# A chart of a $k$-dimensional submanifold maps interior points to $(\mathbb H^k)^\circ$ and boundary points to $\partial\mathbb H^k$

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 $$\mathbb R^k$$.

• At the end of your argument for 1., you wrote "in contradiction to our assumption". What is the contradiction that you have in mind? Jul 12 '20 at 15:40
• @LeeMosher The contradiction is that we assumed $x\in\partial M$ and we know that $M^\circ$ and $\partial M$ are disjoint. Jul 12 '20 at 15:44
• And how do you know that $M^\circ$ and $\partial M$ are disjoint? Jul 12 '20 at 15:48
• @LeeMosher By the argument given in this question: math.stackexchange.com/q/3746357/47771. (See my last comment below the answer.) Jul 12 '20 at 16:01
• 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. Jul 12 '20 at 16:15