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Jun 11, 2022 at 19:00 comment added Neil Strickland Let's consider the case where $X$ is compact. Given $p\in U$, we know that $p\:[0,1]\to p([0,1])$ is a homeomorphism so we can apply the Tietze Extension Theorem to $p^{-1}$ to give $u\colon X\to [0,1]$ with $u\circ p=1$. We can then put $F=\{q\in U:u\circ q\text{ is injective }\}$ and $F_1=\{q\in U:u\circ q=1\}$ and we find that $F=G\times F_1$. I think that $F$ is closed in $U$ with empty interior so it is not clear whether this is useful.
Jun 11, 2022 at 8:15 comment added Philippe Gaucher @BenoîtKloeckner Yes I have edited my question.
Jun 11, 2022 at 7:50 history edited Philippe Gaucher CC BY-SA 4.0
$X\to X/G$ replaced by $U\to U/G$
Jun 11, 2022 at 7:48 comment added Benoît Kloeckner I do not understand your question: to define $X/G$, you would need to give an action of $G$ on $X$. Do you mean $U/G$?
Jun 11, 2022 at 7:36 history asked Philippe Gaucher CC BY-SA 4.0