Open annulus covers any orientable non simply-connected surface without boundary.

Question

$\textbf{Problem:}$ Let $\Sigma$ be an orientable surface without boundary, possibly non-compact, with the non-trivial fundamental group. Then there is a covering map $p:\Bbb S^1\times \Bbb R\to \Sigma$.

$\textbf{Attempt:}$ Let $\alpha\in \pi_1(\Sigma)$ be a non-trivial element and consider the covering $p:\Sigma_\alpha\to \Sigma$ corresponding to the subgroup $\langle \alpha\rangle$ of $\pi_1(\Sigma)$. Then, the fundamental group of the surface $\Sigma_\alpha$ is $\Bbb Z$. Note that $\Sigma_\alpha$ has no boundary, as $p$ is a local homeomorphism. Also, $\Sigma_\alpha$ can not be compact surface by classification theorem. So, $\Sigma_\alpha$ is an open surface. Now, every connected open surface with finitely generated fundamental group is the interior of some compact surface. Also, every manifold is homotopically equivalent to its interior. So, $\Sigma_\alpha$ can be open Möbius strip or open annulus.

$\bullet$ I don't know how to show open Möbius strip can not cover an orientable surface. More generally, is it true that any non-orientable manifold can not cover an orientable manifold?

$\bullet$ Are the lines in the Attempt portion correct?

$\bullet$ Is there any quick proof of the original problem?

Answer 1

Your attempt looks fine, although you may want to justify why $\langle\alpha\rangle \cong \mathbb{Z}$ (as opposed to $\mathbb{Z}_k$). It is true that any covering space of an orientable manifold is necessarily orientable.

Let $p : M \to N$ is a continuous covering map of topological manifolds. Equip $N$ with a smooth atlas $\{(U_{\alpha}, \varphi_{\alpha}\}_{\alpha \in A}$ such that $U_{\alpha}$ is a connected, evenly covered neighbourhood for all $\alpha \in A$. Denote the connected components of $p^{-1}(U_{\alpha})$ by $\{V_{\alpha}^i\}_{i \in I}$; note that $p|_{V_{\alpha}^i} : V_{\alpha}^i \to U_{\alpha}$ is a homeomorphism. Then the smooth atlas on $N$ induces a smooth atlas on $M$ given by $\{(V_{\alpha}^i, \varphi_{\alpha}\circ p|_{V_{\alpha}^i})\}_{\alpha \in A, i \in I}$; with respect to these smooth structures on $M$ and $N$, the map $p$ is smooth. Now suppose $V_{\alpha}^i\cap V_{\beta}^j \neq \emptyset$, then

$$(\varphi_{\alpha}\circ p|_{V_{\alpha}^i})\circ(\varphi_{\beta}\circ p|_{V_{\beta}^j})^{-1} = \varphi_{\alpha}\circ p|_{V_{\alpha}^i}\circ p|_{V_{\beta}^j}^{-1}\circ \varphi_{\beta}^{-1} = \varphi_{\alpha}\circ\varphi_{\beta}^{-1}.$$

In particular, the atlas on $M$ is orientable if and only if the atlas on $N$ is. Therefore, if $p : M \to N$ is a covering map and $N$ is orientable, then so is $M$. Note, the above assumes that $N$ admits a smooth structure (which is always the case in dimension two). The conclusion is also true without this hypothesis, but the proof relies on a different characterisation of orientability (for example, in terms of homology).