# Let $U$ be a simply connected open subset of ${\Bbb S}^2$, is the complement of $U$ also simply connected?

I was looking into particular cases for the Poincaré-Bendixson theorem and I came across a topological problem about simply connectivity.

If $$\gamma$$ is a Jordan curve in $${\Bbb S}^2$$ then using Jordan-Schoenflies, we have that $${\Bbb S}^2\setminus \gamma = U\sqcup V$$ with $$U$$ and $$V$$ being simply connected (s.c.). Moreover, as the sphere minus $$\gamma$$ is homeomorphic to the sphere minus the equator, we also get that $$\overline{U}$$ and $$\overline{V}$$ are s.c.

1. Let $$U$$ be a s.c. open subset in $${\Bbb S}^2$$. We note that $$\overline{U}$$ need not be s.c. (take an open annulus where you make a transversal cut). On the other hand, the complement $$F=S^2\setminus U$$ appears to me to be s.c.

2. $${\rm Int} F$$ need not be connected, but a connected component of $${\rm Int} F$$ appears to me again to be s.c.

Do the above two claims hold in general? Or have I missed some obvious counter-examples?

$$F=S^2\backslash U$$ need not be path connected, e.g. $$F$$ could be homeomorphic to the closed topologists sine curve. However, every path component of $$F$$ must be simply connected.
By identifying $$U$$ with the open unit disk (Riemann mapping), you can realize the compact set $$F=S^2\backslash U$$ as an intersection $$\bigcap_{n\in\mathbb{N}}V_n$$ where each $$V_n$$ is homeomorphic to the closed unit disk. This implies that $$F$$ has trivial shape, i.e. is cell-like.
The path-components of a cell-like continuum don't have to be contractible, e.g. if $$F$$ is the Knaster buckethandle continuum. However, it is known that every cell-like subset of a 2-dimensional manifold is simply connected. See Corollary 6 of:
Various parts this paper could allow you to verify a positive answer to both of your questions without appealing to shape theory. For instance, Lemma 13 is a particularly handy result that would imply that once you know each path-component of $$F$$ is simply connected, then every component of $$int(F)$$ is simply connected.