Let $f:U \subset \mathbb{R}^n \to \mathbb{R}^n$, $U$ open, be a diffeomorphism, i. e. $f$ is bijective and $f,f^{-1}$ are continuously differentiable. Why does the Jacobian $f'(x)$ has full rank for all $x \in U$?
(It's clear that if $f$ is continuously differentiable and $f'(x) \neq 0$ then f is locally invertible (inverse function thm.))
Best Answer
Suppose $f$ and $f^{-1}$ are smooth. Then $f^{-1}\circ f = \mathrm{id}$, and the chain rule shows that \begin{align} \forall x \in U~ \mathrm{d}f^{-1}(f(x))\circ \mathrm{d}f(x) = \mathrm{id} \end{align} thus, $\mathrm{d}f(x)$ is invertible.