# $X$ space satisfying hypotheses used to construct a universal cover, $A$ abelian, $C^*(X, A) \cong \text{Hom}_{\mathbb{Z}[\pi]}(C_*(\tilde{X}), A)$?

Let $X$ be a space that satisfies the hypotheses used to construct a universal cover $\tilde{X}$ and let $A$ be an abelian group. What is the most elementary way to see that$$C^*(X, A) \cong \text{Hom}_{\mathbb{Z}[\pi]}(C_*(\tilde{X}), A)?$$

I do it using singular cochains. Functions from $\tilde\sigma:\Delta^n\rightarrow\tilde X$ to $A$ that are invariant under $\mathbf Z[\pi]$ are precisely those functions that are invariant under deck transformations, which is how $\pi$ acts on $\tilde X$. Hence there is a natural map collapsing orbits under the deck action that puts such functions in bijection with functions sending $\sigma:\Delta^n\rightarrow X$ to $A$. Hence

$$\operatorname{Hom}_{\mathbf Z[\pi]}(C_\ast(\tilde X),A)\cong\operatorname{Hom}_{\mathbf Z}(C_\ast(X),A)=:C^\ast(X,A).$$

As soon as you have a cellular structure, it gets even easier. Pick a cellular structure on $X$, then $\bar X$ inherits a cellular structure. You get the chain complexes as $C_k(X)=\mathbb Z^{i_k}$ and $C_k(\bar X) = \mathbb Z[\pi]^{i_k}$. Considering that we have free $R$-modules for $R=\mathbb Z,\mathbb Z\pi$, the result gets pretty obvious.

You can also look at this recent answer for more details and references on this special case of twisted cohomology.