By the Jordan Curve Theorem we know that the complement of an $S^{n-1}$ embedded into the $S^n$ has exactly two connected components.
What if -- instead of a sphere -- we embed an annulus, i.e. $S^{n-1}\times [-1,1]$ into $S^n$. Intuitively I would say that the complement of this annulus should also have two components, but I couldn't think of or find an easy proof for this statement.
Can anyone here come up with a simple solution maybe deducing that assertion from Jordans theorem as a corollary? If not: What techniques could be used to proof the statement or is it even false?
Note: One idea would be to use the Schoenflies theorem which would allow me to show that the component of the complement of the image of $S^{n-1}\times \{1\}$ that contains the image of $S^{n-1}\times \{-1\}$ is homeomorphic to an open $n$-cell and thus to $\mathbb{R}^n$, allowing me to use the Jordan Curve theorem again on that component. However, I am actually trying to proof exactly that theorem using the above statement, so I cannot use it here.