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Let's define $H^{n} = \left\lbrace x \in \mathbb{R}^{n} : x_n \geq 0 \right\rbrace$

Trying to answer my own question on why "the boundary of the boundary of a manifold is empty", here I don't know how to conclude that the boundary is empty :

By restriction of chart I have parametrized a manifold, which is $\partial M$, with open sets of $\mathbb{R}^{n-1}$, those open are infact $U \cap \partial H^{n}$; I'd like to state that the manifold has boundary simply from this arguing by contraddiction : if it has boundary, I could find at the same time a neighborhood of a point $p \in \partial \partial M$ which is diffeomorphic to some $U \cap \partial H^{n}$ open in $\partial H^{n}$ with the endowed topology (as point of $\partial M$) and to some $U' \cap \partial H^{n-1}$, open in $\partial H^{n-1}$ with the endowed topology (as point of $\partial \partial M$) as well.

I don't how to conclude from here since it should be true that such open can't be diffeomorphic, they shouldn't be even homeomorphic, unless that the open $U \cap \partial H^{n}$ lies in $\left\lbrace x_{n-1} = 0\right\rbrace$ too. This should generate a contradiction which I'm unable to see or state, but I feel like that at that point $U \cap \partial H^n$ won't be open in $\partial H^n$ anymore.

Any help on how to formalize the senteces above would be appreciated, feels necessary to me given the freedom which the boundary has been defined, otherwise I could have just simply stopped to "the boundary has been parametrized with open sets of $\mathbb{R}^{n-1}$", but if this is true, it should be also follow that $U \cap \partial H^n$ can't lie entirely in $\left\lbrace x_i = 0\right\rbrace$, right ?

jacopoburelli
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    If they were diffeomorphic, a diffeomorphism between them would induce an isomorphism between the tangent spaces. Now look at the dimensions. – leoli1 Jan 03 '21 at 09:16
  • Arguing that they are not homeomorphic is a bit harder and requires (usually) some kind of algebraic topology. – leoli1 Jan 03 '21 at 09:17
  • @leoli1 It seems to me that this argument would look fine even in a manifold with boundary, what fails there that we have here ? – jacopoburelli Jan 03 '21 at 09:18
  • Why should it be wrong in the general case? – leoli1 Jan 03 '21 at 09:29
  • @leoli1It seems to me that you stated that a manifold can't have boundary, I don't why in the case where a manifold has boundary taking a point of the boundary I couldn't repeat this argument – jacopoburelli Jan 03 '21 at 09:30
  • My point was that if we have diffeomorphic manifolds (with or without boundary) of dimensions $n,m$ then $n=m$. Here we apply this to open (I assume non-empty) subsets of $\partial H^{n-1}$, $\partial H^n$, which we can identify with open subsets of $\mathbb{R}^{n-2},\mathbb{R}^{n-1}$. – leoli1 Jan 03 '21 at 09:34
  • @leoli1 My question is precisely this, why $\mathbb{R}^{n-2}$ could lie in ${x_{n-1} = 0}$ for example, i.e the $\mathbb{R}^{n-2}$ you're writing ? Be open there as well ? Because if that was the case such a diffeomorphism could exists – jacopoburelli Jan 03 '21 at 09:40
  • @leoli1 Could you explain ? – jacopoburelli Jan 03 '21 at 10:19
  • $\partial H^{n-1}$ is the set of tuples $(x_1,\dots,x_{n-1})$ with $x_{n-1}=0$, so by just looking at the first $n-2$ coordinates we can identify $\partial H^{n-1}$ with $\mathbb{R}^{n-2}$ (at this point this is just an identification of the underlying sets, but you can verify that the topology will also be the same) – leoli1 Jan 03 '21 at 10:26
  • Let us continue this discussion in chat. – jacopoburelli Jan 03 '21 at 10:33
  • @leoli1Yes maybe I didn't write well, was I meant was that why $\partial {H}^n$ couldn't lie in $x{n-1} = 0$ as well ? It this case a diffeomorphism could exists – jacopoburelli Jan 03 '21 at 10:48
  • A non-empty open subset $U$ of $\partial H^n$ cannot lie entirely in $x_{n-1}=0$. If you have $x\in U$ then there is some $r>0$ with $B(x,r)\cap\partial H^n\subseteq U$, in particular $x+\frac{r}{2}e_{n-1}\in U$ where $e_{n-1}$ is the vector with only one 1 at the position $n-1$. Now one of $x$ or $x+\frac{r}{2}e_{n-1}$ will have $x_{n-1}$-coordinate not equal to $0$. – leoli1 Jan 03 '21 at 12:08

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