$\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\RP}{\mathbb{RP}}$ Let $S^\infty$ be given the canonical $\pi_1(\RP^2)=Z/2$-equivariant cell structure.
I am trying to see why the map $S^\infty \times_{\mathbb{Z}/2} \tilde{RP^2} \to pt \times_{\Z/2} S^2=\RP^2$ is a homotopy equivalence.
My motivation is in finding a more explicit model for the map $X \to K(\pi_1(X),1)$ for $X$ a CW complex.(see Explicit description of $X \to K(\pi_1(X),1)$)
I am posting the question because it does not follow from $S^\infty$ being contractible. This is because there is no action of $\pi_1$ on $r_t(S^\infty)$ where $r_t$ is the $t$th stage of the retraction.
You have a fiber bundle $S^\infty \to (S^\infty \times_{\Bbb Z/2} S^2) \to (* \times_{\Bbb Z/2} S^2)$. The fiber is contractible, so the last map is at least a weak homotopy equivalence; indeed it's an actual homotopy equivalence by Whitehead's theorem.