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The question arises from the fact that each topological manifold $X$ is homeomorphic to its universal cover $X_0$ quotiented by the action of the fundamental group $\pi_1(X)$. It is natural to ask wether two spaces with the same universal covering and with isomorphic fundamental group are homeomorphic. The answer is affirmative in the case of compact surfaces. In general we ask wether two isomorphic groups $G$ and $G'$ can act properly discontinously and freely on a simply connected space $X$ in two different ways i.e. the corresponding quotient spaces $X/G$ and $X/G'$ are not homeomorphic.

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  • $\begingroup$ By "properly discountouous" I mean that for each $x\in X$ there is an neighborhood $U$ of $x$ such that $g(U)\cap U=\varnothing$ unless $g=1\in G$. (Munkres, Topology page 490) $\endgroup$ – Veraguth May 13 at 20:25
  • $\begingroup$ Do you have reference for the claim that two (connected) compact manifolds with homeomorphic universal covers and isomorphic fundamental groups are homeomorphic? Surprising (to me) if true! $\endgroup$ – Mike F May 13 at 20:28
  • $\begingroup$ Oh yeah OK for surfaces it's believable, I missed that word. $\endgroup$ – Mike F May 13 at 20:31
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  • The group $\mathbb Z$ acts on $\mathbb R^2$ by the formula $n \cdot (x,y) = (x+n,y)$, with quotient homeomorphic to an open cylinder $S^1 \times \mathbb R$.
  • The group $\mathbb Z$ also acts on $\mathbb R^2$ by the formula $n \cdot (x,y) = (x+n,(-1)^n y)$ with quotient homeomorphic to an open Möbius band.
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    $\begingroup$ The second one is not an action as written, but I think you mean $n(x,y)=(x+n,s(n)y)$, where $s(n)\in\{1,-1\}$ is the parity of $n$ ($1$ for even, $-1$ for odd). $\endgroup$ – Jackozee Hakkiuz May 13 at 20:35
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    $\begingroup$ Oops, good catch, thanks. $\endgroup$ – Lee Mosher May 13 at 20:37
  • $\begingroup$ of course, I forgot about $(-1)^n$ haha $\endgroup$ – Jackozee Hakkiuz May 13 at 20:37

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