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Let $d\in\mathbb N$, $x\in M\subseteq\mathbb R^d$ and $\psi^{(i)}:\Omega_i\to\psi^{(i)}(\Omega_i)$ be a diffeomorphism with $x\in\Omega_i$, $$\psi^{(1)}(M\cap\Omega_1)=\psi^{(1)}(\Omega_1)\cap(\mathbb R^k\times\{0\})\tag1,$$ $$\psi^{(2)}(M\cap\Omega_2)=\psi^{(2)}(\Omega_2)\cap(\mathbb H^k\times\{0\})\tag2,$$ where $\mathbb H^k:=\{u\in\mathbb R^k:u_k\ge0\}$, and $\psi^{(2)}_k(x)=0$.

I want to conclude that both $\psi^{(i)}$ cannot exist simultaneously.

Let $\Omega:=\Omega_1\cap\Omega_2$. One argument that I found started with observing that $\phi^{(1)}(M\cap\Omega)$ is open (in $\mathbb R^d$). But I don't get that. Why is that necessarily the case? By definition of a diffeomorphism, all we know should be that $\Omega_i$ and $\psi^{(i)}(\Omega_i)$ are open.

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  • $\begingroup$ What is $M$ here? $\endgroup$ – mathcounterexamples.net Jul 5 '20 at 18:36
  • $\begingroup$ @mathcounterexamples.net An arbitrary subset of $\mathbb R^d$. $\endgroup$ – 0xbadf00d Jul 5 '20 at 18:50
  • $\begingroup$ I think that $M$ should be open in order to be able to define a diffeomorphism. Or a manifold? $\endgroup$ – mathcounterexamples.net Jul 5 '20 at 18:57
  • $\begingroup$ @mathcounterexamples.net The diffeomorphisms are defined on the open subsets $\Omega_i$. So there's no problem. $\endgroup$ – 0xbadf00d Jul 5 '20 at 19:13
  • $\begingroup$ Just to be sure: When I say that $\psi^{(i)}:\Omega_i\to\psi^{(i)}(\Omega_i)$ is a diffeomorphism, this implicitly involves that $\Omega_i$ and $\psi^{(i)}(\Omega_i)$ are open subsets of $\mathbb R^d$. $\endgroup$ – 0xbadf00d Jul 5 '20 at 19:34
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This has been asked and answered before on MSE.

Why can there not be a diffeomorphism $\phi$ from an open neighborhood of a boundary point—say the origin—in $\Bbb H^k$ to an open neighborhood of a point in $\Bbb R^k$? View $\Bbb H^k$ as a subset of $\Bbb R^k$, and suppose $\phi(0)=a\in\Bbb R^k$. By the inverse function theorem, $\phi^{-1}$ maps some open neighborhood $U$ of $a$ onto a open neighborhood of $0\in\Bbb R^k$. (This follows from the fact that $d\phi^{-1}(a)$ is nonsingular.) So its image cannot be contained in $\Bbb H^k$.

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  • $\begingroup$ Thank you for your answer. Do you know why we can conclude that $\phi^{(1)}(M\cap\Omega)$ is open? $\endgroup$ – 0xbadf00d Jul 6 '20 at 4:26
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    $\begingroup$ You need to explain your notation a lot more carefully. Your intersection with $\Bbb R^k$ comes out of nowhere. $\endgroup$ – Ted Shifrin Jul 6 '20 at 4:38
  • $\begingroup$ Sorry, I don't understand what you mean. I'm intersecting with $\mathbb R^k\times\{0\}\subseteq\mathbb R^d$, where $0\in\mathbb R^{d-k}$. $\endgroup$ – 0xbadf00d Jul 6 '20 at 7:29
  • $\begingroup$ I guess I've figured it out. If we denote the canonical projection of $\mathbb R^d\cong\mathbb R^k\times\mathbb R^{d-k}$ to $\mathbb R^k$ by $\pi$ and $N_1:=\Omega_1\cap M$, then $U_1:=\pi(\psi^{(1)}(\Omega_1)\cap(\mathbb R^k\times\{0\}))$ is open and $\phi_1:=\pi\circ\left.\psi^{(1)}\right|_{\Omega_1\cap M}$ is a homeomorphism from $N_1:=\Omega_1\cap M$ onto $U_1$. Since $\Omega:=\Omega_1\cap\Omega_2$ is open, $\Omega\cap M$ is $N_1$-open and so $\phi_1(\Omega\cap M)=\pi(\psi^{(1)}(\Omega\cap M))$ is $U_1$-open and hence (since $U_1$ is $\mathbb R^k$-open) open. $\endgroup$ – 0xbadf00d Jul 6 '20 at 8:46
  • $\begingroup$ Please take a look at the supplementary answer I provided. Do you agree? $\endgroup$ – 0xbadf00d Jul 12 '20 at 18:32
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Ted Shifrin's answer, but let me rephrase the claim in a version which is more useful for my purposes and adapt Ted Shifrin's arguments for the proof:

What I was trying to show is that if $M$ is a $k$-dimensional embedded $C^1$-submanifold of $\mathbb R^d$ with boundary$^1$ and $(\Omega_i,\phi_i)$ is a $k$-dimensional $C^1$-chart$^2$ of $M$, then

  1. $\psi_2(x)\in\partial\mathbb H^k=\mathbb R^{k-1}\times\{0\}$; and
  2. $\phi_1(\Omega_1)$ is $\mathbb R^k$-open

cannot hold simultaneously.

Proof: $\Omega:=\Omega_1\cap\Omega_2$ is $M$-open and $\phi_i$ is an open map. Thus, $V_i:=\phi_i(\Omega)$ is $\mathbb H^k$-open. Assume $U_1:=\phi_1(\Omega_1)$ is $\mathbb R^k$-open. Since $U_1\subseteq\mathbb H^k$ is $\mathbb R^k$-open and $V_1\subseteq U_1$ is $\mathbb H^k$-open, $V_1$ must be $\mathbb R^k$-open. By assumption, $$f:=\phi_2\circ\phi_1^{-1}:V_1\to V_2$$ is a $C^1$-diffeomorphism from $V_1$ onto $V_2$. In particular, it is an immersion.

This will yield that $V_2$ is $\mathbb R^k$-open. In fact, let $v_2\in V_2$ so that there is a $y\in\Omega$ with $v_2=\phi_2(y)$. Let $v_1:=\phi_1(y)$. Then $f(v_1)=v_2$ and since ${\rm D}f(v_1)$ is injective, the inverse function theorem implies that there is an $\mathbb R^k$-open neighborhood $W_i\subseteq V_i$ of $v_i$ with $f(W_1)=W_2$. Since $f(W_1)\subseteq V_2$, this shows that $V_2$ is $\mathbb R^k$-open.

If now $u_2:=\phi_2(x)\in\partial\mathbb H^k$, then there is no $\mathbb R^k$-open neighborhood of $u_2$. But since $u_2\in V_2$, this implies that $V_2$ cannot be open.


$^1$ i.e. each point of $M$ is locally $C^1$-diffeomorphic to $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$.

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 $\mathbb H^k$.

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