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The mapping class group $\Gamma_n$ of a Riemann surface with genus $g$ acts properly over the Teichhmuller space $\mathcal T_n$, and this action has finite stabilizers. It is said that this implies the existence of an isomorphism in rational cohomology rings $H^*(\mathcal T_n/\Gamma_n;\mathbb Q)\cong H^*(B\Gamma_n;\mathbb Q)$.

By doing a bit of research in the web, I think that the following more general statement is true:

Let $G$ be a discrete group acting properly to the left on a contractible space $X$, such that all stabilizers are finite. Then the projection map $EG\times_G X \to X/G$ induces isomorphisms in rational cohomology.

If this is the case, the contractibility of $X$ implies that the projection $EG\times_G X\to BG$ is a weak equivalence and thus we have an isomorphism $H^*(B\mathcal G;\mathbb Q)\cong H^*(X/G;\mathbb Q)$. This is often used as a well-known fact, but I am not sure how to approach it. Any reference or idea would be appreciated.

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    $\begingroup$ 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. $\endgroup$
    – mme
    Commented Nov 11, 2017 at 18:21
  • $\begingroup$ Johannes Ebert's answer to this question might help: mathoverflow.net/q/51993 $\endgroup$
    – Arnaud Mortier
    Commented Jan 1, 2018 at 12:22

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