Equivariant cohomology

Remarks

Up to isomorphism, $H^\ast_G(X; R)$ does not depend on the choice of $EG$. The projection $X_G \to BG$ makes $H^\ast_G(X; R)$ an algebra over $H^\ast(BG; R) = H^\ast_G(\mathrm{pt}; R)$.

Example: let $G = S^1$ act on a point. One can take $EG = S^\infty$, the unit sphere in $\C^\infty$, so $X_G = BS^1 = \C P^\infty$ and $H^\ast_{S^1}(\mathrm{pt}; \Z) \cong \Z[u]$ with $u$ of degree $2$.

Non-example: the ordinary cohomology of the orbit space $X / G$ is not the equivariant cohomology in general. For $S^1$ acting on a point, $X / G$ is a point with cohomology $\Z$ in degree $0$ only, unlike $\Z[u]$. If $G$ is a compact Lie group acting freely on $X$, then the map $X_G \to X/G$ is a fibration with contractible fibre $EG$, and $H^\ast_G(X; R) \cong H^\ast(X/G; R)$.