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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.

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    Perhaps I'm misunderstanding your question, but if $S^{n-1} \times [-1,1] \to S^n$ is an embedding can't you restrict it to $S^{n-1} \times \{0\}$ for an embedding of $S^{n-1}$ in $S^n$, which has two sides as you've already pointed out?2011-09-06
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    Either your question has an immediate answer, or perhaps you meant to ask about $S^{n-2} \times [-1,1] \to S^n$ ?2011-09-06
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    I really do talk about an embedding of $S^{n-1}$ and yes, the answer is probably obvious but I'm still somehow missing something. With your first comment, @Ryan, I do see that the "collar" splits the $S^n$ _at least_ into two components, as the complement of the embedded collar lies completely inside of the two components of the complement of the embedded $S^{n-1}\times \{0\}$. But how do we assert that there are no more than two components? Again, it seems obvious, but I don't see any formal proof of the collaring not creating any additional connected components. :-/2011-09-06
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    Ah, okay, the complement of the embedded $S^{n-1}\times[-1,1]$ and the complement of the embedded $S^{n-1}\times \{0\}$ have the same homotopy-type. In the smooth case they're diffeomorphic. Try proving the complement of the embedded $S^{n-1}\times \{0\}$ and the complement of the embedded $S^{n-1}\times [-1/2,1/2]$ are diffeomorphic, and then use an isotopy extension argument to extend to the situation you're trying to deal with.2011-09-06
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    Sorry, but I'm not really well versed in differential topology... But you gave me an idea there with the homotopy type argument: Let $h\colon S^{n-1}\times [-1,1] \to S^n$ be our embedding. Wouldn't it suffice to give a deformation retraction from $S^n \setminus h(S^{n-1} \times \{0\}$ to $S^n \setminus h(S^{n-1}\times [-1,1]$ by radially pushing the two halves of the remaining collar into their respective boundary? Since the number of connected components is a homotopy invariant and using Jordans theorem, the latter space then has to have exactly two components.2011-09-07
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    You can also use Alexander duality, though that is sort of overkill.2012-01-08

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