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Let $U$ and $V$ be two bounded open regions in $\mathbb{R}^n$, and let us further assume that their topological boundaries are nice enough that they are homeomorphic to finite simplicial complexes.

Assume $\partial U$ is homeomorphic to $\partial V$.

  1. Is $U$ be homeomorphic to $V$?
  2. A weaker question: if $U$ is contractible, is $V$ contractible?

For $n=1$, the answer to both is yes, since $\partial U\simeq \partial V$ are sets of $2i$-many points, which both divide $\mathbb{R}^1$ into $i$-many open intervals.

For $n=2$, the answer to both is yes, since $\partial U\simeq \partial V$ are sets of $i$-many topological circles, which both divide $\mathbb{R}^2$ into $i$-many open discs.

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2 Answers 2

up vote 11 down vote accepted

No for $n=2$. Consider a disk with two circular holes (red in the picture below) versus a disk inside an annulus (blue) versus three disks (green). In all cases the boundary consists of three disjoint circles.

enter image description here

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Does the disk have to be inside the annulus? Alternatively, is it sufficient to consider an annulus vs. two disks? –  Rahul Jul 30 '12 at 20:43

For $n=3$ you could consider a ball minus a knot. In all cases the boundary is $S^2 \coprod S^1$, but inequivalent knots will yield non-homeomorphic sets. These have the added advantage of being connected.

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