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It is a known fact, for example well explained here, that the boundary of an $n$ manifold is an $n-1$ manifold, and the proof shows essentialy that properly restricted charts which parametrize $M$ parametrize also $\partial M$, with open set of a lower dimension, $n-1$.

My question is : I have to show that $\partial M$ has no boundary, or is implicit somehow having shown that $\partial M$ is parametrizable with open sets of $\mathbb{R}^{n-1} ?$ If so, why ?

Because for my definition of boundary I don't see how this should be implicit.

My definition is $p \in \partial M \iff p \in \phi(U \cap \left\lbrace x_n = 0\right\rbrace)$ i.e is the image of $\partial \mathbb{H}^n \cap U$, (where $U$ is the open set of $\mathbb{R}^n$ such that the intersection with closed half space $U \cap \mathbb{H}^n$ is a diffeomorphism with an open set of the manifold $V$).

How should I prove it ? It could be useful using this fact ?

I'd like to understand the answer given in this previous question, which seems the way I should prove, it but I don't understand how the fact follows easily from the definitions.

Any help would be appreciated.

jacopoburelli
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1 Answers1

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By definition, you $\partial M$ is endowed with charts that "look like" an open subset $U\cap \Bbb R^{n-1}$.

Hagen von Eitzen
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  • The set $U \cap \mathbb{R}^{n-1}$ is open in $\partial \mathbb{H}^n$, right ? But my question is, why parametrizing $\partial M$ with open set of $\mathbb{R}^{n-1}$ doesn't require to show that $\partial \partial M$ is empty ? (with the definition given above) – jacopoburelli Dec 31 '20 at 14:49
  • @jacopoburelli A neighbourhood cannot look like $U\cap \Bbb r^{n-1}$ and $U\cap \partial \Bbb H^{n-1}$ at the same time – Hagen von Eitzen Dec 31 '20 at 15:27
  • I don't think $U$ is the same; Besides, are you using that open sets of different dimension can't be diffeomorphic ? – jacopoburelli Dec 31 '20 at 16:45
  • I can use the definition of $\partial S$ as the subset of $S$ whose points don't have neighboorhoods homeomorphic to open sets in $\mathbb{R}^{n}$? Taking $S = \partial M$ we have that $\partial \partial M$ is the set of the points that don't have neighboorhoods homeomorphic to open sets in $\mathbb{R}^{n-1}$? And from here because $\partial M $ has been parametrized with open set of $\mathbb{R}^{n-1}$ is there was a point in $\partial \partial M$ we would have a contradiction ? – jacopoburelli Jan 01 '21 at 09:07