Consider $M$ an $n-$manifold with boundary $\partial M$. In showing that $\partial M$ is an $(n-1)-$manifold a crucial step is not clear to me:

Let $x\in\partial M$, then there exists an open neighborhood $U_x$ of $x$ and a homeomorphism $f:U_x\to \mathbb R^n_+$ where $$\mathbb R^n_+=\{(x_1,\cdots,x_n)\;|\; x_n\ge 0\}$$ I want to know why $f(U_x\cap \partial M)$ should be $\partial \mathbb R^n_+=\{(x_1,\cdots,x_n)\;|\; x_n=0\}$ and $f(U_x\setminus U_x\cap \partial M)$ should be $\mathrm{Int}\; (\mathbb R^n_+)=\{(x_1,\cdots,x_n)\;|\; x_n>0\}$. Thank you for your help!

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    $\begingroup$ The topological case is delicate, because you basically need to know that a set $U\subseteq \Bbb R^n_+$ is homeomorphic to the $n$-dimensional unit ball only if $U$ is open in $\Bbb R^n$. This is an important result known as invariance of domain theorem. If the manifold is at least $C^1$, the proof is more elementary, and it amounts to showing that a $C^1$ homeomorphism $B^n\to U$ that sends $0\mapsto (x,0)$ must send some curve $\gamma$ with $\gamma(0)=0$ to a curve that goes out of $\Bbb R^n_+$. $\endgroup$ – user228113 Nov 24 '17 at 18:21

The definition of boundary of a manifold with boundary is exactly that. The boundary are the points such that, given any chart, are mapped to $\partial \mathbb{R}^n_+$ (to be formal, $f(U_x\cap\partial M)=f(U_x)\cap \partial \mathbb{R}^n_+$, not all $\partial \mathbb{R}^n_+$).

Then you prove that the definition of boundary points do not depend on what chart are you taking, and that is a topological argument:

Take two charts having a point $x\in\partial M$, $\phi$ and $\psi$,and take $\rho:=\phi^{-1}\circ\psi$, which is a homeomorphism (I will omit the domain and codomain). If you prove that points of $\text{int}(\mathbb{R}^n_+)$ are mapped to $\text{int}(\mathbb{R}^n_+)$ then it is proved. To simplify notation I will take $n=2$ but the idea is the same in more dimensions.

To do so suppose that $x\in\text{int}(\mathbb{R}^n_+)$ is mapped to $\partial\mathbb{R}^n_+$, take an open neighborhood of $x$, $U$, and $\rho(U)$. The idea is to find a set that divides $\rho(U)$ in two connected components but not $U$. Take a straight line $H$ (hyperplane) on $\rho(U)$ but not on $\partial\mathbb{R}^n_+$. If $\rho^{-1}(L)$ does not divide $U$ in two components we are done. If it does then take a curve/set/line $H_0$ starting at $x$ and dividing ONE of the two components of $U-\rho^{-1}(L)$ (*). Then $\rho(H_0)$ divides $\rho(U)$ into two connected components as $\bar{H_0}$ contained a point on the frontier of $U$, $H_0$ is connected and $\rho$ is a homeomorphism, but $H_0$ does not divide $U$ into two components. Absurd.

(*) E.g. if $U$ is a disc, $H_0$ could be a radius.

If you do not see clear what is the argument with $n>2$ take an inductive argument using this as a base case, or even the easier one-dimensional case (that is proven by taking connected components). If you have the result proved for some $n$ then, taking projections and slices, the result follows.

Finally this is way more easy if the manifold is at least $C^1$, using the inverse function theorem.


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