Suppose that two compact topological manifolds with boundary have homeomorphic interiors. Can we conclude that the two manifolds are homeomorphic? What happens in the smooth category?



Manifolds with homeomorphic interiors
Gjergji Zaimi's answer gives a strong positive conclusion: the product of the boundaries with $\mathbb{R}$ are necessarily homeomorphic. I just want to add a couple of explicit examples illustrating that the boundaries themselves need not be homeomorphic.


The first example is given by Milnor in his article "[*Two complexes which are homeomorphic but combinatorially distinct*](http://www.maths.ed.ac.uk/~aar/papers/milnhaup.pdf)", partially stated there as theorem 3. Consider the 3-dimensional lens spaces $L\_1=L(7,1)$ and $L\_2=L(7,2)$. In that article, Milnor proves (section 2) that for any $n>1$, the interior of $L\_1\times D^{2n+1}$ is diffeomorphic to the interior of $L\_2\times D^{2n+1}$. However, Milnor also proves (at the end of section 4) that their boundaries, $L\_1\times S^{2n}$ and $L\_2\times S^{2n}$, have different Reidemeister torsions. Consequently, the boundaries cannot be homeomorphic, by the topological invariance of Reidemeister torsion.


To obtain other examples, we can apply the result stated by Oscar Randall-Williams' in his comment to Gjergji Zaimi's answer. It suffices to find $h$-cobordant closed manifolds $M$, $N$ of dimension greater than 3 which are not homeomorphic: then $M\times I$, $N\times I$ will have diffeomorphic interiors by Oscar's argument, but their boundaries are not homeomorphic. As Milnor proves in the article cited above, we can take $M=L\_1\times S^{2n}$ and $N=L\_2\times S^{2n}$ for $n>0$. I wrote a more detailed [discussion concerning examples of this type](https://mathoverflow.net/questions/125935/cancellation-law-for-mn-times-mathbb-r-nn-times-mathbb-r/125961#125961) in an answer to a related mathoverflow question.


