Isomorphism involving Eilenberg-Maclane space, Tors. 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 $q \neq 1$. Does there exist an isomorphism between $H_*(K(\pi, 1); A)$ and $\text{Tor}_*^{\mathbb{Z}[\pi]}(A, \mathbb{Z})$?
 A: As Justin Young commented, there are many different ways of proving this (and which one is most comprehensible to you will depend on your background); here's one.  Write $T_n(A)=H_n(K(\pi,1),A)$; then $T_n$ is a functor from the category of $\mathbb{Z}[\pi]$-modules to the category of abelian groups.  Moreover, these functors together canonically have the structure of a (homological) $\delta$-functor, meaning that given any short exact sequence $0\to A\to B\to C\to 0$ of $\mathbb{Z}[\pi]$-modules, there are natural maps $\delta_n:T_n(C)\to T_{n-1}(A)$ which give a long exact sequence $$\dots\to T_n(A)\to T_n(B)\to T_n(C)\to T_{n-1}(A)\to\dots\to T_1(C)\to T_0(A)\to T_0(B)\to T_0(C)\to 0.$$
By a theorem of Grothendieck, such a $\delta$-functor is naturally isomorphic to the derived functors of the functor $T_0$ iff for each $n>0$, the functor $T_n$ is coeffaceable, meaning that for any $\mathbb{Z}[\pi]$-module $A$, there exists a $\mathbb{Z}[\pi]$-module $P$ and an epimorphism $P\to A$ such that the induced map $T_n(P)\to T_n(A)$ is $0$.
I now claim that the functors $T_n$ are indeed coeffaceable.  Indeed, for any $\mathbb{Z}[\pi]$-module $A$, let $P$ be a free $\mathbb{Z}[\pi]$-module with an epimorphism $P\to A$.  Then $T_n(P)=0$ for all $n>0$: we can compute $T_n(P)=H_n(K(\pi,1),P)$ as the homology of the chain complex $C_*(E)\otimes_{\mathbb{Z}[\pi]} P$, where $E$ is the universal cover of $K(\pi,1)$.  Since $P$ is free over $\mathbb{Z}[\pi]$, this chain complex is just a direct sum of copies of $C_*(E)$.  But by definition of $K(\pi,1)$, the space $E$ is contractible, so the homology of $C_*(E)$ vanishes in degree $>0$.  Thus $T_n(P)=0$ for $n>0$, and it follows that $T_n$ is coeffaceable.
Finally, we note that the functor $T_0(A)=H_0(K(\pi,1),A)$ is naturally isomorphic the functor $A\otimes_{\mathbb{Z}[\pi]}\mathbb{Z}$; this is easy to prove by a direct computation (this uses only the fact that $K(\pi,1)$ is a path-connected space with fundamental group $\pi$).  Since we have shown that the functors $T_n$ are naturally isomorphic to the derived functors of $T_0$, we conclude that they are also naturally isomorphic to the derived functors of $A\mapsto A\otimes_{\mathbb{Z}[\pi]}\mathbb{Z}$, i.e. the functors $\operatorname{Tor}_*^{\mathbb{Z}[\pi]}(A,\mathbb{Z})$.
A: Using the Davis-Kirk framework:
The "algebraic definition" of $H_*(X;A)$ is the homology of $S_*(\widetilde X)\otimes_{\mathbb Z [\pi]} A$ where $\pi = \pi_1(X)$. They outline a proof (using lifts of maps $\Delta^n \to X$ to $\widetilde X$) that, as a $\mathbb Z[\pi]$ module, $S_*(\widetilde X) \cong \mathbb Z[\pi]\left [ S_*(X)\right ]$, where $S_*(X)$ here refers only to the set of basis elements given by all singular simplicies $\Delta^n \to X$. Thus, if $X = K(\pi,1)$, then $\widetilde{X}$ is contractible, and by the above $S_*(\widetilde X)$ is a $\mathbb Z[\pi]$-free resolution of $\mathbb Z$. It follows immediately that, using this definition, $H_*(X;A) = \text{Tor}^{\mathbb Z[\pi]}_* (\mathbb Z, A)$ . 
