Let $X$ be a locally compact, Hausdorff space, and $G$ a locally compact, Hausdorff group acting continuously and freely on $X$. If we put on $\text{Homeo}(X)$ the compact open topology it can be shown that the natural map $G \rightarrow \text{Homeo}(X) $ is continuous. Since the action is free, in particular this map is injective, but in general is not an homeomorphism on its image.

If we assume that the action is proper (that is the map $G \times X \rightarrow X \times X$ given by $(g,x) \mapsto (gx,x)$ is proper) is it true that this map is an homeomorphism on its image?

I'm asking this because one sometimes finds theorems stated as "Let $G \subseteq \text{Homeo}(X) $ be a group of homeomorphism with the compact open topology, acting properly and freely on a space X...". I would like to know if some generality is lost by considering the group which acts as an embedded subgroup of the group of homeomorphisms.


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