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Jun 20, 2015 at 15:05 comment added Danny Ruberman The N that you construct is $S^{d+1}$ in general. For N is really $\Delta \cup -\Delta$, which is the boundary of $\Delta \times I$. Puncturing this gives an h-cobordism between N and a sphere, which is a smooth product by your dimension hypothesis. This argument breaks down if d=3, but (starting with Mazur's original construction) there are lots of homology 3-spheres that bound contractible manifolds, and for all that I know, the double N is $S^4$.
Jun 20, 2015 at 15:04 comment added foliations Thanks! Do you know what can happen for $d=3$ (i.e. $n=4$ in the original question)?
Jun 20, 2015 at 15:02 vote accept foliations
Jun 20, 2015 at 14:54 history edited Oscar Randal-Williams CC BY-SA 3.0
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Jun 20, 2015 at 14:44 history answered Oscar Randal-Williams CC BY-SA 3.0