Equivariant Coefficient ring action on singular cohomology

Question

Let $X$ be a manifold acted on by a Lie group $G$. The $G$-equivariant cohomology of $X$ with coefficients in a ring $\mathcal{R}$ is defined as the cohomology ring $$ H_G^*(X; \mathcal{R}) := H^*(X_G; \mathcal{R}), $$ where $X_G := (X \times EG) / G$ is the homotopy quotient, $EG \to BG = EG / G$ being the universal principal bundle of the group $G$.

The natural projection $$ X_G \to BG $$ obtained by collapsing the elements of $X$ gives rise to a ring homomorphism $$ H^*(BG; \mathcal{R}) \to H_G^*(X; \mathcal{R}), $$ or equivalently to an action of the ring $H^*(BG; \mathcal{R})$ on the equivariant cohomology $H_G^*(X; \mathcal{R})$ of $X$.

Suppose now that $G$ acts freely on $X$. In this case, the cohomology groups $H_G^*(X; \mathcal{R})$ and the singular cohomology $H^*(X/G; \mathcal{R})$ of $X/G$ agree.

My question is the following: how does $H^*(BG; \mathcal{R})$ act on the singular cohomology of $X/G$ (if it can make things easier, one might take $\mathcal{R}=\mathbb{C}$)?

Answer 1

$EG$ is also the universal free $G$-space, meaning that, if $X$ is a free $G$-space (let's assume of the $G$-homotopy type of a $G$-CW complex), there is, up to $G$-homotopy, a unique $G$-map $X\to EG$. Taking quotients, you get a map $X/G \to BG$ which induces the action of $H^\ast(BG)$ on $H^\ast(X)$ with any coefficients.

Edited to add: I really should have mentioned that $X/G \to BG$ is the classifying map of the bundle $X\to X/G$. I think this makes it a little less mysterious where the map comes from.

Second edit to add (simultaneously with Mike Miller's comment): To give an example where the action is nontrivial, let $G = {\mathbb Z}/2$ and $X = S^n$ with $G$ acting as $-1$. Then $X/G = {\mathbb R}P^n$, $BG = {\mathbb R}P^\infty$, and $X/G \to BG$ is the inclusion. $H^\ast(BG;{\mathbb Z}) = {\mathbb Z}[x]$ then acts in the obvious, nontrivial way on $H^\ast(X/G;{\mathbb Z}) = {\mathbb Z}[x]/x^{n+1}$.