I'm looking for examples of the following phenomena. Let $X$ be a reasonable space (say, a CW complex) and $G$ be a finite group acting on $X$. For all $k \geq 1$, the projection map $X \rightarrow X/G$ induces a map $H_k(X;\mathbb{Q}) \rightarrow H_k(X/G;\mathbb{Q})$ which factors through the $G$-coinvariants $(H_k(X;\mathbb{Q}))_G$; let $\psi_k : (H_k(X;\mathbb{Q}))_G \rightarrow H_k(X/G;\mathbb{Q})$ be the resulting map. I want examples of $X$ and $G$ and $k$ such that $\psi_k$ is not an isomorphism.

If $G$ acts freely, then the map $X \rightarrow X/G$ is a finite regular covering map and $\psi_k$ is an isomorphism by (for instance) the Cartan-Leray spectral sequence (Theorem VII.7.9 in Brown's book on group cohomology). But I have no idea what happens for non-free actions. My guess is that if it were true that $\psi_k$ were always an isomorphism, then I would have seen it somewhere, so I expect that there is a counterexample. However, I have not managed to come up with one.


1 Answer 1


The maps $\psi_k$ are all isomorphisms; this is a simple application of the transfer ("averaging") construction. See Theorem 2.4, Chapter III, of Bredon's book "Introduction to compact transformation groups".

  • 1
    $\begingroup$ Great! I was of course familiar with the transfer for free actions, but somehow I never realized that it could be generalized to non-free actions. $\endgroup$
    – Lisa
    Jun 19, 2014 at 22:23
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    $\begingroup$ If I'm not mistaken, you meant Chapter III, not II. $\endgroup$ Dec 18, 2019 at 13:14
  • $\begingroup$ @BrunoStonek: You are right, the chapter number is now corrected. $\endgroup$
    – Misha
    Dec 19, 2019 at 21:13

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