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.
- 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.