# Homotopic but not equivariantly homotopic maps

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?

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.
$$\newcommand{\RP}{\mathbb{RP}}$$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.