# Quotient spaces and equivariant cohomology

Consider a $G$-equivariant map $\pi:X\to Y$ for $G$ an affine algebraic group, such that $\pi$ is a good categorical quotient. Is there any relationship between $H^*_G(X)$ and $H^*(Y)$? Is there if $\pi$ is a good geometric quotient, or if the quotient space is smooth?

EDIT: A categorical quotient is an equivariant map $\pi:X\to Y$ that is constant on $G$-orbits. It's good if the topology on $Y$ is induced by $X$ ($\pi$ is a surjective open submersion) and the map from the functions on any affine $V \subset Y$ to $G$-invariant functions on $\pi^{-1}(V)$ is an isomorphism. It's a good geometric quotient if the $G$-orbits closed in $Y$.

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The general philosophy for this sort of question is that for the equivariant cohomology to coincide with the cohomology of the quotient, then you need the $G$-action to be free. (See e.g. Jim Conant's answer below.) In general, the equivariant cohomology is computing the cohomology of the stacky quotient $[X/G]$, whereas $Y$ (the good categorical quotient) is probably more like the coarse moduli space of this stack, or something similar. –  Matt E Nov 15 '10 at 13:44
I'm not very familiar with stacks, or with the details of GIT if $G$ is not a torus, but at least when it is the torus the GIT procedure removes the non-free points from $Y$ before quotienting. $X//G$ is supposed to give a scheme $Y$, and I assume that it is constructed thus to avoid having the quotient be a stack. –  James Davidoff Nov 15 '10 at 14:48

I don't know what a good categorical quotient is, but if $X$ is a CW complex, and the $G$ action is free and cellular, then $H^*_G(X)\cong H^*(X/G)$. See Brown's "Cohomology of Groups," p.173.