In “locally contractible” space, cycles are boundaries.

Question

Assume $X$ is a locally finite simplicial complex such that any set $A$ of finite diameter in $X$ is contractible to a point in a given $R$-neighborhood $\mathcal{N}_R(A)$ of itself ($R$ may depend on $A$ but is always finite), all in the realization of the complex (assuming a metric agreeing with the simplicial structure is used, or I guess with just talking about neighborhoods of sets as approximations thereof by simplices).

How do I show that for any cycle $c$ (in the chain complex) in $X$, $c$ is the boundary of some $b$ with $\operatorname{supp} b$ in the $R$-neighborhood of $\operatorname{supp} c$ ?

I think the steps would be to:

1. Fix a retraction from the support (?) of $c$ to a point, all lying in this neighborhood.
2. Use the simplicial approximation theorem to make it into a simplicial map…
3. Whose support will be the support of $b$, somehow.

Visually, it all seems very reasonable, but I can't very well grasp how to make it into an actual argument. For instance, I'm troubled by the fact that a homotopy works on a set, but our chains correspond to “weighted” sets somehow, thus my inclusion of the “support” wording.

Edit: I want to stress out the fact that the question is not about showing that homology is zero or that $X$ is contractible, but that any homological cycle is a boundary, with control over the support of the chain which boundary we consider.

Answer 1

Here is my "progress" to an answer so far:

Fix some cycle $c$ of finite support $C$. By the contractibility assumption, there exists a homotopy: $$ H: C \times I \to B \subseteq |X|, $$ homotoping $C$ to a point in $B$ (wlog assume $B = \mathrm{Im}(H)$). Assuming $B$ is contractible, we would get that $c$ is the boundary of some $b$ with support in $B$, and be done (but that's using the fact that simplicial homology is equal to singular homology, somewhat…)

So I think the goal would be to show that $B$ is essentially a subdivision of a cone.

The first step is to use simplicial approximation to turn $H$ into some: $$ H' : C' \times I' \to B $$ with $C'$ and $I'$ subdivisions of $C$ and $I$ respectively. Now we should check that the simplicial map $H'$ we get is still a homotopy of $C'$ to a point.

Then, what?