# Diferentiable inverse of a vector valued function

Let $f:\mathbb R^2\to \mathbb R^2$ be a continuously differentiable function and let there exists $(x_0,y_0)\in \mathbb R^2$ such that $\det[f^{\prime}(x_0,y_0)]=0$. I have to prove that $f$ can not have a differentiable inverse $f^{-1}:\mathbb R^2\to \mathbb R^2$.

The main issue is I do not know anything regarding global inverse. I know only that $f$ may be an invertible function even though there exists $(x_0,y_0)\in \mathbb R^2$ such that $\det[f^{\prime}(x_0,y_0)]=0$. But how to comment on the differentiability of $f^{-1}$? Any help is appreciated.

Let $g=f^{-1}$ be the invers of $f$, then $g$ is not differentiable at $f(x_0 , y_0) \in R^2$. Because otherwise if $g$ is differentiable at $f(x_0 , y_0)$ considering composition function which is identity , i.e., $gof =I$ then differentiating from both sides at $(x_0 , y_0)$ using chain rule we get:
$$g' (f(x_0 , y_0)) f' (x_0 , y_0) = I_{2 \times 2}$$
Note the determinant of left side is zero, but right side is $1$, Contradiction !