves the residual word in order $YXVU$. Previously extracted contiguous blocks stay intact because the endpoints of the selected new letters cannot cut their interiors. Each extraction processes two new pairs; finite repetition yields ${\prod }_{i=1}^g[a_i,b_i]$. The terminal digon is a sphere, as follows by splitting it into two disks with their whole boundary circles identified. The one-handle square is the usual opposite-side torus. The cell counts are $(1,2g,1)$ for $g\ge 1$ and $(2,1,1)$ for the sphere, giving $\chi =2-2g$; F9 makes this invariant and hence makes $g$ unique. No general triangulation-existence or full-AC classification theorem has been used.

(Free groups and essential boundary words.) Collapse a finite spanning tree of $B$. Each remaining edge becomes one circle, giving a homotopy equivalence with a finite wedge; the finite tree contraction and its homotopy extend over the adjacent edge intervals by their endpoint parameters. F5 makes ${\pi }_1(N)$ free of rank $r=\mathrm{\mid }E(B)\mathrm{\mid }-\mathrm{\mid }V(B)\mathrm{\mid }+1$. If $B$ is a tree, its thickening is a disk: remove a leaf disk and its incident band, which is a disk attached along one arc, and induct to one vertex disk. Conversely, if $B$ has a cycle, repeatedly delete leaf edges and their end disks; this leaves a nonempty core whose vertex degrees are at least two, and does not change the boundary-loop classes except for deletion of immediate edge-and-inverse excursions. A boundary circle of the thickening follows an edge band and, at its next vertex disk, takes the next germ in cyclic order. In the core this is never the germ that would immediately reverse the incoming edge, because there are at least two germs. Every boundary circuit therefore gives a nonempty cyclically nonbacktracking closed edge path. Such a path has no null positive power: the covering graph whose vertices are reduced edge paths from a fixed vertex and whose edges append an edge and cancel an immediate backtrack is a tree (every nonroot path has its unique shorter prefix as parent). Its local edge stars map bijectively onto those of $B$, so it is a covering. A nonbacktracking path of positive length lifts from the root to its distinct path vertex; its repeated cyclically nonbacktracking powers have the same property. By F5, a nullhomotopy would lift and make their endpoints agree, a contradiction. Thus every boundary circle of a connected compact boundary surface other than a disk generates an injected infinite cyclic subgroup. This verifies the boundary injections that will be used below.

(A null simple curve separates the finite subsurface.) Include $c$ and a compact nullhomotopy in a connected $N$ as in step 1.1, and use F1 with $c$ as a prescribed graph. A regular embedded compact $C^2$ circle in an oriented surface has a two-sided $C^1$ collar: the tangent is $C^1$, and the induced normal orientation selects the positive transverse cone. Patch finitely many local transverse fields with chart bumps to a $C^1$ field near $c$; its short flow in F12 gives a map $c\times (-\delta ,\delta )\to N$. Its differential on the zero section is invertible, and compactness plus the local inverse in F12 excludes collisions after a common shrink. This proves the collar, with no $C^2$ collar-flow assertion needed. If $N\setminus c$ were connected, choose a simple path there joining opposite sides of a small crossing segment. Close it across that segment to obtain a dual circle $d$ meeting $c$ once. Choose $d$ as a finite normal path through triangles: the triangle dual graph of the connected cut surface is connected by its vertex links, so it joins the two triangles beside a selected interior point of a $c$-edge without crossing another $c$-edge. The joining path inside those triangles closes across that one edge, misses vertices, and meets the other edges transversely. Define an integer cochain on oriented triangulation edges by their signed crossings with $d$. On every triangle the entering and exiting crossings cancel, so this cochain annihilates its boundary. Its evaluation on the edge cycle $c$ is $\pm 1$. It therefore detects a nonzero cellular homology class of $c$, hence a nonzero singular class by F9. The supplied nullhomotopy makes that class zero: triangulate the parameter disk and push its finite singular two-chain into $N$; its boundary is the subdivided curve cycle. This contradiction shows that $c$ separates $N$. The collar has two connected sides, and every component of $N\setminus c$ has frontier on one of them (otherwise it would be open and closed in connected $N$); hence there are exactly two components. Their closures $N_1,N_2$ are compact connected boundary surfaces, each with the distinguished boundary $c$.

(Boundary genus and the separating handle curve.) For a connected boundary surface with $b$ circles, cap them by $b$ abstract disks, triangulated by finite fans along the existing boundary subdivisions. The capped surface is an oriented closed topological surface with a supplied finite triangulation, so the finite reduction of step 1.3 applies, with no need for a differentiable smoothing of the capped charts. Its genus $g$ defines the genus of $N$. Each cap adds one face in the disk-cell count and no new boundary cells, so F9 gives $\chi (N)=2-2g-b$. Since the spine of step 1.2 is a connected graph, its free rank is $1-\chi (N)=2g+b-1$. The closed normal word is also the connected sum of $g$ tori: in two normal polygons remove small interior disks and identify their boundary circles with reversed orientations. Cut the resulting polygonal annuli along a bridge between their outer marked vertices. The resulting disk word is $WtV{t}^{-1}$; the bridge is an embedded edge with distinct endpoints, and the contraction in step 1.3 gives precisely $WV$. Iterating proves this assertion from the actual finite disk and annulus gluings. For $g\ge 2$, the seam splitting the first torus from the other $g-1$ tori is therefore an embedded separating circle. Its two sides are compact boundary surfaces that are not disks: their spines have ranks $2$ and $2g-2$. Step 2.1 supplies their injective infinite cyclic boundary subgroups.

(The choice cost of amalgam normal form.) When two finite-rank free groups are amalgamated over such injected boundary circles, order each finite free alphabet and list all reduced words by length and then lexicographically. For every left coset of the cyclic boundary subgroup, take its least word in this well-order. This defines all representatives at once by a formula, with the identity representing the subgroup; it chooses no element from an arbitrary-index family. The coefficient in the boundary subgroup is unique because its generator has infinite order by step 2.1. Thus the fixed transversal data needed by F7 actually exist without full AC. Use the factor actions on these data and the uniqueness/nonidentity conclusion of F7; the general supplier's preliminary appeal to AC to find unspecified transversals is not a premise here. Every amalgam element is conjugate either into a factor or to a cyclically reduced alternating word of syllable length at least two: if the first and last syllables are in the same factor, conjugate by the first syllable and merge the new terminal pair, shortening the finite word; repeat. When the two ends are in different factors, concatenating any positive number of copies has no merging seam, so F7 makes it nonidentity. Consequently finite-order elements are conjugate into a factor. If both factors are torsion-free by F6, so is this amalgam. F7 also makes the common cyclic subgroup inject into the amalgam, including its nonidentity length-zero coefficients.

(Torsion-freeness for compact and arbitrary surfaces.) A compact boundary surface has a free group by step 2.1, hence is torsion-free by F6. A closed genus-zero surface has trivial group, and a genus-one surface has ${\mathbb{Z}}^2$, by F10 and step 1.3. For genus at least two, enlarge the two sides of the seam in step 3.1 by open annular collars; these are open path-connected sets with connected overlap retracting onto that circle. F8 identifies the group with the amalgam of the two free groups over the injected cyclic boundary group; step 3.2 proves torsion-freeness. Finally, if a loop in an arbitrary component of $S$ has a positive power nullhomotopic, the loop and an actual nullhomotopy have compact connected image. Step 1.1 places that image inside a connected compact subsurface $N$. The power is null in $N$, whose group has just been proved torsion-free; hence the original loop is null in $N$ and therefore in $S$. No injectivity of ${\pi }_1(N)\to {\pi }_1(S)$ has been assumed.

(One side is an embedded disk.) If neither $N_i$ were a disk, step 2.1 would make the distinguished circle inject as an infinite cyclic subgroup in both free fundamental groups. Enlarge the two sides by open collars of $c$; their overlap is a connected annulus. F8 and step 3.2 then give an amalgam in which the common cyclic subgroup is injective, so the loop $c$ is nontrivial in ${\pi }_1(N)$. This contradicts the actual nullhomotopy placed in $N$. At least one $N_i$ is therefore a disk, and its inclusion in $S$ is the required embedded compact disk region. Its boundary is the original regular $C^2$ circle, so the region's half-space charts are $C^2$ by the local graph inverse coordinates; its interior carries the given $C^2$ surface structure. The proof produces an embedded region, rather than an immersed null cap or a claim that a nullhomotopy is already embedded.

(Uniqueness and components.) If $D\subset S$ is any such disk region, $D\setminus c$ is connected and open in $S\setminus c$, and is closed there because compact $D$ is closed in Hausdorff $S$. Hence its interior is an entire component of $S\setminus c$. There are at most two components in the connected component of $S$ containing $c$, by the same two-sided collar and frontier argument as in step 2.2. Two disk regions on the same side must consequently have the same interior and closure. If both sides are disks, their union fills a neighborhood also at every point of $c$, so it is an open and closed compact surface in that ambient connected component, hence equals the whole component. Gluing two disks by their boundary-circle homeomorphism gives a sphere: extend that homeomorphism radially across one disk and identify the resulting pair with the two hemispheres. Thus two distinct disk regions are possible only when that connected component is a sphere; conversely, if the ambient component is a sphere, take the disk already supplied by step 4.2. The closure of its other side is a compact connected boundary surface. Additivity of the finite cell count along their common circle gives $2=1+\chi (N_2)$, since a disk has Euler characteristic one and a circle zero. Thus $\chi (N_2)=1$, its graph spine has rank zero by step 3.1, and step 2.1 makes it a disk too. These constructions prove all clauses. Every geometric choice and word reduction was finite, and the only stated background choice is F11; the canonical coset formula of step 3.2 adds no full AC.