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Jan 1, 2018 at 12:22 comment added Arnaud Mortier Johannes Ebert's answer to this question might help: mathoverflow.net/q/51993
Nov 11, 2017 at 18:21 comment added mme The fibers of your projection are $BH$, where $H$ is the stabilizer of the corresponding point. Because $H$ is finite this has trivial rational cohomology, which is why you should think that map is an isomorphism. If I generally had a map with acyclic fibers, I would want to do something sheafy to show it's an isomorphism in homology. Consider the presheaf $\mathcal H$ on the $X/G$ which assigns to $U$ the Q-cohomology of $f^{-1}(U)$, sheafify, take the corresponding spectral sequence from $H^*(X/G;H^*(\mathcal H))$ to $H^*(EG \times_G X)$. Now use that $\mathcal H$ is acyclic.
Nov 11, 2017 at 17:51 history asked user109300 CC BY-SA 3.0