# 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?

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I think the answers to this question and this MO thread should yield a positive answer. –  Martin Jan 26 '13 at 16:31

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.

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