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.

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    $\begingroup$ Hi Laura, the complex $D^\bullet$ doesn't coincide on the nose with the compactly supported cochains. E.g. taking $\Bbb R$ with its action of $\Bbb Z$, the complex $D^\bullet$ contains the 1-dimensional cochain $f$ such that: $f(\sigma) = 1$ if $\sigma(t) = a t$ for some $a$, and $f(\sigma) = 0$ otherwise. I haven't been able to determine whether it still computes compactly supported cohomology; I suspect not, but haven't been able to solve it. $\endgroup$ – Tyler Lawson Mar 10 at 14:52
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    $\begingroup$ Nothing to do with your question, but doesn't the (co)homology of the complex $D^\bullet$ rather compute the cohomology of $X/G$ with coefficients in a local system locally isomorphic to $\mathbf{Z}[G]$, instead of constant coefficients $\mathbf{Z}[G]$? $\endgroup$ – Johannes Huisman Mar 12 at 15:25
  • $\begingroup$ @TylerLawson: That's a good example -- I thought I had worked out $\mathbb{Z}$ acting on $\mathbb{R}$, but looking back over my notes I had screwed it up. Still very interested if you ever figure this out. $\endgroup$ – Laura Mar 13 at 19:29
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    $\begingroup$ @JohannesHuisman: That's what I meant, and I hope it was clear from the context. It would be perverse to take a trivial local system that also happened to be a module over the fundamental group... $\endgroup$ – Laura Mar 13 at 19:30
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    $\begingroup$ @Laura This previous question appears to indicate that this might be VIII.7.5 in Ken Brown's "Cohomology of groups"? mathoverflow.net/questions/102654/… I don't currently have access to the book to check. $\endgroup$ – Tyler Lawson Mar 13 at 20:24

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