3
$\begingroup$

Let $U \in \mathbb{R}^n$ be a open set and $f:U \rightarrow \mathbb{R}^n$ be a continuously differentiable function such that $Df(x_0)$ is an isomorphism for every $x_0 \in U$. I'm trying to use the inverse function theorem to show that $f(U)$ is a open set.

What I've got so far: we have that $det(Df(x_0)) \neq 0$ so, by the inverse function theorem, I know that there are open sets $V_{x_0} \subset U$ and $W_{x_0} \subset \mathbb{R}^n$ such that $x_0 \in V_{x_0}$, $f(x_0) \in W_{x_0}$ and the restriction $f|V_{x_0} \rightarrow W_{x_0}$ is invertible and the inverse continuously differentiable. So, $W_{x_0}$ is an open set as it is the inverse image of an open set ($U_{x_0}$) by continuous function ($f^{-1}$). But how can I prove that the all set $f(U)$ must be open?

Thank you!

$\endgroup$
8
$\begingroup$

You know that $f(V_{x_0}) = W_{x_0}$ is open, by the inverse function theorem. hence

$$f(U) = \bigcup_{x\in U} f(V_x)$$

is open.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.