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Take a Hausdorff topological space $X$. Take two distinct points $x$ and $y$ of $X$. Consider a set $U$ of continuous paths $p$ from $[0,1]$ to $X$ equipped with the compact-open topology such that: $p(0)=x$, $p(1)=y$, $p$ is one-to-one and such that if $p\in U$, then $p\phi\in U$ for any nondecreasing homeomorphism $\phi:[0,1]\to [0,1]$ (let us call this group $G$).

Is the map $U\to U/G$ always a weak homotopy equivalence ? Or could someone provide a counterexample ?

I know cases where it is true. I am unsure that it is true in full generality and I'd be interested in seeing a counterexample.

Note: in fact, I work with $\Delta$-Hausdorff $\Delta$-generated spaces $X$ ($\Delta$-Hausdorff meaning that any continuous map $[0,1]\to X$ has a closed image), but I do not think that it is relevant here.

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  • $\begingroup$ I do not understand your question: to define $X/G$, you would need to give an action of $G$ on $X$. Do you mean $U/G$? $\endgroup$
    – Benoît Kloeckner
    Commented Jun 11, 2022 at 7:48
  • $\begingroup$ @BenoîtKloeckner Yes I have edited my question. $\endgroup$
    – Philippe Gaucher
    Commented Jun 11, 2022 at 8:15
  • $\begingroup$ 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. $\endgroup$
    – Neil Strickland
    Commented Jun 11, 2022 at 19:00

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