# Show that these two diffeomorphisms cannot exist simultaneously

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.

• What is $M$ here? – mathcounterexamples.net Jul 5 '20 at 18:36
• @mathcounterexamples.net An arbitrary subset of $\mathbb R^d$. – 0xbadf00d Jul 5 '20 at 18:50
• I think that $M$ should be open in order to be able to define a diffeomorphism. Or a manifold? – mathcounterexamples.net Jul 5 '20 at 18:57
• @mathcounterexamples.net The diffeomorphisms are defined on the open subsets $\Omega_i$. So there's no problem. – 0xbadf00d Jul 5 '20 at 19:13
• 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 $\mathbb R^d$. – 0xbadf00d Jul 5 '20 at 19:34

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 $$\Bbb H^k$$.

• Thank you for your answer. Do you know why we can conclude that $\phi^{(1)}(M\cap\Omega)$ is open? – 0xbadf00d Jul 6 '20 at 4:26
• You need to explain your notation a lot more carefully. Your intersection with $\Bbb R^k$ comes out of nowhere. – Ted Shifrin Jul 6 '20 at 4:38
• Sorry, I don't understand what you mean. I'm intersecting with $\mathbb R^k\times\{0\}\subseteq\mathbb R^d$, where $0\in\mathbb R^{d-k}$. – 0xbadf00d Jul 6 '20 at 7:29
• 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. – 0xbadf00d Jul 6 '20 at 8:46
• Please take a look at the supplementary answer I provided. Do you agree? – 0xbadf00d Jul 12 '20 at 18:32

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 $$\mathbb H^k$$.