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Let $G$ be a (discrete) group, $X$ a topological space that works as a classifying space for $G$, and $\mathcal{L}$ a local system on $X$ with stalk $L$. It is a fairly standard result that

$$ H^i(G, L) \cong H^i(X,\mathcal{L})$$

where the left hand side is the group cohomology and the right hand side is sheaf cohomology. This follows from the equivalence of categories of local systems on (nice enough) $X$ and representations of the fundamental group $G$, but I can't seem to find a paper or a book where this isomorphism of cohomologies is explicitly proven/stated, and I would like to be able to provide a reference. Any suggestions?

On a similar note, is there any sense in which this isomorphism holds for $X$ an orbifold with orbifold fundamental group $G$?

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    $\begingroup$ What definition are you using for $H^i(G,L)$? $\endgroup$
    – Ryan Budney
    Commented Jan 17 at 20:54
  • $\begingroup$ The shortest definition would be the derived functors of the functor taking a representation to its invariants, though it can also be defined in terms of a cochain complex $\endgroup$
    – Aidan
    Commented Jan 18 at 10:51

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For CW-complexes, this is roughly spelled out in Ken Brown's bible: Proposition II.4.1 (which is the combination of Proposition I.4.2 and Proposition II.2.4), assuming the algebraic-definition of group cohomology in Section II.3 (effectively we have explicit maps between their explicit $\mathbb{Z}G$-free resolutions of $\mathbb{Z}$ which you can then twist with local coefficients as clarified in Chapter III.1). For general spaces, this is Exercise II.4.2, which he gives hints and furthermore gives the foundational reference that writes it out (Eilenberg--Maclane's 1945 "Relations between homology and homotopy groups of spaces").

More generally, for any path-connected topological space X we can construct a discrete group G and K(G,1) inducing (co)homology isomorphisms (reference Kan–Thurston theorem) and then apply the above.

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    $\begingroup$ What exactly is the relevance of the Kan–Thurston theorem here? The OP explicitly states that X is a classifying space for a discrete group G, which by definition makes it weakly equivalent (and not merely homology equivalent) to K(G,1). $\endgroup$
    – Dmitri Pavlov
    Commented Jan 18 at 0:55
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    $\begingroup$ What does the Kan-Thurston theorem have to do with this? We already know that $X$ is a $K(G,1)$. $\endgroup$
    – Andy Putman
    Commented Jan 18 at 0:55
  • $\begingroup$ You're right, updated response. $\endgroup$
    – Chris Gerig
    Commented Jan 18 at 2:15

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