This question might be silly; if so, let me know, I will delete it. 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.

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.