Let $X$ be a contractible space and let $G$ be a group acting freely, properly discontinuously, and cocompactly on $X$. We get an induced action on the singular chain complex $C_{\bullet}(X)$.

Define $D^{\bullet}$ to be the subcomplex of the singular cochain complex $C^{\bullet}(X) = \text{Hom}(C_{\bullet}(X),\mathbb{Z})$ whose $n$th term consists of all $f\colon C_n(X) \rightarrow \mathbb{Z}$ such that for all $\sigma \in C_n(X)$, the set $$\{\text{$g \in G$ $|$ $f(g \cdot \sigma) \neq 0$}\} \subset G$$ is finite.

**Question**: Is $D^{\bullet}$ the same as the singular cochain complex with compact support? Or at least is the cohomology of $D^{\bullet}$ the same as the cohomology of $X$ with compact support?

It is clear that every singular cochain with compact support lies in $D^{\bullet}$ (this just uses the fact that $G$ acts properly discontinuously).

Here's the reason I am interested in this question. We can identify $D^{\bullet}$ with the complex $\text{Hom}_G(C_{\bullet}(X),\mathbb{Z}[G])$, namely $f \in D^{n}$ is identified with the $G$-invariant map from $C_n(X)$ to $\mathbb{Z}[G]$ taking $\sigma \in C_n(X)$ to $\sum_{g \in G} f(g^{-1} \sigma) g$. The condition in the math display above is precisely what we need for this to be a finite sum for all $\sigma$. So the cohomology of $D^{\bullet}$ is the same as the cohomology of $X/G$ (a $K(G,1)$) with coefficients in $\mathbb{Z}[G]$.

**Note**: This would be a much easier question is I were using simplicial complexes and simplicial homology; however, for technical reasons I really want to use singular homology.