Can one do without a classifying space when showing vanishing of cohomology Let $G$ be a discrete group and $A$ an abelian group, then $H^n (G,A)$ can be defined as $$ H^n (G,A) = H^n (B_G, A)$$
Where $B_G$ is the classifying space of $G$, i.e. $B_G = E_G / G$ where $E_G$ is a contractible space on which $G$ acts s.t. $\pi_1 (B_G) = G$.
Now say you have an action of $G$ on a simplicial complex $Y$ and that the action is simplicial, cocompact and free, but $Y$ is not necessarily contractible. Say you know that (for some fixed $n$) $H^n (Y / G,A) = 0$ can you deduce that $H^n (G,A) = 0$?
(I've tried to ask this question a couple of days ago, but my formulation was bad so I had to delete and re-ask) 
 A: There are certain cases where the answer is yes.
If $Y$ is a free $G$-complex, then the Cartan-Leray spectral sequence of the regular cover $Y\to Y/G$ is of the form
$$ H^p(BG;H^q(Y;A))\Rightarrow H^{p+q}(Y/G;A).$$
For instance, in the extreme case that $Y$ has the $A$-cohomology of a point, one has $H^\ast(BG;A)\cong H^\ast(Y/G;A)$. 
There should be other cases where it is possible to draw conclusions about vanishing of group cohomlogy by working backwards through this spectral sequence.
Reference: Ken Brown, Cohomology of groups, Section VII.7.
Added: In particular, if $Y$ is a $\mathbb{Z}$-acyclic complex with a free $G$-action, then the cohomological dimension of $G$ is less than or equal to $\mathrm{dim}(Y)$. See the MR of 
Howie, James, Bestvina-Brady groups and the plus construction. 
Math. Proc. Cambridge Philos. Soc. 127 (1999), no. 3, 487–493
for an example application.
A: You get a useful result if $Y$ is not contractible, but highly connected. If $\tilde{H}_i(Y)=0$ for $i \leq m$, then the (up to homotopy) natural map $Y/G \to BG$ is a homology equivalence in degrees less than $n$.
A: To highlight the assumption of high connectivity in Johannes' answer, I should point out that for any  $G$, you can always take $Y = G$ (a simplicial complex with only vertices, seeing as $G$ is discrete).  Then you have a simplicial, cocompact, free action, and it is true that $H^n(Y/G, A) = 0$ for $n>0$. But that tells us nothing about $H^n(G, A)$.
A: No.  Let $G = \mathbb{Z}/p$ and $Y = S^1$.  The group $G$ acts on $Y$ in the usual way (the generator acts as rotation by $2\pi/p$), and we have $Y/G \cong S^1$.  Thus all the homology groups of $Y/G$ above degree $1$ vanish, but the homology groups of $G$ are nonzero in infinitely many degrees.
