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 $X$) ?
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.