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When a free action gives rise to a $G$-principal bundle

Let a (topological) group $G$ act freely on a (topological) space $X$. Assume that
$G \backslash X$ is Hausdorff. (equivalently the image of the map $G \times X \to X\times X$ is closed).

Does it mean that $X \to G \backslash X$ is a $G$-principal bundle?

We are interested in the case when both $G$ and $X$ are l-spaces and moreover when $G=\mathbb Z^n$. However, I'll be happy to hear about other contexts too: locally compact spaces, topological manifolds, $C^\infty$-manifolds, algebraic varieties, etc.

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    $\begingroup$ The answer will depend on your definition of a principal bundle. Section 2 in Chapter 4 of Husemoller's "Fibre bundles", 3rd edition addresses your question for Husemoller's definition. Also see mathoverflow.net/questions/57015/…. $\endgroup$ – Igor Belegradek May 18 at 18:52
  • $\begingroup$ So you are not quite asking for the action to be proper? This implies $C\times X\to X\times X$ is a closed map, not just that the image is closed. $\endgroup$ – David Roberts May 18 at 21:28
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    $\begingroup$ In the case of $C^\infty$ manifolds a smooth Lie group action on a manifold $X$ is the principal action of a (unique) principal bundle structure $X \to X/G$ iff $G$ acts freely and properly. This seems to be folklore and can be found e.g. in Duistermaat-Kolk "Lie groups" in the first chapter. $\endgroup$ – Stefan Waldmann May 19 at 15:32

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