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?
Best Answer
The argument in the quoted paper is a bit too sketchy. An actual proof will have two parts:
Let $f:X \to Y$ be map. Assume that for each $y \in Y$, $\tilde{H}^{\ast} (f^{-1}(y);\mathbb{Q})=0$ (this is what the phrase ''$\mathbb{Q}$-acyclic'' means). Find conditions that guarantee that $f$ induces an isomorphism in cohomology. Essentially, one needs to provide assumptions that the Leray spectral sequence of the map is as nice as claimed in Marks answer.
Provide conditions for group action of a topological group $G$ on a space $X$ such that $EG \times_G X \to X/G$ has the properties found in 1 (and the answer in 1 depends on the example you want to consider).
ad 1. Just having finite stabilizers is not enough. Consider the action of $G=\mathbb{Z}$ on $X=S^1$ by an irrational rotation. Then $EG \times_G S^1 \simeq S^1 \times S^1$. The quotient space is an uncountable set with the trivial topology. Because each map into this space is continuous, it is contractible and has trivial (Cech or singular) cohomology.
One needs criteria that show that a map $f:X \to Y$ with contractible/acyclic/$\mathbb{Q}$-acyclic fibres is itself a weak equivalence/acyclic/$\mathbb{Q}$-acyclic. Having contractible fibres does not suffice: look at the identity map $[0,1]^{\delta} \to [0,1]$, where $\delta$ means ''discrete topology''.
Such problems are solved by what I would call ''fibre theorems''. You wish to know how close the natural map $f^{-1}(y) \to hofib_f (y)$ from the point-preimages to the homotopy fibres is to be a weak homotopy equivalence. What one needs is a suitable local condition and then a local-to-global result. Many technical results are of this type: ''fibre bundles are Serre-fibrations'', ''local Hurewicz fibrations are Hurewicz fibrations'', Dold-Thom's result on quasifibrations, the Vietoris-begle mapping theorem, Quillens Theorems A and B and McDuff-Segals formulation of the ''group-completion'' theorem. A variation of the latter solves your problem:
Assume, as above, that for each $y \in Y$, $f^{-1}(y)$ is $\mathbb{Q}$-acyclic. Assume moreover that $Y$ is Hausdorff and for each $y \in U \subset Y$, there is $y \in V \subset U$ ($U$ and $V$ open) such that $V$ is contractible and $ f^{-1} (V)$ is $\mathbb{Q}$-acyclic (let us call these sets ''good''). Then, I claim, $f$ induces an isomorphism in rational cohomology.
Proof: Let $P:=\coprod_U U$, where $U$ runs through all good sets. This is a topological poset: $(U,x) \leq (V,y)$ iff $x=y$ and $U \subset V$. Take the nerve $|N_{\bullet} P|$ and consider the map $g:|N_{\bullet} P| \to Y$. It has contractible point-preimages: the preimage of $x$ is the poset of all good sets containing $x$, ordered by inclusion and the assumption says that this has contractible realization. In their recent paper arXiv:1201.3527, Galatius and Randal-Williams prove a lemma that under the present assumptions, $g$ is a Serre fibration with contractible fibres, in particular a weak homotopy equivalence.
Apply the same argument to get a poset $Q$, formed not from good subsets of $Y$, but their preimages in $X$. Thus up to weak equivalence, we can replace $f:X \to Y$ by $|N_{\bullet} Q| \to |N_{\bullet} P|$. By assumption, on the $p$th simplicial stage, the map $N_p Q \to N_p P$ is a rational homology equivalence (because this is just a disjoint union of the maps $f^{-1}(U) \to U$). The spectral sequence of simplicial spaces gives then that $|N_{\bullet} Q| \to |N_{\bullet} P|$ is a rational homology equivalence.
ad 2. There are several instances where the above argument can be applied. Example 1: let $G$ be a discrete group that acts simplicially on a simplicial complex, with finite stabilizers. Example 2: let $G$ be a Lie group that acts properly on a smooth manifold with finite stabilizers. By some standard results, there is a $G$-invariant Riemannian metric and $G$-equivariant tubular neighborhoods of orbits $Gx\cong G/G_x$. The image of such a tubular neighborhood $U \subset M$ is a contractible subset of $M/G$. The preimage of $U/G$ under $EG \times_G M \to M/G$ is $EG \times_G U \simeq EG \times_G G/G_x \simeq BG_x$, which is $\mathbb{Q}$-acyclic.