= 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,
$$
DF(x,y)=\begin{pmatrix}3x^2-1&-2y\\0&1\end{pmatrix},\qquad
\det DF(x,y)=3x^2-1.
$$
Thus its local smooth-invertibility set is
$$
\boxed{\{(x,y):x\ne\pm1/\sqrt3\}.}
$$
A 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, $y^2=x^3-x=x(x-1)(x+1)\ge0$. 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 $p(y)\ge1$ with $p(y)^3-p(y)=y^2$.
At every point of this second component, $3x^2-1\ge2$, 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 \b[$p$ is smooth on all of $\mathbb R$]. For example,
$$
p'(y)=\frac{2y}{3p(y)^2-1},
$$
whose nonzero denominator also makes the smoothness mechanism transparent.
Back to article page
