Let $X$ be a locally compact metric space and $G$ be a discontinuous group of homeomorphisms of $X$. I need to show that the orbit map $p :$ $X \rightarrow X/G$ has the path lifting property.

Defn : A group $G$ of homeomorphisms of a topological space $X$ is called discontinuous if for each point $x\in X$

  1. $G_x :=$ the stabilizer of $x$, is finite.

  2. $x$ has an open neighbourhood $U$ such that $gU \cap U = \emptyset$ for all $g \notin G_x$

Defn : A map $f:X \rightarrow Y$ between topological spaces is said to have the path lifting property if for each path $\alpha : I \rightarrow Y$ and each point $x\in f^{-1}(\alpha (0))$ there is a path $\tilde{\alpha}:I \rightarrow X$ such that $f\tilde{\alpha} = \alpha$ and $\tilde{\alpha}(0)=x$.

If $X$ is a compact metric space then I have shown that the orbit map has the path lifting property. Now when $X$ is locally compact for each point $x\in X$ I can find a compact neighbourhood $U_x$ which is $G_x$ - stable and mapped outside itself by any element not in $G_x$.

For $p: U_x \rightarrow U_x/G_x $ I can apply the path lifting property. Is it possible to glue these lifts so that I get a lift from all of $X$?

Thank you.

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    $\begingroup$ You may have a look at Brown's Topology and Groupoids, page 414ff. According to the statement there, you only need a Hausdorff space. $\endgroup$ – Stefan Hamcke Oct 21 '15 at 11:15
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    $\begingroup$ This path lifting result goes back to a 1972 book of Bredon on compact transformation groups. What is in T&G and nowhere else is the result on p416ff on the fundamental groupoid of $X/G$ as, under useful conditions, the orbit groupoid of the fundamental groupoid of $X$. $\endgroup$ – Ronnie Brown Oct 21 '15 at 20:29
  • $\begingroup$ @RonnieBrown: See my answer to math.stackexchange.com/questions/656858/… and the reference to Armstrong's paper (he proved the result few years before Bredon, in the setting of locally compact metrizable spaces, as OP needs). $\endgroup$ – Moishe Kohan Oct 21 '15 at 22:17
  • $\begingroup$ @studiosus: good point. But one does want to know the most general result, and what it leads to. $\endgroup$ – Ronnie Brown Oct 22 '15 at 9:45

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