Consider a group $G$ having a finite, free resolution $C_*(G)\to \mathbb Z$ over the group ring of $G$.
I want to understand why this resolution may be viewed as the equivariant chain complex of the universal cover $\widetilde X$ of the Eilenberg-MacLane space $X=K(G, 1)$.
I found this problem when I was computing the homology group of the circle with local coefficients, I read that the chain complex of the universal cover of $X=S^1$ which is $\widetilde X=\mathbb R$ is the same as a free resolution of $\mathbb Z$. So I wanted to understand this relation between the free resolution of the group and the equivariant chain complex of the universal cover of its corresponding $K(G,1)$. Thank you for your help!
It is actually other way around: every classifying space gives rise to a free resolution. Why is so? Take your favourite CW structure on this space. Then chain complex of universal cover under monodromy action becomes a resolution of a trivial module. Modules will be free because cellular action is free, and homology will be $\Bbb Z$ in degree $0$ as for any contractible space.
And it is actually only in this way. Not every free resolution of a trivial module can be obtained as a chain complex of contractible free $G$-space. For example, there are groups which have free resolution of finite rank in each dimension, but are not finitely presented; not-finitely presented group cannot have a classifying space with finite 2-skeleton, so this resolution does not come from topology.