Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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? –  Ryan Budney Sep 6 '11 at 15:53
Either your question has an immediate answer, or perhaps you meant to ask about $S^{n-2} \times [-1,1] \to S^n$ ? –  Ryan Budney Sep 6 '11 at 15:54
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. :-/ –  JulianKniephoff Sep 6 '11 at 16:56
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. –  Ryan Budney Sep 6 '11 at 18:59
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. –  JulianKniephoff Sep 7 '11 at 11:54
show 1 more comment

1 Answer

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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.