The complement of Jordan arc If $A$ is the image of a Jordan arc in $S^2$, that is, $A$ is the image of an injective continuous map from $[0,1]$ to $S^2$, is $S^2-A$ necessarily a simply-connected set?
 A: Yes, $S^2 - A$ is simply connected.
Let $C$ be any closed loop in $S^2 - A$. Since $A$ and $C$ are compact,
the distance between them: 
$$d(A,C) = \inf \{ d(x,y) : x \in A, y \in C \} > 0$$
Start with any triangulation of $S^2$ into geodesic triangles, repeat
subdivide it until the diameter of all triangular faces are smaller than
$\frac{1}{3} d(A,C)$. 
Let $K$ be the union of all closed triangular faces which intersect $A$.
The boundary of $K$, $\partial K$, is a union of geodesic arcs. 
Every vertex of $\partial K$ is connected to even number of arcs of $\partial K$.
For any vertex connected to more than 2 arcs, connect the mid points of the arcs
to form a geodesic polygon and expand $K$ to completely fill the interior and edges
of this polygon.
It is easy to see the final $K$ satisfies:


*

*$K$ is connected

*$A \subset K - \partial K$

*$\forall x \in K, d(x,A) < \frac{2}{3} d(A,C) \implies K \cap C = \emptyset$ .

*Every vertex of $\partial K$ is connected to 2 arcs $\implies \partial K$ is a disjoint union of Jordan curves.


Since $C$ is connected, $C$ lives in a connected component of
$S - K$. Let's call it $L$ and its boundary $\partial L$ consists of
a subset of Jordan curves from $\partial K$. We claim that $\partial L$
is a single Jordan curve. Otherwise, let $L'$ and $L''$ be two Jordan curves 
on $\partial L$. Since $L$ is a connected component, $L'$ and $L''$ cannot lies
outside of each other. If one of them, say $L''$ is "inside" of another $L'$,
$L'$ will separate the portion $K$ outside of $L'$ to those inside of $L''$. 
This contradicts with the fact $K$ is connected.
Apply $S^2$ version of Schoenflies theorem to the Jordan curve $\partial L$, 
we find $C$ can be contracted to a point in $L$ and hence in $S - A$.
Since $S^2 - A$ is connected (usually proved as a lemma when one proves
Jordan Curve theorem) and every loop in it can be contracted to a point, 
$S^2 - A$ is simply connected.
