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We have a theorem that says that if a group $G$ acts on a path-connected space $Y$ properly discontinuously, then $\pi: Y \rightarrow Y/G$ is a covering map. Especially, $G$ is isomorphic to the group of deck transformations.

Now, I would like to understand this:

1.) Does the group of deck transformations to a given covering map always act properly and discontinuously on the covering space?

2.) If I have a covering map $p:X \rightarrow Y$ and I look at the group of deck transforms $G(X,p)$. Does it then follow that $Y$ is isomorphic to $X/G(x,p)$?

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    It is not true that if $G$ acts on $Y$ properly discontinuously then $\pi$ is a covering map. For example, every finite group action is properly discontinuous, but if $G=\mathbb{Z}/2\mathbb{Z}$ acts on $Y=\mathbb{R}$ by reflecting across $0$ then $\pi$ is not a covering map. – Lee Mosher Jul 31 '14 at 16:16
  • interesting. so the theorem should maybe be more like: if $\pi$ is a covering map, then $G$ is isomorphic to the group of deck transformations. (That's what I just found in the internet). –  Jul 31 '14 at 16:21
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    Yes, in that direction the theorem holds as asked. The opposite direction will become true if you add an additional hypothesis; see my comment to the answer of @msteve. – Lee Mosher Jul 31 '14 at 18:28

1 Answers1

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For (1), if $G$ is assumed to be a group of homeomorphisms of $X$, then the fact that $\pi \colon X \to X/G$ is a covering map is equivalent to $G$ acting on $X$ freely and properly discontinuously (this is Thm 81.5 in Munkres).

For (2), there is indeed a homeomorphism $X/G \simeq Y$ if we assume that the covering map $p \colon X \to Y$ is regular (that is, the action of $G$ on the fibers is transitive).

msteve
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  • but concerning (1). if $G$ is the group of all deck transformations, then $G$ acts always properly discontinuously, right? –  Jul 31 '14 at 16:31
  • Yes, the group of deck transformations acts properly discontinuously on the cover. This discussion is more or less chapter 81 of `Topology' by Munkres, you should check it out. – msteve Jul 31 '14 at 16:34
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    (1) is wrong. In order for $\pi : X \to X/G$ to be a covering map, in addition to $G$ acting properly discontinuously you must also require that $G$ acts freely, in other words each nontrivial element of $G$ has no fixed points. See my counterexample in the comments. @user159356 – Lee Mosher Jul 31 '14 at 18:21
  • Also, in (1) it is unnecessary to hypothesize that $G$ acts by homeomorphisms. An action of a group $G$ on a topological space $X$ is by definition a homomorphism from $G$ to the group of self-homeomorphisms $\text{Homeo}(X)$. – Lee Mosher Jul 31 '14 at 18:23