The boundary of a manifold has empty boundary 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.
By definition, you $\partial M$ is endowed with charts that &quot;look like&quot; an open subset $U\cap \Bbb R^{n-1}$.
By definition, you $\partial M$ is endowed with charts that &quot;look like&quot; an open subset $U\cap \Bbb R^{n-1}$.