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Suppose $X$ is a topological space and $G$ is a topological group, and $G$ acts on $X$. Here is my question: If $G$ acts freely on $X$, then what are the maps showing $(X \times EG)/G$ is homotopy equivalent to $X/G$?

(Some background information: Let $EG$ be some contractible space with a free $G$-action. The $G$-equivariant cohomology $H^i_G(X)$ of $X$ is the singular cohomology $H^i(X_{hG}, \mathbb{Z})$ of the homotopy orbit space $$X_{hG}:=(X \times EG)/G$$ where the $G$-action on $X \times EG$ is given by $g\cdot (x, e)=(gx, ge)$.

Now suppose $G$ acts freely on $X$, i.e. $gx=gy$ implies $x=y$. Then it is often asserted that in this case the $G$-equivariant cohomology of $X$ is the cohomology of $X/G$, but I couldn't figure out why.

What are the maps showing $(X \times EG)/G$ is homotopy equivalent to $X/G$ (when $G$ acts freely on $X$) ?

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If the action is free then the projection $(X\times EG)/G\to X/G$ is a fibration with contractible fibres (homeomorphic to $EG$), and so it is a homotopy equivalence.

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    $\begingroup$ Can you give more details why the action being free implies $(X\times EG)/G\to X/G$ is a fibration? And why is the fiber homeomorphic to $EG$? $\endgroup$
    – ykm
    Mar 23, 2014 at 21:21
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    $\begingroup$ Given a point $x \in X$, projecting to $xG \in X/G$, the orbits in $(X \times EG)/G$ in that project to it are $[x,e]_G$, the orbits of points $(x,e) \in X \times EG$. Any such class $[x,e]_G$ has a unique representative with first coordinate $x$, so it follows $$ EG \overset{\approx}\longrightarrow \{x\} \times EG \hookrightarrow X \times EG \twoheadrightarrow (X \times EG)/G $$ is injective, and surjects onto the fiber over $xG$. $\endgroup$
    – jdc
    Jan 18, 2015 at 0:34
  • $\begingroup$ @jdc I've run into a relevant question and was wondering if you could be kind enough to respond. I've read in a set of lecture notes that $(X\times EG)/G\to X/G$ is a fiber bundle, however it's not clear to me how one can define a local trivialization for it in a well-defined way. Does the author mean a (Serre) fibration by a fiber bundle? $\endgroup$
    – Philip M
    Mar 7, 2017 at 3:03
  • $\begingroup$ @PhilipM: sorry for the delay. I got distracted, probably because this isn't immediately obvious to me either. I think there is probably some dependence on properties of $X$ and the action necessary to make it true. Have you found anything in the intervening time? $\endgroup$
    – jdc
    Mar 25, 2017 at 23:23

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