Let $M^n$ be a smooth closed embedded hupersurface in $\mathbb R^{n+1}$. Denote by $D$ the bounded connected component of $\mathbb R^{n+1}\backslash M$. We assume that $\mathbb R^{n+1}\backslash D$ is simply connected. Let $B=B_R^n$ be the ball centered at the origin with radius $R>>1$ whose boundary is denoted by $S$.

Question: Does there exist a smooth map $$F:M\times [0,1]\rightarrow \bar B\backslash D$$ such that:

1) $F(x,0)=x,$, $f(x, 1)\in S, \forall x\in M$,

2) For any fixed $x\in M$, $F(x,\cdot):[0,1]\rightarrow \bar B\backslash D$ is injective.

The motivation of this question comes from several complex variables, where $M$ is taken to be the boundary of a bounded domain.


It seems that once you let your ball $B$ be large enough to contain $M$ with some its neighborhood $M\times[0,1]$, the question is about keeping the bottom copy $M\times \{0\}$ (boundary of $D$) fixed and deforming the top copy $M\times\{ 1\}$ to the sphere $S$. You can look at a compact surface in $\mathbb{R}^3$: choose an open disk on it and organize a flow outside of this disk so that it shrinks the complement of the disk to a point. This could be impossible to do if you want to deform staying always in $M\times [0,1] \subset \mathbb{R}^{n+1}$ but if you don't require this, it seems that it's easy.

  • $\begingroup$ NB: In the case that $M$ is a surface in $\mathbb R^3$, $\mathbb R^3\backslash D$ is not simply connected unless $M$ has genus 0. Thanks a lot! $\endgroup$ – Entaou Oct 22 '13 at 11:15
  • $\begingroup$ By the way, I don't see how being/not being simply connected can be an obstacle. $\endgroup$ – N B Oct 22 '13 at 13:36
  • $\begingroup$ It's subbtle. I am not sure, I just think that being not simply connected may cause some monodromy phenominen. $\endgroup$ – Entaou Oct 22 '13 at 14:16

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