# Open map as a corollary of the inverse function theorem

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!

You know that $f(V_{x_0}) = W_{x_0}$ is open, by the inverse function theorem. hence
$$f(U) = \bigcup_{x\in U} f(V_x)$$