I am confused about this "homology with local coefficients" business. First, I am confused by its name. What is "local" about it?
Second:
Let $X$ be a topological space with a universal cover. Let $G=\pi_1(X)$. Let $M$ be a $G$-module, that is, a left $\mathbb{Z}[G]$-module: an abelian group with a left action of $G$. Then we can do two things:
1) Take the homology with local coefficients in $M$, which I take it as defined in Hatcher, through the action of $G$ by deck transformations on the universal cover of $X$,
2) Do the following procedure (which occurred to me and I don't know if it is relevant). If $R$ is a ring, define $C_n^R(X)$ as the free $R$-module with basis the singular $n$-chains, and a boundary map defined as for the usual $C_n(X)$. Then you can take homology, which gives you "$R$-modules of homology". Just as you do when you define homology with coefficients in an abelian group, you can define homology with coefficients in an $R$-module. Now, take $R=\mathbb{Z}[G]$. Consider this "homology with coefficients in $M$".
Do these two procedures yield isomorphic modules?
Best Answer
To answer your first question, one way to define (co)homology with local coefficients is the following.
Let $X$ be a space, let $\Pi(X)$ be the fundamental groupoid of $X$, ie. a category with objects points of $X$ and morphisms $x \rightarrow y$ given by homotopy classes of paths. A local coefficient system $M$ on $X$ is a functor $M: \Pi(X) \rightarrow \mathcal{A}b$ from the fundamental groupoid to the category of abelian groups. In particular, $M$ associates a "group of coefficients" $M(x)$ to every point of $x \in X$.
Associated to $M$ is the singular complex given by
$C _{n}(X, M)= \bigoplus _{\sigma \in Sing_{n}(X)} M(\sigma(1,0,\ldots,0))$,
where $Sing _{n}(X)$ is the set of all maps $\Delta ^{n} \rightarrow X$. The differential can be defined using the fact that the groups $M(x)$ are functorial with respect to paths in $X$. Observe that this is a very similar to the usual definition of singular complex with coefficients in an abelian group $A$, which would be
$C _{n}(X, A) = \bigoplus _{\sigma \in Sing_{n}(X)} A$,
except in the "non-local" case, we count the occurances of any $\sigma: \Delta^{n} \rightarrow X$ in a given chain using the same group $A$ and in the local case, we use $M(\Delta^{n}(1, 0, \ldots, ))$, which might be different for different $\sigma$.
This is the locality (localness?) in the name, which should be contrasted with globality of usual homology with coefficients, where the choice of the group $A$ is global and the same for all points.
The definition I have given above is enlightening but perhaps not suitable for computations. Luckily under rather weak assumptions one can use the definition you allude to. Let me explain. Let $X$ be path-connected, let $x \in X$ and consider $\pi = \pi_{1}(X, x)$ as a category with one object and morphisms the elements of the group.
The obvious inclusion $\pi \hookrightarrow \Pi(X)$ is an equivalence of categories and so the functor categories $[\pi, \mathcal{A}b], [\Pi(X), \mathcal{A}b]$ are equivalent, too. But the left functor category is exactly the category of $Z[\pi]$-modules! In particular, we have a bijection between isomorphism classes of local coefficient systems on $X$ and $Z[\pi]$-modules.
If $X$ is nice enough to admit a universal cover $\tilde{X}$, the above allows us to give another definiton of homology with local coefficients, the one you know. Let $M$ be a local coefficient system and let $M^\prime$ be the associated $\mathbb{Z}[\pi]$-module under the above equivalence (which is unique up to a unique isomorphism). Since $\pi$ acts on $\tilde{X}$, it also acts on $C _{\bullet}(X, \mathbb{Z})$ and so the latter is a chain complex of $\mathbb{Z}[\pi]$-modules. We the can define homology with local coefficients to be homology of the complex
$C_{n}(X, M ^\prime) = C_{n}(\tilde{X}, \mathbb{Z}) \otimes _{\mathbb{Z}[\pi]} M^\prime$
One can show that this two definitions I gave agree, that is, for $X$ like above we have an isomorphism $H_{n}(X, M) \simeq H_{n}(X, M^\prime)$.