This is theorem 9.24 in Rudin, known as the inverse function theorem:
Suppose $f$ is a continuously differentiable map of an open set $E
\subseteq \mathbb{R}^n$ into $\mathbb{R}^n$, $f'(a)$ is invertible for
some $a \in E$ and $b= f(a)$. Then(a) There exists open sets $U,V$ in $\mathbb{R}^n$ with $a \in U, b
\in V$ such that $f: U \to V$ is a bijection.(b)$f^{-1}$ is continuously differentiable.
My question: Can we add the following to the conclusion of the inverse function theorem?
(c) $f'(x)$ is invertible for all $x \in U$?
Looking at Rudin's proof, I think we can, but maybe I'm not too sure.
Best Answer
Yes, (although perhaps for a smaller neighborhood than $U$). Note that $f$ is continuously differentiable. Since $f'(a)$ is invertible, $\det f'(a)\neq0$. But then, by continuity, $\det f'(x)\neq0$ when $x$ is close enough to $a$, and this means that, for those $x$'s, $f'(x)$ is invertible.