TITLE: Can one do without a classifying space when showing vanishing of cohomology
QUESTION [9 upvotes]: 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)

REPLY [3 votes]: 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 $H^n(G, A)$.