TITLE: Cohomology of quotient by free action
QUESTION [6 upvotes]: 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.

REPLY [15 votes]: 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.