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Mar 30, 2013 at 15:17 vote accept Igor Khavkine
Mar 30, 2013 at 15:17 comment added Igor Khavkine Having occasionally thought about this question some more, I think this answer is quite close to what I was looking for. In particular, if the hypotheses are strengthened such that the two possible boundaries are $s$-cobordant (rather than just $h$-cobordant), then the two boundaries are not just diffeomorphic, but one can be transformed into the other by a sequence of blow-up and collapse simple homotopy moves. en.wikipedia.org/wiki/…
Nov 24, 2011 at 0:01 comment added Igor Khavkine I see a finer relationship between $\partial\bar{M}_i$ than the mere existence of an $h$-cobordism. If $\mathcal{F}$ is a closed filter in the interior of the $\bar{M}_i$ (identified via $\psi$) that converges to a closed set $F_0$ in $\partial\bar{M}_0$, then it must converge to another closed set $F_1$ in $\partial\bar{M}_1$. Thus, closed sets in $\partial\bar{M}_0$ can be put into correspondence with closed sets in $\partial\bar{M}_1$. How complicated could the correspondence be? Could it be induced by a homeo- or diffeomorphism up to expansion or collapse of some points (simple homotopy?)?
Nov 23, 2011 at 23:50 comment added Igor Khavkine Thanks! This is very close to the kind of information I was looking for. I guess I should try to understand the properties of $h$-cobordisms better now. It would be ideal for my purposes if a diffeomorphism $\psi$, like in your answer, would induce by continuity a homeo- or diffeomorphism $\partial\bar{M}_0\to\partial\bar{M}_1$, but that's obviously impossible. In fact, I think $\psi$ does not even always extend to a map. But I'd like to see something as close to such an extension of $\psi$ as possible.
Nov 23, 2011 at 17:25 history answered Oscar Randal-Williams CC BY-SA 3.0