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Suppose $X$ and $Y$ are path-connected and locally path-connected, and a group $G$ acts freely on both spaces discretely by homeomorphisms. Let $p_X$ and $p_Y$ be the projections onto the quotient spaces $X/G$ and $Y/G$. If $f:X\to Y$ is a continuous $G$-equivariant map, then by the universal property of quotient spaces there is a continuous map $\overline{f}:X/G\to Y/G$ such that the following diagram commutes: $\require{AMScd}$ \begin{CD} X @>f>> Y\\ @V p_X V V @VV p_Y V\\ X/G @>>\overline{f}> Y/G \end{CD}

Conversely, if $\overline{f}:X/G\to Y/G$ is a continuous map between quotient spaces, then it lifts to a continuous map $f:X\to Y$ iff $$(\overline{f}\circ p_X)_\ast(\pi_1(X)) \le (p_Y)_\ast(\pi_1(Y)).\tag{1}\label{1}$$

This follows from Proposition 1.33 in Hatcher applied to $\overline{f}\circ p_X:X\to Y/G$, since freeness implies $p_Y$ is a covering map. Moreover this lift can be chosen to be $G$-equivariant.

In the more general case when the action of $G$ is discrete but not necessarily free, the universal property still means that equivariant maps descend to maps on quotients.

Question: What conditions on $\overline{f}:X/G\to Y/G$ is equivalent to the existence of an equivariant lift in this case?

In addition to the condition on fundamental groups \eqref{1} above, it is easy to check that the following condition is necessary:

Compatibility condition: for any $\overline{x}\in X/G$ let $x\in X$ be a lift of $\overline{x}$, and let $y\in Y$ be a lift of $\overline{f}(\overline{x})$. Then the stabiliser $\textrm{Stab}_G(x)$ is conjugate in $G$ to a subgroup of $\textrm{Stab}_G(y)$.

I suspect that this condition (or something very similar to it) is also sufficient. In fact I could probably prove it with enough effort, but I feel like this is something which has almost certainly already been done, so my real question is: where can I find this written down?

Some extra information

The path lifting property (a key ingredient in Hatcher's proof of Proposition 1.33) holds for non-free group actions (see, for example, 11.1.4 in Topology and groupoids by Brown). Also, a lifting property is proved under very strong conditions on $\overline{f}$ in Lemma 23 of Orbifolds as diffeologies by Iglesias, Karshon, and Zadka, but this is far too strong for my needs.

The situation I'm primarily interested in is when $G$ is finite and acting orthogonally on finite dimensional vectors spaces, so I wouldn't mind extra conditions being imposed (eg $G$ acting with finite stabilisers, or acting by isometries), but I would also like to have the statement in something like the generality I have given above.

David Sheard
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  • Hi did you figure this out? Also, why can the lift be taken to be G-equivariant? – monoidaltransform Jan 09 '24 at 12:33
  • @monoidaltransform Never to my satisfaction, as I say, some hints of it in "Orbifolds and diffeologies", but I don't have a clearly written out proof in this setting. The question of why/when the lift is G-invariant is exactly what I'm asking for clarity on. If you assume equivariance, it is easy to verify that the pushforward of the fundamental group of $X$ is a subgroup of the push forward of the fundamental group of $Y$, and also the compatibility condition I state, but I cannot see a proof of the sufficiency of these conditions. – David Sheard Jan 09 '24 at 14:57
  • I mean right after (1) you say: 'since freeness implies is a covering map. Moreover this lift can be chosen to be -equivariant' I wondering about everything after 'Moreover' – monoidaltransform Jan 09 '24 at 15:00
  • @monoidaltransform A geometric way to think about it is to assume that you have fundamental domains for $G$ acting on $X$ and $Y$, say $\mathcal{F}_X$ and $\mathcal{F}_Y$. Then you can certainly lift $\bar{f}$ to a map $f:\mathcal{F}_X\to\mathcal{F}_Y$. For any point $x\in X$, there exists $g\in G$ such that $g.x\in\mathcal{F}_X$. Now extend $f$ to the whole of $X$ by setting $$f(x)=g^{-1}.f(g.x).$$ You can easily check that this defines a $G$-equivariant lift. – David Sheard Jan 09 '24 at 16:23

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