Let $U, V ⊆ \mathbb R^n$ be open sets and $f : U → V$ a differentiable bijection with a differentiable inverse. Show that $\det[D f (x)] \neq 0$ for any $x ∈ U$.
This shows that the converse of the Inverse Function Theorem holds, but how should I prove it? Thank you.
Since $f$ is invertible and say its inverse is $g$ then we see that \begin{align} g(f(x)) = x \end{align} which means \begin{align} Dg(f(x))Df(x) = 1. \end{align} Likewise, you have that \begin{align} f(g(y)) = y \end{align} where $x=g(y)$ and \begin{align} Df(g(y))Dg(y)=Df(x)Dg(f(x)) = 1. \end{align} Hence $Df(x)$ is invertible which means $\det [Df(x)] \neq 0$.