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I am looking for an example of a topological group $G$ acting by homeomorphisms on a metrizable space $X$ such that the orbit map $X\to X/G$ doesn't have the path lifting property, that is, there is a path in $X/G$ that cannot be lifted to $X$.

EDIT: Let's also assume that every $G$-orbit is a closed subset.

Non-example: If $G$ is a compact Lie group, then the orbit map has the path lifting property (in fact, there is a a slice). Palais gave a generalization to proper Lie group actions.

YCor
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Igor Belegradek
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    How about $\mathbb{Z}$ acting on $S^1$ by an irrational rotation? The induced topology on $S^1/\mathbb{Z}$ is indiscrete, so we can make a path that just jumps from one point to another. This cannot be lifted to $S^1$ because every path in $S^1$ is either constant or its range is all of $S^1/\mathbb{Z}$ under the quotient map. – Robert Furber Apr 24 '20 at 01:58
  • Thank you! Dense orbit is enough. I was focused on the case of closed orbits. – Igor Belegradek Apr 24 '20 at 02:12
  • I think, there are even examples even when $G$ is compact and the action is free, but, of course, $G$ is not a Lie group. Check the example I gave in my answer here. – Moishe Kohan Apr 24 '20 at 04:07
  • @MoisheKohan: I don't know why a path lifting property (for the orbit map of a free action by a compact group) has to imply the map is a principal bundle (or fibration). Ruling our principal bundles is much easier, see Remark 2 in https://mathoverflow.net/questions/57015/which-principlal-bundles-are-locally-trivial. – Igor Belegradek Apr 24 '20 at 12:27
  • Remark: the question doesn't refer to any topology on $G$ (still, the "non-example" says that the lifting property holds assuming that $G$ carries some group topology such that the action is continuous, plus some extra-assumptions). – YCor Apr 24 '20 at 14:04
  • @YCor: not sure what you are saying. As stated, the question is about any topological group, and non-example says that the desired group cannot be a compact Lie group, or more generally Lie group that acts Palais-properly. Naturally, a simpler example would be better. All I want is to gain intuition on paths that don't lift. – Igor Belegradek Apr 24 '20 at 15:28
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    This is not what I am saying: I am saying that there is a good chance that this example does not have a path-lifting property. In this example, a totally disconnected group acts on a 1-dimensional space such that the quotient is 2-dimensional. – Moishe Kohan Apr 24 '20 at 16:56

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