Finite generation from cap with a finite fundamental cycle

Fix $0\le q\le n$, and put $p=n-q$. Let $S_q$ be the submodule of $C_q(M;R)$ freely spanned by the distinct back faces ${\sigma }_i[p,\dots ,n]$ occurring in $z$. It is free of rank at most $m$: these are a subset of the specified singular simplex basis, and repetitions are removed. For every degree-$p$ cochain $a$, the cap formula [F2] gives $a\cap z={\sum }_{i=1}^m{r}_{i}a({\sigma }_i[0,\dots ,p]){\sigma }_i[p,\dots ,n]\in S_q\mathrm{.}$ In particular every cocycle caps to a cycle lying in $S_q$.

Put $T_q=S_q\cap \ker {\mathrm{\partial }}_q$. This is a submodule of the finite free module $S_q$, so [F3] makes it finite free. Its map to $H_q(M;R)$, sending a cycle to its homology class, is onto: any homology class is $D_M[a]$ by [F1], and a cocycle representative of $a$ gives the cycle in step 1.1. The images of a finite basis of $T_q$ therefore generate $H_q(M;R)$. This uses a finite generating module of cycles, not the unsupported claim that the individual back faces are cycles.

For $q>n$, [F1] identifies $H_q(M;R)$ with $H^{n-q}(M;R)=0$, and negative homology degrees are zero by convention. For $0\le p\le n$, [F1] identifies $H^p(M;R)$ with the finitely generated $H_{n-p}(M;R)$ of step 2.1. If $p>n$ its target is a negative homology group, and if $p<0$ the cochain complex is zero, proving the stated vanishings. Since only finitely many degrees survive, even the direct sums over all degrees are finitely generated.

Empty $M$ has $z=0$, all $S_q=T_q=0$, and zero homology. A PID is nonzero by definition, so the zero ring is not a hypothesis here. At $q=0$, ${\mathrm{\partial }}_0=0$ and $T_0=S_0$; at $q=n$, cap uses degree-zero cochains and retains the original simplices. For $n=0$ the same proof uses only those vertices. A one-simplex support gives rank at most one for $T_q$. Degenerate simplices are legitimate basis elements, and duplicate back faces were removed explicitly. Beyond the AC inherited from [F1] for the countable atlas and local UCT, this proof chooses only a single representative cycle and a finite basis in one finite free module; it introduces no further infinite selection.