Let $W$ be a finite group acting on a space $X$. In what generality is it true that $H^*_W(X) = H^* (X)^W$?  We always have a map $H^*_W(X) \rightarrow H^* (X)^W$, but it is certainly not an isomorphism with $\mathbb{Z}$-coefficients (take $X = E_W$, the total space of a universal bundle). But is it true with $\mathbb{Q}$-coefficients? Perhaps if we invert the order of the group?



This is a follow up to my [prior question](https://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers) about equivariant cohomology. Again, I am referring to the notes <http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf> on equivariant cohomology by Michel Brion.

I am interested in understanding the proof of Proposition 1 on pages 6 and 7. Let $G$ be a compact Lie group, let $T$ be a maximal torus, let $N$ be the normalizer of $T$ in $G$, and let $W = N/T$ denote the Weyl group. In part (i), we have the $W$-bundle $G/T \rightarrow G/N$, from which the author claims $H^* (G/N) = H^* (G/T)^W$ when using $\mathbb{Q}$-coefficients.

A few sentences later, a similar statement is made for a more arbitrary $W$ bundle. So it seems like the above statement about $W$-invariants of cohomology is true in some generality. Could someone explain why this is true or give a reference? Or perhaps I am mistaken in interpreting this argument.