Proof: The complement of an annulus embedded in a sphere has two connected components 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.

Let $f: S^{n−1}×[−1,1]\to S^n$ be our embedding. My goal is to show $S^n-f(S^{n-1}\times\{-1,1\})$ has three connected components.
First look at the two connected components of $S^n-f( S^{n−1}×{\{−1\}})$
Take two points in one of the components. 
If they both lie in the annulus we can choose a path  in the annulus using our domain as a chart. 
Suppose both are outside the annulus, though. Then there is a path between them in   $S^n-f( S^{n−1}×{\{−1\}})$, but it may go through the annulus. There must be time the path first enters the annulus and a time it finally leaves. Moreover, we know these points must be on $f(S^{n-1}\times \{1\}$ We can then use our chart to homotope this path using radial projection to be either on $f(S^{n-1}\times \{1\}$ our completely off the annulus.
Now use small open balls and compactness to homotope the path completely off $f(S^{n-1}\times \{1\}$.
Thus, these points are still in the same connected component when we remove the boundaries of the annulus. Clearly, upon removing the boundaries of the annulus the interior is a connected component (any path would have to enter the annulus). So we have shown there are three connected components upon removing the boundaries so when we remove the interior of the annulus we are left with two.
