3
$\begingroup$

Consider a submanifold of $\mathbb{R}^n$ that is homeomorphic to the $(n-1)$-sphere. By the Jordan-Brouwer separation theorem, removing that submanifold from $\mathbb{R}^n$ partitions what remains into two connected components, one of which is bounded. Let $B$ be the bounded component, and let $\overline{B}$ be its closure. Once $n>2$, additional assumptions are needed to ensure that $\overline{B}$ is homeomorphic to the closed $n$-ball, as discussed in the question DMiz.

I originally asked --- or, at least, I intended to ask --- whether requiring $\overline{B}$ to be contractible would imply that it was a closed $n$-ball. In fact, Bing showed that $\overline{B}$ is always contractible; so adding that requirement wouldn't change anything.

Lots of study has gone into this area recently, though mathematicians typically consider topological embeddings of the $(n-1)$-sphere in the $n$-sphere, rather than in $\mathbb{R}^n$, so that the closures of the two components can be studied on an equal footing. Those closures are called "crumpled $n$-cubes", and they can be either closed $n$-balls or "wild". Both of them can be wild. If a crumpled $n$-cube is a manifold-with-boundary, then it is an $n$-ball.

The notes of Andrew Putman that Lee Mosher suggested are helpful, and this posting on MathOverflow has clear explanations and further pointers.

$\endgroup$
3
  • 1
    $\begingroup$ Regarding the linked questions, there may be some confusion. If one requires the $(n-1)$ sphere to be smoothly embedded in $\mathbb{R}^n$, then it satisfies the "locally flat" hypothesis of the Brown-Masur theorem, by the tubular neighborhood theorem of differential topology. Therefore the conclusion you want does hold. $\endgroup$
    – Lee Mosher
    Commented Oct 23, 2017 at 16:05
  • 1
    $\begingroup$ You may have been confused by one of the comments in the linked question: the h-cobordism theorem for 3-dimensional manifolds remains unknown. However, that theorem is not necessary to prove what you want for the bounded complementary component of a smoothly embedded 3-sphere in $\mathbb{R}^4$. It's the unbounded component which could potentially give one trouble. $\endgroup$
    – Lee Mosher
    Commented Oct 23, 2017 at 16:07
  • $\begingroup$ You might also want to look at Andy Putman's notes, which cites a theorem of Bing: the closure of the bounded complementary component $C$ is always contractible. But that does not imply contractibility of $C$ itself, as is seen from the "inside-out Alexander horned sphere", so this does not answer your question. www3.nd.edu/~andyp/notes/Schoenflies.pdf $\endgroup$
    – Lee Mosher
    Commented Oct 23, 2017 at 16:12

1 Answer 1

Reset to default
1
$\begingroup$

I assume that your question is the one in the header:

Suppose that $f: S^{n-1}\to E^n$ is a homeomorphism to its image $\Sigma$ and the bounded component $B$ of $E^n - \Sigma$ is contractible. Is it true that $B$ is homeomorphic to $E^n$?

The answer to this question is already negative for $n=3$. Namely, take a wild arc $a$ in $E^3$ connecting points $p, q$, which is only wild at $q$. See the figure below:

enter image description here

Then $S^3 -a$ is simply connected and, hence, contractible (for $i\ge 2$, $\pi_i(E^3-a)=0$ by Alexander Duality plus Hurewicz theorem). However, it is not "simply connected at infinity" (since $a$ is wild at $q$) and, hence, not homeomorphic to $E^3$. Now, "thicken" $a$ by replacing it with a 3-dimensional wild ball $\bar{A}$ with a single wild point, again at $q$. The complement $B=S^3 - \bar{A}$ is homeomorphic to $S^3- a$ and, hence, contractible but not homeomorphic to $E^3$. The boundary of $\bar{A}$ is a topological sphere $\Sigma$. If you place a point, called $\infty$, in the interior $A$ of $\bar{A}$ then $B$ becomes the bounded component of $E^3- \Sigma$, where $E^3= S^3- \{\infty\}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .