Let $\Gamma$ be some discrete subgroup of isometries on $\mathbb{R}^n$ acting on $\mathbb{R}^n$ properly discontinuously so that $\mathbb{R}^n/\Gamma$ becomes a Riemannian manifold.
Let $U\subset \mathbb{R}^n/\Gamma$ be an open and simply connected set. Is it true that there exists some $V\subset \mathbb{R}^n$ such that $U$ and $V$ are isometric? Maybe someone knows a reference where I can find an answer?
I know that $ \mathbb{R}^n/\Gamma$ is locally isometric to $\mathbb{R}^n$ and that $\mathbb{R}^n$ is the universal cover of $ \mathbb{R}^n/\Gamma$. So the situation is clear if $U$ is somewhat 'small'. But what if it is 'big'?
Best regards
If $\pi:\widetilde{X} \to X$ is a covering map, and if $U \subset X$ is open, the restriction of $\pi$ to each path component of the preimage $\pi^{-1}(U)$ is a covering of $U$. Fix one such, $V$. If $U$ is simply-connected, $\pi:V \to U$ has one sheet, hence is a homeomorphism.