Group cohomology as cohomology of a K(G,1) CW-complex I am trying to understand the topologycal interpretation of group chomomology.
I am familiar with the algebraic definition of cohomology for a group $G$, i.e. $H^i(G; A) := \text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$.
I have read on the internet that this should be tha same as the singular or cellular cohomology of a $CW-$complex $K(G,1)$, but this is not clear to me. Wherever I found this assertion, they would do the following construction:
Let $Y$ be a cellular $K(G,1)$ and let $\widetilde{Y}$  be its universal covering space. I can lift $Y$'s cell structure to $\widetilde{Y}$, then $G$, being the fundamental group of $Y$, will permute freely the cells of $\widetilde{Y}$ and the covering map will be exactly the projection on the orbit space. So the augmented cellular chain complex of $\widetilde{Y}$ is a free $G-$complex, which is exact because $\widetilde{Y}$ is contractible. Thus, this complex is a free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$, which I can use to compute $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$.
My problem is that, from this construction, it seems to me that I would be getting the cohomology of $G$ not as the cellular cohomology of the $K(G,1)$ $Y$, but as the cellular cohomology of its universal cover $\widetilde{Y}$.
Where am I mistaken? Is the construction for getting a free $G$-resolution of not the one I encountered? Do you have any references where they prove explictly why the cohomology of $G$ is exactly the same as the cellular cohomology of a $CW-$complex $K(G,1)$, or could you explain to me how this works?
 A: You're skipping the step where you go from the free resolution of $\mathbb{Z}$ over $\mathbb{Z}G$ to actually computing the Ext.  The free resolution is the cellular chain complex of $\widetilde{Y}$.  However, to compute $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$, you have to Hom this resolution into $A$ before taking cohomology.  If $F$ is a free $\mathbb{Z}G$-module with basis $B$, then $\operatorname{Hom}_{\mathbb{Z}G}(F,A)$ is just $A^B$.  Now, if $C_n(\widetilde{Y})$ denotes the $n$th cellular chain group of $\widetilde{Y}$, the $n$-cells of $\widetilde{Y}$ form a basis for $C_n(\widetilde{Y})$ over $\mathbb{Z}$.  But over $\mathbb{Z}G$, a basis would instead be a set of representatives of the $G$-orbits of the $n$-cells of $\widetilde{Y}$.  The $G$-orbits of the $n$-cells of $\widetilde{Y}$ just correspond to the $n$-cells of $Y$, so you can identify a basis for $C_n(\widetilde{Y})$ with the set of $n$-cells of $Y$.  This means that when you Hom into $A$ over $\mathbb{Z}G$, you just get the group of functions from the set of $n$-cells of $Y$ to $A$, which is exactly the $n$th cellular cochain group of $Y$ with coefficients in $A$.  So, the chain complex that computes $\text{Ext}^i_{\mathbb{Z}G}(\mathbb{Z}, A)$ has the same groups in it as the cellular cochain complex of $Y$ with coefficients in $A$.  (Of course, you have to do some additional work to check that the differentials in these chain complexes are also the same.)
