Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 13F / Solution

The inverse function theorem says: if $U\subseteq {\mathbb{R}}^2$ is open, $F:U\to {\mathbb{R}}^2$ is smooth, $a\in U$, and $DF(a)$ is invertible, then there are open neighbourhoods $V\subseteq U$ of $a$ and $W$ of $F(a)$ for which $F:V\to W$ is a bijection with a smooth inverse. Its derivative is $D(F^{-1})(F(x))=[DF(x){]}^{-1}$ after shrinking $V$ if necessary. A smooth function here has derivatives of every order.

For the given map,Thus its local smooth-invertibility set isA differentiable local inverse is impossible at either excluded line by the chain rule. Even a merely continuous local inverse is impossible there: at fixed $y$, the cubic first component has a strict local extremum at these $x$ values, giving equal values at two nearby $x$ coordinates and preventing local injectivity.

On the curve, $y^2=x^3-x=x(x-1)(x+1)\ge 0$. Its sign intervals give exactly $x\in [-1,0]\cup [1,\infty )$, so the two prescribed subsets cover the curve and are disjoint. For fixed $y$, the function $x^3-x$ is strictly increasing on $[1,\infty )$, starts at zero and tends to infinity. The intermediate value theorem therefore gives a unique $p(y)\ge 1$ with $p(y{)}^3-p(y)=y^2$.

At every point of this second component, $3x^2-1\ge 2$, so the inverse function theorem applies. Locally its smooth inverse evaluated at $(0,y)$ has the form $(p(y),y)$, by the proved uniqueness. These local descriptions agree on overlaps, proving $p$ is smooth on all of $\mathbb{R}$. For example,whose nonzero denominator also makes the smoothness mechanism transparent.