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There are many references which will show that for (say) a finite discrete group $G$, you can construct the classifying space $BG$ which $G$ as its fundamental group and the same cohomology groups, that is $H^n(G; A) \cong H^N(BG; A)$ where the first is the group cohomology for a $G$-module $A$ where $G$ is acting on $A$ with the trivial action, and the second is singular or other such cohomology for the topological space $BG$.

However, in the case that $M$ is a non-trivial $G$-module, that is, $G$ has some action other than the trivial one on $M$, it appears that it's necessary to consider cohomology with local coefficients on $BG$ in order to account for the action of $G$. It appears to be "common knowledge" that in this case as well, $H^n(G; M) \cong H^n(BG; M)$ where the former is cohomology with so-called "twisted" coefficients (taking into account the action of $G$ on $M$) and the latter is local coefficients on the space. It makes sense that this would be so, but I've been unable to find anywhere an explicit proof of this. Wikipedia, for example, makes this claim and cites chapter II of Cohomology of Finite Groups by Adem and Milgram, but as far as I can tell that book never even mentions local coefficients, only briefly describing the classifying space in regard to group cohomology with the trivial action and then proceeding to describe any other group cohomology in terms of free resolutions.

Is there any reference in English, such as a book, which develops and specifically points out the nature of this particular isomorphism? Or if anyone can give a simple description of it here, that would also be sufficient.

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    $\begingroup$ The same claim appears without proof on pg 59 of Brown's book Cohomology of Groups. Brown does, however, give a hint to prove the statement yourself: If $\mathcal{M}$ is the local coefficent system corresponding to the $G$-module $M$, then on the chain level we have $C^*(BG;\mathcal{M})=Hom_G(F_*,M)$, where $F_*$ is the chain complex of the universal cover of $BG$. Once you write this down, it seems fairly clear how things go through. $\endgroup$
    – Tyrone
    Commented May 25, 2020 at 17:01
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    $\begingroup$ It depends on your definition of cohomology with local coefficients and almost have nothing to do with BG. Let X be a space and M be a Z[\pi_1(X)] module, then we may define cohomology in M as complex of equivariant cochains on universal covering \tilde X with values in M. Then all you need to know about EG is that on cochain level it is some Z[G]-free resolution of Z and this is enough to naturally compute group-cohomology. So all definitions (group-cohomology, BG-cohomology, cohomology with locall coefficients) now means the same thing $\endgroup$
    – Bad English
    Commented Jun 2, 2020 at 21:44

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