"Thickening" an arc on a 2-manifold What is the argument for the fact that each arc in the interior of a 2-manifold can be "thickend" to obtain a 2-cell containing the arc in its interior and being disjoint from any preassigned compact set in the complement of the arc? Thanks in advance!
 A: Suppose that $S$ is the surface, $\alpha$ is the arc, and $K$ is the compact set.
Proof 1: Find an arc $\alpha'$ so that $\gamma = \alpha \cup \alpha'$ is a Jordan curve (+).  The Jordan/Schoenflies [J/S] theorems provide an annulus (or Möbius) neighbourhood $A$ of $\gamma$ and a homeomorphism on $A$ making $\gamma$ "straight".  Shrink $A$ as needed to avoid $K$ and then use $A$ to find the desired two-cell.
Proof 2: Let $T$ be the double branched cover of $S$, branched over the two points of $\partial \alpha$.  Let $\beta$ and $\beta'$ be the two preimages of $\alpha$ in $T$.  So the union $\gamma = \beta \cup \beta'$ is a Jordan curve.  Apply [J/S] to find an annulus neighbourhood $A$ of $\gamma$.  Shrinking $A$ as needed, we may assume that it is disjoint from the preimage of $K$.  We further arrange matters so that $A$ is invariant under the deck transformation, as are two of its cutting arcs (meeting $\gamma$ in exactly the points $\partial \beta$). So the image of $A$ is the desired two-cell.

Of course, in Proof 1, sentence (+) is saying that endpoints of $\alpha$ are "accessible".  Showing such things is part of various proofs of [J/S]. The branched cover trick in Proof 2 is designed to avoid (+).  However, I'll guess that no proof can avoid using [J/S], or at least some of its "working parts".
