All non-compact simply connected $2$-manifolds with boundary

Question

There are two corresponding posts MSE and MSE by me without any answers.

Problem: Let $\Sigma$ be a non-compact simply-connected $2$-dimensional manifold, with boundary. Then, up to homeomorphism $\Sigma$ is of the form: delete a closed subset from the boundary $\Bbb S^1$ of the closed unit disc.

Motivation: The lemma 1.8., on page 30 of the book A Primer on Mapping Class Group by Benson Farb and Dan Margalit, says the following:

Two simple closed curves on a surface having finitely many intersection points and having no bi-gon, whenever lifted(assuming the existence of liftings) to the universal cover intersects at most one point.

Now, the lemma has been proved for closed hyperbolic surfaces. One crucial step in proving this lemma is: there are two arcs of two liftings together bound a disc in the universal cover $\Bbb H$, which is possible by the plane Jordan curve theorem. In other words, as long as the universal cover is $\Bbb S^2$ or a convex subset of $\Bbb R^2$, the argument of the lemma 1.8. works fine.

My Thoughts: I am trying to use collaring the boundary of $\Sigma$ to get a non-compact simply connected surface without boundary. I know there is a technique for proving any non-compact simply connected surface without boundary is homeomorphic to $\Bbb R^2$, and this technique is straight forward, in the sense that it does not use the classification theory(considering genus, number of compact boundary components, orientability, isomorphic diagram) of all $2$-dimensional manifolds. Also, $\Sigma$ is contractible.

Any help in proving the problem will be appreciated. Thanks in advance.

Answer 1

Here is one proof, using the Uniformization Theorem. This proof will be easier in the setting of the "Primer" since the authors are considering universal covering spaces of complete hyperbolic surfaces with geodesic boundary.

Start with a simply-connected surface with boundary $S$ and let $DS$ denote the double of $S$ along its boundary. Then $DS$ admits an involution $\tau$ fixing $\partial S\subset DS$ pointwise. This all can be done smoothly. Now, put a $\tau$-invariant Riemannian metric on $DS$; this defines a conformal structure on $DS$ with respect to which $\tau$ is an antiholomorphic involution. Let $X$ denote the universal covering space of $DS$ with lifted conformal structure. Then $\tau$ lifts to antiholomorphic involutions on $X$. By the Uniformization Theorem, $X$ is conformal to the unit disk (which I will equip with the Poincare metric) or complex plane or $S^2$. I will consider the first case since the proof in the two other cases is similar but simpler. The surface $S$ lifts diffeomorphically to a subsurface $Y\subset X$ (since $S$ is simply connected). Each boundary component $c$ of $Y$ in $X$ is fixed by a lift $\sigma_c$ of $\tau$. Since $\sigma_c$ is an antiholomorphic involution of the unit disk, it is a hyperbolic isometry, hence, its fixed-point set is a geodesic in the hyperbolic plane ${\mathbb H}^2$. Thus, $Y$ is a closed convex subset in ${\mathbb H}^2$. Now, switch to the projective (Klein) model of the hyperbolic plane. Every convex subset of ${\mathbb H}^2$ then becomes a convex subset of the Euclidean plane. It is an elementary exercise to see that each compact convex subset (with nonempty interior) of the Euclidean plane is homeomorphic to the closed unit disk ${\mathbb D}$ in ${\mathbb R}^2$. Under this homeomorphism $cl_{{\mathbb R^2}}(Y)\to {\mathbb D}$, $Y$ maps to the complement to a closed subset of the boundary of ${\mathbb D}$. Since $Y$ is homeomorphic to your surface $S$, you get the statement you are after.