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Does anybody know a reference to the following fact?

If $G$ is a topological group acting transitively and effectively on a space $X$, then the evaluation map $G \rightarrow X$, $g \mapsto g \cdot x_0$ for some fixed $x_0$ is a fibration. (It's fiber is then given by the stabilizer of $x_0$.)

Thanks!

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    $\begingroup$ Do you mean "fibration" or "fiber bundle"? en.wikipedia.org/wiki/Fibration en.wikipedia.org/wiki/Fiber_bundle $\endgroup$
    – Tom Church
    Jul 17, 2015 at 17:26
  • $\begingroup$ Fibration is good enough for me, but I don't know what the general proposition states. By fiber I mean the fiber over $x_0$, or simply homotopy fiber, whichever people prefer. $\endgroup$
    – user76162
    Jul 21, 2015 at 6:36
  • $\begingroup$ If $X$ is a homogeneous real tree not reduced to a line, I think its isometry group $G$ is totally disconnected but $X$ is geodesic, so the action will not be a locally trivial fibre bundle. ($G$ is totally disconnected because its unit component has to act trivially on the boundary, and the action on the boundary is faithful) $\endgroup$
    – YCor
    Jul 30, 2015 at 15:00
  • $\begingroup$ If the action is effective, it's a bijection. $\endgroup$
    – Fernando Muro
    Aug 29, 2015 at 13:49
  • $\begingroup$ The evaluation map certainly need not be locally trivial: consider $\mathbb Z$-action on $S^1$ rotating by an irrational angle, and let $X$ be any orbit. Then the evaluation map $e:\mathbb Z\to X$ is not a locally trivial fiber bundle because no point of $X$ has a discrete neighborhood and $e$ is a bijection. This $e$ does have the homotopy lifting property, but I suspect playing with the same idea should lead to examples where the evaluation map is not a fibration. $\endgroup$
    – Igor Belegradek
    Aug 29, 2015 at 14:13

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This question is discussed in Steenrod's "The Topology of Fibre Bundles", part I, sections 7.3 and 7.4. Note that if the action of $G$ on $X$ is proper (eg, if $G$ is compact), then your map factors through a homeomorphism $G/G_{x_0}\to X$. Then the question is just whether the orbit map $G\to G/G_{x_0}$ is a fibration. I have a feeling this is not always true (but it is certainly true if $G$ is a Lie group, as noted in section 7.5 of Steenrod's book).

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