This question might be silly; if so, let me know, I will delete it. There are two corresponding posts [MSE][1] and [MSE][2] by me without any answers. 

> **Problem:** Let $\Sigma$ be a non-compact simply-connected $2$-dimensional manifold, 
> with boundary. Then, up to homomorphism $\Sigma$ 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.






[1]: https://math.stackexchange.com/questions/3930167/simply-connected-non-compact-surface-with-boundary
[2]: https://math.stackexchange.com/questions/3931592/embedding-a-disc-in-a-simply-connected-surface