It is well-known that the homology group and cohomology group of the Eilenberg-Maclane space $K(\pi,1)$ are:
$$H_n(K(\pi,1))=\mathrm{Tor}^{\mathbf{Z}\pi}_n(\mathbf{Z},\mathbf{Z}),\quad H^n(K(\pi,1))=\mathrm{Ext}_{\mathbf{Z}\pi}^n(\mathbf{Z},\mathbf{Z})$$
I am trying to prove this and I want to avoid using the spectral sequence since this is an exercise on Davis-Kirk's book before the spectral sequence part.
There is an answer, using some theorem of derived functors. I like that proof. But I was wondering if there is any topological proof? Since we have the fibration (especially, universal covering), we can consider the local coefficient system.
On the other hand, when we use a $\mathbf{Z}\pi$-resolution to compute $\mathrm{Tor}$, we compute the homology of a complex tensored with $\mathbf{Z}$, which would hopefully coincide with the local coefficient system. Does this work? How can I continue?
There is a similar question on MSE. Feel free to close this question if you would like to answer there.
Also, I would like to know other approaches to proof the above result. Any hints, references or answers are welcome!
You're trying to prove this for $\mathbf{Z}$ : the local coefficient system is constant.
But then the proof is easier, for instance by considering a cell-structure on $K(\pi,1)$: take $E$ a universal covering space of your $K(\pi,1)$, and give it a cell-structure induced by the one on $K(\pi,1)$; that makes it into a free $\pi$-complex.
Then $C_*(E)$ (the cellular complex) is a chain complex of free $\pi$-modules; and if you look at what the cells in $E$ look like, it's pretty clear that $\hom_\pi(C_n(E), \mathbf{Z}) \simeq \hom(C_n(K(\pi,1)), \mathbf{Z})$ and similarly $C_n(E)\otimes_\pi \mathbf{Z} \simeq C_n(K(\pi,1))$, with maps induced by the covering map (so they must have the right differential).
You thus immediately get the result. If you have a nontrivial coefficient module, it's a bit trickier.