TITLE: Does every embedded 2-sphere in $\mathbb{R}^n$ bound an embedded ball?
QUESTION [9 upvotes]: Fix $n \geq 3$ and let $S \subset \mathbb{R}^n$ be a smoothly embedded $2$-sphere.  Must there exist a smoothly embedded $3$-ball $B \subset \mathbb{R}^n$ such that $\partial B = S$?  This is true for $n=3$ by the Schoenflies theorem.  Also, it is true for $n \geq 7$; indeed, since $\mathbb{R}^n$ is contractible we can find a continuous map $B \rightarrow \mathbb{R}^7$ taking $\partial B$ homeomorphically onto $S$, and by Whitney's theorem we can jiggle this continuous map a little to make it an embedding.  This might also work for $n=6$, but the homotopy to make $B$ embedded might be complicated and mess up $S$ (I don't know the details of Whitney's strong embedding theorem), so I'm not sure.
What about $4 \leq n \leq 6$?

REPLY [11 votes]: Haefliger proved that $S^k$ in $\mathbb{R}^n$ is unknotted if $2n > 3(k+1)$. Therefore, any 2-sphere embedded in $\mathbb{R}^n$ with $n\ge 5$ is unknotted. On the other hand, as Mark Grant points out in his comment, there are knotted 2-spheres in $\mathbb{R}^4$ (see this paper by Andrews and Curtis, for example).