What shown below is a reference from Analysis on Manifolds by James Munkres.
So using the last lemma I ask to prove that the boundary and the interior of a manifold are disjoint. In particular I attempted to show this using reductio ad adsurdum but it seems I didn't be able to do this. Anyway reading the above definition it seems that the boundary and the interior are disjoint by them own definition. Indeed there are two possible cases: either there exist a coordinate patch $\alpha$ about $p$ with domain an open set in $\Bbb R^k$ or there not exist a such coordinate patch. Perhaps is my last argument incorrect? So could someone help me, please?
Best Answer
There is definitely something to show: A priori it might be that for a given point $p\in M$ there exist two different charts - one that takes $p$ to the boundary of $\mathbb{H}^k=[0,\infty)\times \mathbb{R}^{k-1}$ and one that takes it to the interior. We have to show that this is impossible.
Suppose we had such charts, denoted with $\varphi_i:U\rightarrow \mathbb{H}^k$ ($i=1,2$) for which $\varphi_1(p)\in \mathrm{int}\mathbb{H}^k$ and $\varphi_2(p)\in \partial\mathbb{H}^k$. Then there would be a small neighbourhood $V\subset \mathrm{int}\mathbb{H}^k$ around $x=\varphi_1(p)$ such that the composition $f = \varphi_2\circ\varphi_1^{-1}\vert_V:V\rightarrow f(V)\subset\mathbb{H}^k$ was a diffeomorphism onto its image. Then $f(x)\in \partial \mathbb{H}^k$, which is forbidden by the following Lemma:
Lemma. Suppose $V\subset \mathrm{int}\mathbb{H}^k$ is open and $f:V\rightarrow f(V)\subset \mathbb{H}^k$ is a diffeomorphism onto its image. Then $f(V)\cap \partial \mathbb{H}^d = \emptyset$.
This Lemma, depending on your point of departure, is definitely non-trivial. E.g. in Lee's smooth manifold book it is Excercise 7.7 and follows from the earlier Proposition 7.16, which asserts that $f$, viewed as map into $\mathbb{R}^k$, is open. But $f(V)$ is only an open subset of $\mathbb{R}^k$ if it does not intersect $\partial \mathbb{H}^k$.
Another way of proving the Lemma is using algebraic topology - and this is what I alluded to in my comment about 'poking holes'. This might be a somehwat heavy gun for the situation at hand, but has the advantage that one can relax the requirement from $f$ being a diffeomorphism to it merely being a homeomorphism. The idea is as follows: Without loss of generality you can assume that $V$ is an open ball around $x$, but then $f\vert_{V\backslash x}:V\backslash x \rightarrow f(V)\backslash f(x)$ is a homeomorphism between a space which is homotopic to $S^{k-1}$ and a space which is contractible. But this is impossible, as one can check e.g. by looking at homology groups.