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.

3more comments