Let $G$ be a topological (or simplicial) group, let $X$ and $Y$ be $G$-spaces, and let $f,f':X\to Y$ be $G$-maps which are homotopic as maps of spaces. In general, $f$ and $f'$ may (of course) fail to be equivariantly homotopic. For instance, we can just take $X$ to be a point and $Y$ to be any path-connected $G$-space having two distinct fixed points of the $G$-action, with $f$ and $f'$ given by the inclusion of two fixed points. But is there any example of such $f$ and $f'$ if we assume that $X$ and $Y$ are total spaces of principal $G$-bundles?


2 Answers 2


For any $G$-space the $G$-equivariant maps $[EG,X]_G$, also known as the homotopy fixed points $X^{hG}$, are a Borel homotopy invariant of $X$ meaning that it is an invariant of $G$-equivariant maps for which the underlying map is a weak equivalence. If $Z$ has the trivial action, this means we can compute $[EG,Z \times EG]_G$ as $[EG,Z]_G=[EG/G,Z]=[BG,Z].$ On the other hand, the nonequivariant maps $[EG,Z] \cong \pi_0(Z)$ because $EG$ is contractible. So it suffices to find a space, $Z$ such that $[BG,Z] \not= \pi_0(Z)$. The universal example (if $G$ is not contractible) is $Z=BG$.

Explicitly, if $q:EG \rightarrow BG$ is the quotient, this produces homotopic maps $EG \xrightarrow{q \times \mathrm{Id}} BG \times EG$ and $EG \xrightarrow{* \times \mathrm{Id}} BG \times EG$ which are not equivariantly homotopic.

  • $\begingroup$ Ah, this is such a nice example! Thank you! $\endgroup$
    – Ken
    Commented Jun 16, 2023 at 0:39

$\newcommand{\RP}{\mathbb{RP}}$Connor Malin's answer is excellent. Derived from that, here is a small example: Let $G = C_2$, the group with two elements, let $X = S^1$ with antipodal action, and let $Y = \RP^2\times S^1$ where $G$ acts trivially on $\RP^2$. Let $f\colon X\to Y$ be $q\times\text{Id}$, where $q$ is the composite $S^1\to \RP^1\hookrightarrow \RP^2$, the first map being the quotient, and let $f' = *\times\text{Id}$ for some point $*\in \RP^2$. The two maps are nonequivariantly homotopic because the map $S^1\to \RP^2$ is twice the generator of $\pi_1(\RP^2) \cong \mathbb{Z}/2$, but $S^1\to \RP^2$ is equivariantly essential because its quotient by $G$ is the inclusion $\RP^1\hookrightarrow \RP^2$, which is essential.


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