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Let $M$ be a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary$^1$, $M^\circ$ and $\partial M$ denote the manifold interior and boundary of $M$, respectively, and $(\Omega,\phi)$ be a $k$-dimensional $C^1$-chart$^2$ of $M$.

I would like to show a few very simple results$^3$:

  1. $\phi(\Omega\cap M^\circ)=\phi(\Omega)\cap(\mathbb H^k)^\circ$.
  2. $\phi(\Omega\cap\partial M)=\phi(\Omega)\cap\partial\mathbb H^k$.
  3. The restriction of $\phi$ to $\Omega\cap M^\circ$ is a $k$-dimensional interior$^4$ $C^1$-chart of $M^\circ$.
  4. Let $\pi$ denote the canonical projection of $\mathbb R^k$ onto $\mathbb R^{k-1}$ with $\pi(\mathbb R^{k-1}\times\{0\})=\mathbb R^{k-1}$. Then the restriction of $\pi\circ\phi$ to $\Omega\cap\partial M$ is a $(k-1)$-dimensional interior $C^1$-chart of $\partial M$.

Let $U:=\phi(\Omega)$.

1.: $U\cap(\mathbb H^k)^\circ$ is $\mathbb R^k$-open. If $x\in\Omega\cap\partial M$ with $\phi(x)\not\in\partial\mathbb H^k$, then there is an $\mathbb R^k$-open neighborhood $V$ of $\phi(x)$ with $V\subseteq U\cap(\mathbb H^k)^\circ$. Since $V$ is $U$-open and $\phi$ is continuous, $\tilde\Omega:=\phi^{-1}$ is an $M$-open neighborhood of $x$. Since $\phi$ is a homeomorphism from $\omega$ onto $U$, $\left.\phi\right|_{\tilde\Omega}$ is a homeomorphism from $\tilde\Omega$ onto $\phi(\tilde\Omega)=V$ and hence $x\in M^\circ$; in contradiction to our assumption.

2.: I'm not sure how to approach this. I think we need to distinguish whether $U$ is an open subset of $\mathbb R^k$ or an open subset of $\mathbb H^k$ with $U\cap\partial M\ne\emptyset$. Let $x\in\Omega$ and $u:=\phi(x)$.

Let $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$. I would like to show that $x\in\partial M$ if and only if $u\in\partial\mathbb H^k$.

Assume $U$ is $\mathbb R^k$-open. Then there is a $\varepsilon>0$ with $$B_\varepsilon(u)=\{v\in\mathbb R^k:\left\|u-v\right\|<\varepsilon\}\subseteq U.$$ If $x\in\partial M$, then there is a $k$-dimensional boundary $C^1$-chart $(\tilde\Omega,\tilde\phi)$ of $M$ with $x\in\tilde\Omega$ and $\tilde\phi(x)\in\partial\mathbb H^k$. Now we should need to assume $u\not\in\partial\mathbb H^k$ and show that this is a contradiction.

How can we do that and how do we obtain (3.) and (4.) from that?


$^1$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k$.

If $E_i$ is a $\mathbb R$-Banach space and $B_i\subseteq E_i$, then $f:B_1\to E_2$ is called $C^1$-differentiable if $f=\left.\tilde f\right|_{B_1}$ for some $E_1$-open neighborhood $\Omega_1$ of $B_1$ and some $\tilde f\in C^1(\Omega_1,E_2)$ and $g:B_1\to B_2$ is called $C^1$-diffeomorphism if $g$ is a homeomorphism from $B_1$ onto $B_2$ and $g$ and $g^{-1}$ are $C^1$-differentiable.

$^2$ A $k$-dimensional $C^1$-chart of $M$ is a $C^1$-diffeomorphism from an open subset of $M$ onto an open subset of $\mathbb H^k$.

$^3$ $(\mathbb H^k)^\circ=\mathbb R^{k-1}\times(0,\infty)$ and $\partial\mathbb H^k=\mathbb R^{k-1}\times\{0\}$.

$^4$ i.e. it is a $C^1$-diffeomorphism onto an open subset of $\mathbb R^k$.

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  • $\begingroup$ At the end of your argument for 1., you wrote "in contradiction to our assumption". What is the contradiction that you have in mind? $\endgroup$
    – Lee Mosher
    Jul 12 '20 at 15:40
  • $\begingroup$ @LeeMosher The contradiction is that we assumed $x\in\partial M$ and we know that $M^\circ$ and $\partial M$ are disjoint. $\endgroup$
    – 0xbadf00d
    Jul 12 '20 at 15:44
  • $\begingroup$ And how do you know that $M^\circ$ and $\partial M$ are disjoint? $\endgroup$
    – Lee Mosher
    Jul 12 '20 at 15:48
  • $\begingroup$ @LeeMosher By the argument given in this question: math.stackexchange.com/q/3746357/47771. (See my last comment below the answer.) $\endgroup$
    – 0xbadf00d
    Jul 12 '20 at 16:01
  • $\begingroup$ The actual answer about why $M^\circ$ and $\partial M$ are disjoint is given not in your comment to that answer, but in the answer itself (which you should accept). And that generalizes easily to answer question 2. $\endgroup$
    – Lee Mosher
    Jul 12 '20 at 16:15

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