Why is $H^*(K(\pi, 1); A) \cong \text{Ext}^*_{\mathbb{Z}[\pi]}(\mathbb{Z}, A)$?

Question

As the question title suggests, what is the easiest way to see that there is an isomorphism$$H^*(K(\pi, 1); A) \cong \text{Ext}^*_{\mathbb{Z}[\pi]}(\mathbb{Z}, A)?$$

Answer 1

Take $CW$-decomposition for your space $K(\pi,1)$. For universal covering $B_\pi\to K(\pi,1)$ you can construct according $CW$-decomposition.

Note that $\pi\,\,$ free acts on the $B_\pi$, and the factor $B_\pi/\pi$ will be homeomorphic to $K(\pi,1)$. This action respects $CW$-complex structure, so cochain complex $C^\bullet(B_\pi,A)$ is complex of free $\mathbb Z[\pi]$-modules with zero $n$-th homology, $n>0$, and $0$-th homology $=A$. The $\pi$-invariant elements of $C^\bullet(B_\pi,A)$ is just $C^\bullet(K(\pi,1),A)$, therefore

$$H^\bullet(K(\pi,1),A)=H(\{\text{$\pi$-invariant subcomplex of } C^\bullet(B_\pi,A)\})=:\mathrm{Ext}^\bullet_{\mathbb Z[\pi]}(\mathbb Z,A).$$