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Let $G$ be a finite group. Let $G$ act freely on a CW-complex $X$. I heard that the following fact is true.

Claim. The canonical map $H^*(X/G,F)\to H^*(X,F)^G$ is an isomorphism, where $F$ is a field so that $|G|$ is invertible.

I am looking for a reference to this fact.

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    $\begingroup$ Hatcher, Proposition 3G.1. The proof is short and the essential content is the construction of a transfer map going the "wrong way", so that one of the composites is obviously multiplication by |G|. $\endgroup$
    – mme
    Commented Jul 11, 2021 at 12:31
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    $\begingroup$ In the characteristic zero case, see Proposition III.2.4 of Bredon's "Introduction to compact transformation groups" $\endgroup$
    – John Klein
    Commented Jul 11, 2021 at 16:41

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This is true even if the group does not act freely. See Proposition 1.1 of my notes here. I deal with simplicial complexes and work over the rationals, but the statement you give can be proved the same way.

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    $\begingroup$ Couldn't you simplify the argument by ignoring representation theory and just observing that if R is a common ring with $|G|$ invertible $e=1/|G|\sum_{g\in G}g$ is an idempotent with $eV=V^G$ for any RG-module V. Since multiplication by an idempotent is obviously an exact functor taking invariants commutes with homology. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 11, 2021 at 15:00
  • $\begingroup$ @BenjaminSteinberg Sorry, I don't understand, what do you mean? Certainly in general taking invariants is not an exact functor - if it would be, there would be no group cohomology. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 11, 2021 at 15:13
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    $\begingroup$ @მამუკაჯიბლაძე: That idempotent does not exist if the order of G is not invertible. Multiplying by an idempotent is definitely exact. $\endgroup$
    – Andy Putman
    Commented Jul 11, 2021 at 15:16
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    $\begingroup$ @BenjaminSteinberg: I think that is basically the same argument, but expressed in different language. You are just projecting onto the isotypic component of the trivial representation. I find it easier to understand when put in the context of representation theory, but someone like you who does a lot of ring theory might have other preferences. $\endgroup$
    – Andy Putman
    Commented Jul 11, 2021 at 15:19
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    $\begingroup$ A group ring RG cannot be semisimple unless R is semisimple because of the augmentation homomorphism. It is just the trivial module that becomes projective when |G| inverts. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 11, 2021 at 15:49

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