I'm proving this:
"Let $U$ open set in $\mathbb R^n$ and $f:U \to \mathbb R^n$. $f$ is $C^1$ in $U$ and $\det(D(f(x)) \neq 0$. Prove that $f(U)$ is and open set in $\mathbb R^n$"
My proof is the following:
Let $x\in U$. We can apply the Inverse Function Theorem in $x$. So, $\exists V_{x} \subset \mathbb R^n$, open set, with $x\in V_{x}$, and $\exists W_{x} \subset \mathbb R^n$, open set,with $f(x)\in W_{x}$, that $f|V_{x}: V_{x} \to W_{x}$ is bijective.
Taking $\mathfrak W = \bigcup_{(x \in U} W_{x}$, we obtain that $\mathfrak W$ is an open set, and $f(U)=\mathfrak W$. Proven.
My question is: is $f(U)=\mathfrak W$ true?
Thank you all