TITLE: Equivariant Cohomology for actions with finite stabilizers
QUESTION [5 upvotes]: Let $X$ be a reasonable topological space (let's say it has the homotopy type of a CW complex) and let $G$ be a topological group acting on that space. Let $E_G \rightarrow B_G$ be the universal bundle. Then the equivariant cohomology $H^*_G(X)$ is defined as $H^*(E_G \times_G X)$. We call the space $E_G \times_G X$ the homotopy quotient space. We always have a map $E_G \times_G X \rightarrow G \backslash X$, where $G \backslash X$ denotes the naive quotient space. This induces a map $H^* (G\backslash X) \rightarrow H^*_G(X)$. 
I'm curious about the following proof. From now on, we will take cohomology with rational coefficients. Suppose the action of $G$ on $X$ has finite stabilizers. Then the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is in fact an isomorphism.  This is proved in the following notes by Michel Brion (http://www-fourier.ujf-grenoble.fr/~mbrion/notesmontreal.pdf, see p. 4). The proof is as follows: the fibers of the map $E_G \times_G X \rightarrow G \backslash X$ are of the form $E_G/G_x$, where $G_x$ is the stabilizer of a point $x \in X$. Since $G_x$ is a discrete group, we have $\pi_1 (E_G/G_x) = G_x $ and vanishing of all higher homotopy groups. Since $G_x$ is finite, all the homotopy groups vanish when we tensor with $\mathbb{Q}$. I think this is what Brion means by "$\mathbb{Q}$-acyclic".
Then it is claimed that this is sufficient to prove that the map $H^* (G\backslash X) \rightarrow H^*_G(X)$ is an isomorphism. Why is this true? Could someone please explain this argument to me?
Even if the fibers were contractible, it seems to me that you would need this map to be a Serre fibration to prove a homotopy equivalence. Is it true that any surjective map with contractible fibers is a homotopy equivalence?

REPLY [5 votes]: For ease of referencing I'll prove the stated isomorphism under the hypothesis that 
$G$ is a compact Lie group and $X$ is a finite dimensional compact topological $G$-manifold. 
By the hypothesis on $G$ we can choose $EG$ to be a dimension-wise finite CW complex. The 
$(n+1)$-skeleton $E$ is compact, $n$-connected and for fixed $m$ and $n$ large enough, the inclusion $E \hookrightarrow EG$ induces an isomorphism 
$$H^m(EG\times_G X,\mathbb{Q}) \xrightarrow{\cong} H^m(E\times_G X,\mathbb{Q})$$
(that's 4) on p. 4 of the linked paper in the question). Hence it suffices to show that the map 
$E\times_G X \to X/G$ induces an isomorphism in rational cohomology in degree $m$. 
But this follows from the Vietoris-Begle theorem [Q, Corollary A.7] that applies if we have shown that $f$ is closed and the fibres of $f$ are compact, n-acyclic and relative Hausdorff in $E\times_G X$: 
Closedness of $f$ and compactness of the fibres $E/G_x$ are obvious, $n$-acyclity follows 
from $H^i(E/G_x,\mathbb{Q})=H^i(G_x,\mathbb{Q})=0$ for $0< i< n$.  Relative Hausdorff 
means different points in the fibre have disjoint neighborhoods in $E \times_G X$. This 
holds since $E \times_G X$ is Hausorff. QED
[Q] Quillen: The Spectrum of an Equivariant Cohomology ring: I 

The cited Vietoris-Begle theorem states: 

If $f: X \to Y$ is closed and its fibres are compact, relatively Hausdorff in $X$ and $n$-acyclic, i.e. $H^i(f^{-1}(y),k)=H^i(\ast,k)$ for $i< n$ ($k$ any constant coefficients), then $f^\ast:H^i(Y,k) \xrightarrow{\cong} H^i(X,k)$ for $i \lt n$.