Let $ X $ be an $ n+1 $ dimensional manifold. Suppose that $ M_1$ and $ M_2 $ are compact manifolds with boundary such that $ M_1 $ and $ M_2 $ both have interior homeomorphic to $ X $. Let $ Y_1 $ be the boundary of $ M_1 $ and $ Y_2 $ the boundary of $ M_2 $. Is it the case that $ Y_1 $ must be homeomorphic to $ Y_2 $?

As a sort of converse, suppose that $ M_1,M_2 $ are compact manifolds with boundary and moreover their boundaries are homeomorphic. Consider the interior $ X_1 $ of $ M_1 $ and the interior $ X_2 $ of $ M_2 $. Do $ X_1 $ and $ X_2 $ just differ by taking connected sum with closed $ n+1 $ manifolds? By that I mean something like: Is it the case that there exist closed $ n+1 $ manifolds $ Z_1 $ and $ Z_2 $ such that $$ X_1 \# Z_1 \cong X_2 \# Z_2 $$

For the first question, my idea for existence is that you can take any tubular neighborhood of the $ n+1 $ manifold $ X $, then embed that into some $ \mathbb{R}^k $. For any such embedding the boundary of the image of $ X $ under the embedding should give you the same $ n $ manifold $ Y $ (also $ X $ should be noncompact otherwise $ Y $ is just the empty set viewed as an $ n $ manifold).


1 Answer 1


In general, you can only conclude that the boundaries are h-cobordant, but not homotopy equivalent. See Mathoverflow here: https://mathoverflow.net/q/81714

For the second question certainly not, I can do all sorts of surgery on the interior to change the topological type (for instance to change the fundamental group in a more complicated way than just taking a free product). For instance, take any two manifolds $M_1$ and $M_2$ and delete a ball, so that they have sphere boundary; $M_1$ and $M_2$ will rarely just differ by a connected sum.

  • $\begingroup$ For the second question, though, what about something like $X_1 \# Z_1 \cong X_2 \# Z_2$ for closed manifolds $Z_1$ and $Z_2$ as suggested in the question? In the case that the $M_i$ are closed manifolds with a ball removed, you could take the $Z_i$ to be those closed manifolds (in the reverse order). $\endgroup$ Dec 29, 2019 at 16:57
  • $\begingroup$ @ Eric Wofsey Yes, ideally someone would answer that part of my question as well. If not I'll just accept the answer from @ manifold boundary $\endgroup$ Feb 6, 2020 at 13:27

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