Local invertibility of

Question

Please check my understanding.
(a) Let $f:\mathbb{R^2}\to\mathbb{R^2}$ defined by $F(x,y)=(x^2-y^2, 2xy)$. Calculate derivative matrix of $F$ and show $F$ is locally invertible except possibly at the origin.
(b) Let $U \subset{\mathbb{R^2}}$ be an open set containing the point $p=(1,1)$ chosen so that the restriction of $F$ to $U$ is one-to-one and let $G:F(U) \rightarrow{U} $ denote the inverse of $F$. Calculate the derivate matrix of $G$ at $F(p)$.

My attempt:
(a) By derivate matrix is it meant the partial derivatives in a Jacobian, as below?
$ J = \left| \begin{align} & 2x & -2y \\ & 2y & 2x\end{align} \right| $
How do I prove this function is locally invertible except possibly as the origin?
(b) Should I find inverse, then calculate Jacobian and then plug in the $p$ values?
I appreciate your help.

Answer 1

Both parts are applications of the inverse function theorem. I'll take for granted that you know why $F$ is $C^1(\mathbb{R}^2)$.

(a) Your calculation of the Jacobian matrix is correct. Note that its determinant is $4x^2 + 4y^2$. For $(x,y) \ne (0,0)$ we can conclude that $J$ is invertible. Applying the inverse function theorem tells us that at every point other than the origin we are guaranteed that $F$ is locally invertible.

(b) Again, we apply the inverse function theorem. From (a), at $p = (1,1)$, we are guaranteed that there exists some neighborhood $U$ about $F(p)$ for which there exists an inverse $G: F(U) \to U$. We are also told that $J_G(F(p)) = J_F(p)^{-1} = \begin{equation*} \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \end{equation*}^{-1} =$ $\begin{equation*} \begin{bmatrix} 1/4 & 1/4 \\ -1/4 & 1/4 \end{bmatrix} \end{equation*}$