Connected CW complex, isomorphism? Let $\pi$ be a group and let $K(\pi, 1)$ be a connected CW complex such that $\pi_1(K(\pi, 1)) = \pi$ and $\pi_q(K(\pi, 1)) = 0$ for $q \neq 1$. My question is, are $H_*(K(\pi, 1);A)$  and $\text{Tor}_*^{\mathbb{Z}[\pi]}(A, \mathbb{Z})$ necessarily isomorphic?
 A: I've a small worry that it might depend on what category your $A$ lives in.  If $A$ is just an abelian group, then:
From the User's Guide

Theorem (Cartan-Leray) for a free and proper action of $\pi$ on $X$ there is a spectral sequence converging strongly $E \Rightarrow H_*(X/\pi,A)$ with $E^2_{p,q} \simeq H_p(\pi,H_q(X,A)) $

In particular, $H_p(\pi,-)$ is your $\operatorname{Tor}_p^{\mathbb{Z}[\pi]}(-,\mathbb{Z})$.  In the Particular Case of Interest, $X\simeq\hat{K}(\pi,1)$ is contractible, so that $H_*(X,A)$ would seem to be just $A$, in degree zero.  Consequently, the spectral sequence collapses on the second page; there are no extension problems to solve.

I readily admit there is something unsatisfying/disturbing in this argument: the tool is heavier and more blunt than the result we really want. It might be more fun/enlightening to construct $K(\pi,1)$ from $\pi$ as a $\triangle$-complex and then write down its cellular complex of $A$-valued chains, vs. writing a resolution of $\mathbb{Z}$ as a trivial $\pi$-module and tensoring with $A$.  Work on paper, be careful, and consult any mathematical friends in your neighborhood.
