# Derivative of an inverse map from $\mathbb{R}^2 \rightarrow \mathbb{R}^2$

Let $$f$$ be a map from $$\mathbb{R}^2 \rightarrow \mathbb{R}^2$$ be defined as:

$$f: \left( \begin{matrix}x\\y\end{matrix} \right) \mapsto \left( \begin{matrix}x+y-xy\\x^2\end{matrix} \right)$$

How do you determine the set of points in $$\mathbb{R^2}$$ such that $$f$$ is invertible near those points, and compute the derivative of the inverse map. I believe this is done using inverse and/or implicit function theorem. I get the derivative:

$$Df(x,y) = \left( \begin{matrix}1-y&1-x\\2x&0\end{matrix} \right)$$

But according to the inverse function theorem since $$f$$ continuously differentiable we have:

$$Df^{-1}(x,y) = \left[ Df(f^{-1}(x, y)) \right]^{-1}$$

and I computed $$f^{-1}(x, y) = \left( \begin{matrix}\pm \sqrt x\\ \frac{x \mp \sqrt y}{1 \mp \sqrt y}\end{matrix} \right)$$ But this inverse mapping sends one point to two so I'm not sure if I'm on the right track.

• The inverse function theorem's statement now holds for all of the points you specified: $f$ is invertible when restricted to a sufficiently small neighborhood of $(\sqrt x,\frac{x-\sqrt y}{1-\sqrt y})$, but also when restricted to a sufficiently small neighborhood of $(-\sqrt x,\frac{x+\sqrt x}{1+\sqrt y})$. The statements about the derivative of the inverse hold on both neighborhoods. Nov 8 '20 at 16:45

## 1 Answer

Take $$(a,b)\in\Bbb R^2$$. Then $$f$$ is differentable at $$(a,b)$$ and the Jacobian of $$Df_{(a,b)}$$ is$$\begin{bmatrix}1-b&1-a\\2a&0\end{bmatrix}.\tag1$$If $$a=0$$ or $$a=1$$, then $$(1)$$ is not invertible. At all other points of $$\Bbb R^2$$, $$(1)$$ is invertible and its inverse is$$\begin{bmatrix}0&-\frac1{2a}\\\frac1{1-a}&\frac{1-b}{2a(a-1)}\end{bmatrix}.\tag2$$So, at those points $$f$$ is locally invertible, and the Jacobian of $$Df^{-1}$$ at $$(a+b-ab,a^2)$$ is $$(2)$$.