Suppose we have an acyclic group $G$ and let $X$ be a contractible CW-complex such that $G$ acts freely on $X$ (we do not suppose that the action is proper).
Is there a way to understand the homology $\mathrm{H}_{\ast}(X/G, \mathbb{Z})$ ?
We assume that the quotient space $X/G$ is Hausdorff.
For example if the action were free and proper then $$\mathrm{H}_{\ast}(X/G, \mathbb{Z})=\mathrm{H}_{\ast}(pt, \mathbb{Z})$$
Edit: Here is, I think, a more general question. Let $A$ be an abelian group. And let $G$ be a group such that $$\mathrm{H}_{\ast}(BG, A)=\mathrm{H}_{\ast}(pt, A)$$
Suppose that $X$ is a contractible CW-complex such that $G$ acts on $X$ freely. We assume that the orbit space $X/G$ is Hausdorff. What can we say about $\mathrm{H}_{\ast}(X/G, A)$ ?
Edit 2 (2019 January 5-th): May be the initial question sounds wild. I would be curious of an example where $G$ is an acyclic group acting freely on a contractible CW-complex and $$\mathrm{H}_{\ast}(X/G, \mathbb{Z})\neq\mathrm{H}_{\ast}(pt, \mathbb{Z})$$
with $X/G$ Hausdorff.
Best Answer
As per the comments, there is currently a paradox in the first paragraph, somehow depending on $BG$ (where $G$ is a topological group) versus $BG^\delta$ (where $G^\delta$ has the discrete topology). I do not yet know the flaw/resolution.
I do not think properness is needed here, only that the (free) action is continuous. Since $X$ is a free $G$-space, there is a homotopy equivalence $EG\times_GX\simeq X/G$, and since $X$ is contractible there is a weak homotopy equivalence $EG\times_GX\simeq EG\times_G\lbrace pt\rbrace$. So in the notation of equivariant singular homology (the singular homology of the Borel construction) $$H_\ast(X/G)\cong H_\ast^G(X)\cong H_\ast^G(pt)\cong H_\ast(G)\cong 0$$ where the last isomorphism is due to acyclicity of $G$.
Here is a special case: Assume $G$ also permutes the cells of some CW decomposition of $X$ (called a $G$-CW-complex). Then I can work with equivariant cellular homology, the definition in Brown's book Cohomology of Groups (chapter VII.7) which uses chain complexes. Since $G$ acts freely, it is the equivariant homology $H_\ast(X/G)\cong H_\ast^G(X)$. Since $X$ is contractible, $H_\ast^G(X)=H_\ast^G(pt)=H_\ast(G)$. Since $G$ is acylic, $H_\ast(G)=0$. This is a special case of the Cartan-Leray spectral sequence, I believe. (It might be the case here that $X\to X/G$ is a regular cover.)