Past exam of the options II course of the University of Oxford / 2012 / F (Multivariable Calculus)

A $C^1$ diffeomorphism between open subsets of ${\mathbb{R}}^n$ is a bijection $f:U\to V$ for which both $f$ and $f^{-1}$ are continuously differentiable. More generally, a $C^k$ diffeomorphism requires that regularity for both maps; a smooth diffeomorphism has derivatives of every order. In the $C^1$ setting used here,

The inverse function theorem states that if $f:U\to {\mathbb{R}}^n$ is $C^1$ on an open set and $Df(a)$ is invertible, then there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $f(a)$ such that $f:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse satisfiesIf $f$ is $C^k$ or smooth, the inverse has the same regularity.

The planar implicit function theorem states that if $F$ is $C^1$ near $(a,b)$, $F(a,b)=0$, and $F_y(a,b)\ne 0$, then in some product neighbourhood the equation $F(x,y)=0$ is equivalent to $y=h(x)$ for a unique $C^1$ function with $h(a)=b$. Moreover,If instead $F_x(a,b)\ne 0$, exchange the coordinates to solve locally for $x$ as a function of $y$.

The function is smooth, $f(0,0)=0$, andAt the origin both partial derivatives equal one. In particular, $f_y(0,0)=1\ne 0$, so the implicit function theorem supplies a neighbourhood in whichIn fact $h$ is smooth because $f$ is smooth.

By the implicit derivative formula,All sufficiently near points of the curve have $y=h(x)$, and the only such point with $x=0$ is the origin itself. Therefore for the punctured approach along the curve,

The chain rule givesIts determinant is $-1$, so it is invertible. The map $G$ is smooth and $G(0,0)=(0,0)$. The inverse function theorem therefore yields an open neighbourhood $U$ of the origin for which $G(U)$ is open, contains the origin, and

The inverse derivative formula, evaluated at $G(0,0)=(0,0)$, givesMultiplying the two displayed matrices in either order gives the identity element, verifying the total derivative explicitly.