Hello Everyone, 


I have a question about smooth embeddings of spheres into larger spheres (by sphere I mean the usual, non-exotic kind, in case that makes a difference).


Let $h : S^n \rightarrow S^{n+k}$ be a smooth embedding (with $n,k > 0$). It's always possible to extend $h$ (or any continuous map from $S^n$ to $S^{n+k}$ for that matter) to a continuous map $H : D^{n+1} \rightarrow S^{n+k}$ defined on the closed ball $D^{n+1}$. For example, let $\overline{H} : S^n \times [0,1] \rightarrow S^{n+k}$ be any null-homotopy of $h$ and then let $H$ be the induced map on the quotient $D^{n+1} = S^n \times [0,1] / S^n \times \{ 1 \}$ (sorry if I'm being pedantic). 


I'd like to know whether there exists such a continuous extension $H$ having a certain property, namely that


$H(D^{n+1} - S^n) \cap h(S^n) = \emptyset$


That is, the image of the restriction of $H$ to the ball $D^{n+1} - S^n$ should not intersect the image of the original embedding $h$. This is a weakening of the condition that $h(S^n)$ bound an embedded $n+1$-disk in $S^{n+k}$, since the restriction of $H$ to the ball $D^{n+1} - S^n$ needn't be injective.


Such an $H$ exists if $h$ is the standard embedding of $S^n$ into $S^{n+k}$, or more generally if the pair $(S^{n+k}, h(S^n))$ is homeomorphic to the standard pair $(S^{n+k}, S^n)$. 


It follows that $H$ exists when $k=1$, since in this case the Generalized Schoenflies theorem provides a homeomorphism between $(S^{n+1}, h(S^n))$ and the standard pair $(S^{n+1},S^n)$ (this is because $h$ is smooth).


I'd like to know if anything can be said about the existence of $H$ when $k>1$. Sorry for the long preamble if this is totally trivial.


Thanks!



Extending Smooth Embeddings of S^n into S^{n+k}
Your condition can be stated equivalently as the *knot longitude*, as an embedded $S^n$ in the *knot complement* $C\_h = S^{n+k} \setminus h(S^n)$ is null homotopic. 


Generally the answer is no when $k=2$:


* When $n=1$ this follows from the *loop theorem* which you can find in the book Mark Grant cites.
* For all $n$ your condition implies that the fundamental group of $C\_h$ is infinite cyclic. Conjecturally this implies the knot is smoothly trivial, when $n=2$. In higher dimensions this is not quite enough for the knot to be trivial as the complement $C\_h$ can have non-trivial higher homotopy groups. Try Googling "Classification of simple knots" with names like Kearton, Farber and Levine.


When $k>2$ my understanding is generally you can construct your continuous map $H$. For example, see Masamichi Takase *The Hopf Invariant of a Haefliger Knot*. 


