Finite geometric chains and compact manifold chain domination without choice

Here the right-module singular complex is defined intrinsically: $C_n={⨁}_{\sigma }{\mathcal{R}}_{\sigma (v_0)}$, and on a coefficient $a$ its first face has coefficient $T_{01}a$ over $v_1$, while face $i>0$ has coefficient $(-1{)}^{i}a$ over $v_0$. The signs include the first face sign $+1$. The identities $d_i{d}_j=d_{j-1}{d}_i$ pair and cancel faces for $i<j$; the only two potentially different transport composites are along edges $01,12$ and edge $02$, equal because their paths are homotopic in the simplex. Thus $d^2=0$ for arbitrary noncommutative $R$. The same formula on affine simplices defines $F_{\ast }$, and affine inclusion defines a chain map $J$. All formulas are right-linear. Choosing one basis in each of the finitely many vertex fibers makes each $F_n$ finite-free, and $F_n=0$ above the dimension of $L$.

Cover $L$ by its finitely many open vertex stars. A simplex is called small if its image lies in one of them. Apply [F1] to obtain the small chain complex $C_{\ast }^{\mathrm{s}\mathrm{m}}$, inclusion $I$, map $\rho :C_{\ast }\to C_{\ast }^{\mathrm{s}\mathrm{m}}$ and homotopy $D$. The proof of [F1] transports coefficients only along paths within the domain simplex; hence its face cancellations and least-integer subdivision formula apply to the right-module complex of step 1.1, or equivalently to its underlying abelian-group complex. In particular $\rho$ and $D$ are right-$R$ linear and preserve every subcomplex generated by simplices whose images lie in a specified subcomplex of $L$.

For a small singular simplex $\sigma$, let $A_{\sigma }$ be the finite subcomplex generated by the geometric simplices whose relative interiors meet its image. If $v$ is a vertex whose open star contains that image, every such geometric simplex contains $v$. Therefore $A_{\sigma }$ is a cone with apex $v$: joining $v$ to any of its faces remains in it. If $\tau$ is a face of $\sigma$, then $A_{\tau }\subseteq A_{\sigma }$. Choose as apex the least vertex whose open star contains the image. Both its simplicial chains and its singular chains have explicit augmented cone contractions: join to $v$, using straight segments in the geometric cone, and transport coefficients along those segments. For a simplex map $f$, the singular cone is $t_{0}v+(1-t_0)f(t_1\mathrm{/}(1-t_0),\dots )$, extended by $v$ at $t_0=1$; boundedness makes it continuous there. The boundary is the original chain minus the cone on its boundary, with the usual augmentation term in degree zero. The simplicial cone is the oriented join, zero when the vertex already occurs. These identities follow by the finite face cancellations. The coefficient system trivializes on the cone by its straight paths, so the identities apply to all its fibers without choosing trivializations independently at every point.

Define $S:C_{\ast }^{\mathrm{s}\mathrm{m}}\to F_{\ast }$ by induction. For a point simplex, send its coefficient to the least apex vertex of its carrier along their straight segment. For degree $n>0$, the already defined $S(d\sigma )$ is a cycle in $A_{\sigma }$; in degree zero its transported augmentation is zero. Put $S\sigma$ equal to its simplicial cone at the chosen apex. The cone identity gives $dS\sigma =S(d\sigma )$. Images of all faces lie in their nested carriers, so each operation stays in $A_{\sigma }$. Extend linearly on each coefficient fiber. This constructs the entire chain map by a formula on generators, with finite sums and least vertices, rather than an infinite set of individually chosen fillings.

Construct $H:C_{\ast }^{\mathrm{s}\mathrm{m}}\to C_{\ast +1}$ similarly. After lower degrees are defined, the chain $\sigma -JS\sigma -H(d\sigma )$ is a cycle in $A_{\sigma }$, with zero augmentation in degree zero. Cone it singularly at the chosen apex and call the result $H\sigma$. Its boundary is precisely that cycle, so $dH+Hd=I-JS$. Put $P=S\rho$. Then $1-JP=(1-I\rho )+(I-JS)\rho =d(D+H\rho )+(D+H\rho )d\mathrm{.}$ Thus the displayed homotopy is actually right-linear and defined on all singular generators.

For an affine simplex $\tau$ of $L$, image preservation in step 2.1 and the carriers of step 3.1 put $PJ\tau$ in the simplicial chains of $\tau$ itself. On a vertex $PJ$ is the identity. Inducting on dimension, $d(PJ\tau -\tau )=0$. For a positive-dimensional geometric $n$-simplex, its top simplicial chain group is one fiber and its top boundary is injective: each codimension-one face occurs with coefficient the transported original coefficient, so zero boundary forces that coefficient zero. Hence $PJ\tau =\tau$. This proves $PJ=1$, including degenerate singular simplices, which were never discarded from the source complex.

For the manifold consequence, take the actual retraction $e:M\hookrightarrow L$ and $r:L\to M$ from [F2]. Triangulate each grid cube by its coordinate-order simplices: for each permutation of the axes take the successive vertices from its lower vertex by adding those coordinate edges in that order. These finitely many simplices cover the cube, and on shared faces the same ordering of the remaining coordinates gives the same triangulation. Pull the coefficient system to $L$ along $r$. With $F_{\ast }$ its finite simplicial complex, set $i_C=PC(e)$ and $r_C=C(r)J$. The homotopy of step 5.1 gives $r_C{i}_C\simeq C(r)C(e)=1$, since $re=1$ literally. The coefficient maps have these same identities on fibers, so this is domination in right-$R$ chain complexes, not just ordinary homology.

Finally the singular modules on $M$ have actual formula-defined free bases. Take its finite convex ball/half-ball atlas. The finite intersection graph is connected; choose a finite spanning tree, one point for each tree-edge intersection, and coordinate-straight paths through these points joining the finitely many chart centers to the basepoint. For a point $x$, use the first chart containing it and append that chart's straight path from its center to $x$. This specifies a based path ${\gamma }_x$ for every point without pointwise selection. Transport the supplied base vector $1$ along ${\gamma }_x$ to obtain a basis of its fiber. Each singular generator then has its specified first-vertex basis. The differential and homotopies of the preceding steps respect those right-module structures. For disconnected finite complexes the first argument already works componentwise with finitely many vertex bases; empty complexes give zero maps. All manifold choices were finite atlas/tree data, and the only infinite constructions were recursive formulas on generators. This proves the full choice-free assertion.