0
$\begingroup$

Let $E_i$ be a $\mathbb R$-Banach space, $U_2\subseteq E_2$ be open and $f$ be a $C^1$-diffeomorphism from $B_1$ onto $U_2$.

What do we need to assume in order to conclude that $B_1$ is open?

By definition, $$f=\left.\tilde f\right|_{B_1}\tag1$$ for some $\tilde f\in C^1(U_1,E_2)$ for some $E_1$-open neighborhood $U_1$ of $B_1$ and $$g:=f^{-1}\in C^1(U_2,E_1)\tag2.$$ Now, $g(U_2)=B_1\subseteq U_1$ so that the chain rule is applicable and yields $$\operatorname{id}_{E_2}={\rm D\left(\tilde f\circ g\right)(x_2)={\rm D}\tilde f(g(x_2))\{\rm D}g(x_2)\;\;\;\text{for all }x_2\in U_2\tag3.$$ Now let $x_1\in B_1$ and $x_2:=f(x_1)$. By $(3)$, ${\rm D}g(x_2)$ has a left-inverse. Assuming $E_1$ is finite-dimensional, this implies that $$\operatorname{rank}{\rm D}g(x_2)\ge d:=\dim E_1\tag4;$$ hence $\operatorname{rank}{\rm D}g(x_2)=d$.

Now, in order to apply the inverse function theorem, we need that ${\rm D}g(x_2)$ is bijective. This will clearly hold if $\dim E_2=d$. In that case the inverse function theorem yields that $\left.g\right|_{\Omega_2}$ is a $C^1$-diffeomorphism from an $E_2$-open neighborhood $\Omega_2\subseteq U_2$ of $x_2$ onto an $E_1$-open neighborhood $\Omega_1$ of $g(x_2)=x_1$. And since $\Omega_1=g(\Omega_2)\subseteq g(U_2)=B_1$, this shows that $B_1$ is open.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.