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

$$\textbf{Problem:}$$ 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?

• Do you have a reference for the italicised statement? Jan 30, 2021 at 11:55
• Can anyone explain what does "a covering corresponding to the subgroup $\langle \alpha\rangle$" mean? Jan 30, 2021 at 12:16
• @C.F.G see this or Hatcher's Proposition 1.36. Jan 30, 2021 at 12:39
• @MichaelAlbanese This appears 3.3.5 in An Introduction to 3-manifolds written by Peter Scott. I have no other proof of this fact. So, I was looking for alternative proof. Maybe something easier, not using too many things: classification theorem, orientability, etc. Jan 30, 2021 at 12:44
• The hard part is the italicized statement about surfaces with finitely generated fundamental groups. By the way, you should also assume that your surface is connected. Jan 30, 2021 at 16:27

## 1 Answer

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).

• Thanks, for your proof. I think you are talking about(last line) fixing a generator from $H_n(M, M-x)$ for each $x\in M$, $n=\dim M$ and then local consistency as appearing in Hatcher page 234. Sorry, I have forgotten(before asking this question) we need to work locally while considering covering: local consistency works fine as a covering is a local homeomorphism. Am I right? Jan 30, 2021 at 12:32
• Yes, that's the idea. Jan 30, 2021 at 13:41