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. 


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*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.

*${\rm Int} F$ 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? 
 A: $F=S^2\backslash U$ 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:
H. Fischer, A.Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676.
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.
