Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 4 / 13F / Solution

The inverse function theorem says that a continuously differentiable map on an open subset of ${\mathbb{R}}^n$ whose derivative at a point is invertible has a continuously differentiable inverse between neighborhoods of that point and its image. The derivative of the inverse is the inverse derivative matrix.

For the given map the Jacobian matrix and its determinant areHence every point of the nonempty open sethas a local continuously differentiable inverse. At its image, the derivative of that inverse isAt the omitted points a differentiable inverse with differentiable composition cannot exist, since the chain rule would force the singular derivative to be invertible.

To locate the exceptions on the curve, substitute $x=y^2$ into its defining relation. This gives $y^6-2y^3=y^3(y^3-2)=0$. The two exceptional points are thereforeAt any other point $(a,b)$ of the curve, $f(a,b)=(a,0)$ and the local inverse exists. Because the first coordinate of $f$ is $x$, its inverse at $(t,0)$ has the form $(t,h(t))$. The function $h$ is continuously differentiable for $t$ in a sufficiently small interval $I$ containing $a$. Shrink an interval $J$ containing $b$ so that $I\times J$ lies in the inverse neighborhood and $h(I)\subset J$. A point of the curve in that rectangle maps to $(t,0)$, so injectivity forces it to be $(t,h(t))$. Conversely all these inverse images belong to the curve. This proves the requested local graph description by the inverse function theorem, rather than merely asserting an implicit-function conclusion. Differentiating the relation also gives $h^{\prime }(t)=-(t^2-h(t))/(h(t{)}^2-t)$ there.