Question from Munkres Analysis on Manifolds Inverse Function Theorem Section

Question

This is the first exercise in the section on the Inverse Function Theorem (section 8).

Let $f:\Bbb R^2\to \Bbb R^2$ be defined by the equation $f(x,y)=(x^2-y^2,2xy)$.

a) Show that it is one to one on the set containing all $(x,y)$ such that $x>0$.
b) What is the image of the function on that set?
c) If $g$ is the inverse function, find $Dg(0,1)!$

My question is mainly about part c. Here is what I think needs to be done.

Use the fact that $g(f(x,y))=(x,y)$ and differentiate to get $D(g(0,1))\cdot Df(\frac{1}{\sqrt2},\frac{1}{\sqrt2})=I$
Then multiply by the inverse matrix of $Df(\frac{1}{\sqrt2},\frac{1}{\sqrt2})$ to get $Dg(0,1)$. Is this right?

$f(1/\sqrt2,1/\sqrt2)=(0,1)$ because from chain rule, wouldn't it actually be $Dg(f(x,y))*Df(x,y)$? So i need to find (x,y) s.t f(x,y)=(0,1) — Bobby Jones, Sep 07 '14 at 00:34
Yeah I checked it in part a. Just making sure, do we check for injectivity to prove the existence of an inverse function in the first place? — Bobby Jones, Sep 07 '14 at 01:14
Not only in the first place - the subtlety is that the Jacobian can be invertible everywhere while same time the function isn't... — C-star-W-star, Sep 07 '14 at 01:35
Think of rolling up the real line $f:\mathbb{R}\to\mathbb{S}:\alpha\mapsto(\cos \alpha,\sin \alpha)$. Here the Jacobian never vanishes so is invertible but the function itself is not injective and so never had an inverse... — C-star-W-star, Sep 07 '14 at 01:40

score 2 · Answer 1 · answered Feb 15 '16 at 00:45

We know from Theorem 7.4 that, if $g$, the inverse function of $f^{-1}$, is differentiable at a point $f(\vec{a})$, then the matrix $D_g(f(\vec{a}))$ is the inverse of the matrix $Df(\vec{a})$. So first we need to find the point $\vec{a}$ that $f$ maps to $(0,1)$ and show that $g$ is differentiable at $(0,1)$. We can then do the necessary arithmetic.

From the solution of (b) it is clear that the polar coordinates of the point that $f$ maps to $(0,1)$ are $(1,\pi/4)$. Therefore $f(\frac{1}{\sqrt2},\frac{1}{\sqrt2})=(1,0)$. Notice that $\det Df(\frac{1}{\sqrt2},\frac{1}{\sqrt2})=4\neq0$. So, by the Inverse Function Theorem, $g$ is differentiable at $(0,1)$ and we can thus apply Theorem 7.4 to find

$$D_g(0,1)=Df(\frac{1}{\sqrt2},\frac{1}{\sqrt2})^{-1}=\frac{1}{2\sqrt2}\begin{bmatrix} 1 & 1\\ -1 & 1\\ \end{bmatrix}$$

Question from Munkres Analysis on Manifolds Inverse Function Theorem Section

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