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If $M$ is a manifold, then $\partial(\partial M)) = \emptyset.$ I've searched this question here and I did not find any solution. I know that this problem is equivalent to show that $\partial(\partial \mathbb{H}^n)) = \emptyset,$ but I still have no idea how solve.

Thanks

L.F. Cavenaghi
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1 Answers1

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Suppose $M$ is $n$-dimensional and $x$ is an element of the boundary. Then there is an open set $U$ about $x$ and a homeomorphism $\phi \colon U \to [0,2) \times (0,2)^{n-1}$ such that $\phi(x) = (0,1,1, \ldots, 1)$. Observe that $\phi^{-1}(\{0\} \times (0,2)^{n-1} )$ is an open subset of the boundary homeomorphic to $(0,2)^{n-1}$. But since $(0,2)^{n-1}$ is open in $\mathbb R ^{n-1}$ this shows the boundary is an $(n-1)$-dimensional manifold.

Daron
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  • I know that the boundary is a manifold, what I am asking is, what about the boundary of the boundary, it has to be empty! Why? – L.F. Cavenaghi Mar 16 '16 at 23:17
  • Because a "manifold with boundary" is not a "manifold" for the same reason a "false prophet" is not a "prophet". Showing the boundary is a manifold proves it is not a manifold with boundary or equivalently it is a manifold with empty boundary. Of course that requires knowing something cannot simultaneously be a manifold with and without boundary. – Daron Mar 16 '16 at 23:37
  • I got it, you are claiming the same way as Spivak on his book Calculus on Manifolds. Ok, it is a proof. Thank you! – L.F. Cavenaghi Mar 16 '16 at 23:46
  • I'm afraid I have only read little bits of that (series of) book(s), so cannot confirm the method is the same as Spivak's. What does he do and what is he trying to prove? – Daron Mar 17 '16 at 14:41
  • He establish that a candidate to be a manifold satisfy one of each conditions, that are, the condition to be a manifold with boundary and the condition to be just a manifold (no boundary). He shows that the both conditions are exclusionary... Thanks for your answer. – L.F. Cavenaghi Mar 17 '16 at 17:44