# 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 at 11:55
• Can anyone explain what does "a covering corresponding to the subgroup $\langle \alpha\rangle$" mean? Jan 30 at 12:16
• @C.F.G see this or Hatcher's Proposition 1.36. Jan 30 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 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 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 at 12:32
• Yes, that's the idea. Jan 30 at 13:41