Annular cover of a surface?

Question

Say we have a surface $\Sigma$ that is compact, orientable, and has boundary, with $\chi (\Sigma)<0.$ Say $\gamma$ is an essential simple closed curve in $\Sigma$. I'm trying to understand what the annular cover of $\Sigma$ with respect to $\gamma$ looks like. Obviously it is topologically an annulus, and I believe the core curve should be a lift of $\gamma$, but I don't totally see what the covering space map should look like.

Answer 1

It's somewhat unclear to me what your question is other than "How do I see the covering space?", and I don't know how to "see" the covering space other than to apply the classification theorem for covering spaces.

The covering map $$f: \widetilde\Sigma \to \Sigma $$ which comes from application of that theorem has the property that the induced homomorphism $f_* : \pi_1(\widetilde\Sigma) \to \pi_1 \Sigma$ is injective and its image is equal to the infinite cyclic subgroup $Z_\gamma$ which is the image of the inclusion induced homomorphism $\pi_1\gamma \hookrightarrow \pi_1\Sigma$ (which is an injective homomorphism). I am abusing notation here by suppressing base points.

One can prove that an annulus neighborhood $A \subset \Sigma$ of $\gamma$ does indeed lift homeomorphically to an annulus $\widetilde A \subset \widetilde\Sigma$, and that the whole of $\widetilde\Sigma$ deformation retracts to $\widetilde A$, although $\widetilde\Sigma$ will generally be strictly larger than $\widetilde A$. And one can prove that $\widetilde\Sigma$ is itself a manifold with boundary whose interior is an open annulus.

But again, as for how to "see" $\widetilde\Sigma$, you would have to be more explicit about what you are "looking" for. Perhaps some details are contained in the proof of the classification theorem, namely the portion which constructs the covering $f : \widetilde\Sigma \to \Sigma$. Alternatively, one can simply try to deduce properties from the knowledge that $\pi_1(\widetilde\Sigma)$ is infinite cyclic.

Added: It occurred to me that one can say more about $\widetilde\Sigma$ with regard to its boundary, leading to a complete description of the topological type of $\widetilde\Sigma$, although I remain unsure whether this counts as "seeing" $\widetilde\Sigma$ in a sense that you desire.

Since you started with $\Sigma$ being a compact surface with nonempty boundary, it follows that $\widetilde\Sigma$ also has nonempty boundary. But since $\Sigma$ is not itself an annulus, and since $\gamma$ is an essential simple closed curve in $\Sigma$, it follows that each connected lift in $\widetilde\Sigma$ of each component of $\partial \Sigma$ is homeomorphic to $\mathbb R$. Thus, $\partial\widetilde\Sigma$ is a union of copies of $\mathbb R$. From this, together with the fact that $\widetilde\Sigma$ has infinite cyclic fundamental group, we can deduce that $\widetilde\Sigma$ is homeomorphic to a noncompact surface obtained from a compact annulus by removing a nonempty, nowhere dense, closed subset of each component of $\partial\widetilde\Sigma$ (there's some work buried here). Furthermore, the set removed cannot contain any isolated points (here's some more work).

Therefore $\widetilde\Sigma$ is obtained from a compact annulus by removing a Cantor subset of each component of the boundary of that annulus.