This is a visualization, and proof-by-picture, of the counterexample to the Jacobian conjecture recently found by Fable (prompted by Levent Alpöge). $\Gamma$ is a standard parabola and $R$ is a rescaled and flipped parabola (the rescaling factor is not essential, any rescaling will do). The counterexample to the Jacobian conjecture is the function $(\pi,\kappa,\sigma)\mapsto(e_1,e_2,e_3)$, which can be accessed in “Map” mode. However, the construction is clearer by starting in “Point” mode, and we suggest starting there first. Please expand the corresponding section below for explanation.
Personally, this suggests to me how Fable stumbled upon the example, since it is good at IMO problems. I suspect that if this Jacobian problem had been asked in 1830 rather than 1930, it would have been solved long ago, since mathematicians in the 1800s were much better at this sort of thing than those of the 20th century.
Drag $x$ and $Q$. $\Gamma(x)$ is the point of $\Gamma$ so that its tangent line $D_x$ intersects the horizontal axis at $x$; $h(x)$ is obtained by rescaling and flipping $\Gamma(x)$ onto $R$. One computes $\pi(x,Q)$ by looking at the other intersection (called $\rho$) of the line $\ell_{h(x),Q}$ through $Q$ and $h(x)$ with $R$, and projecting to the horizontal axis (labelled $S$). $R$ is precisely the family of intersections of the lines $E_x$ and $D_x$.
This fixes the value of $\pi$, which determines $\rho$. Then $Q$ determines $x$, unless $Q=\rho$, in which case $x$ moves freely and determines the line $\ell_{h(x),Q}$.
Let $U$ denote the region of points where $Q$ lies on neither $E_x$ nor $D_x$. You can see that $Q$ leaves $U$ when $Q$ crosses $R$, which is the case when $Q$ lies on $E_x$; or when $Q$ crosses the dashed line $L_\rho$, which is the line where $\ell_{h(x),Q}$ and $D_x$ agree, i.e. when $Q$ lies on $D_x$.
The set of points $(x,Q)$ in $U$ with a fixed value of $\pi$ (or equivalently, $\rho$) turns out to be parameterized by a pair of numbers $(\sigma,\kappa)$. $\sigma$ is essentially the slope of the line $\ell_{h(x),Q}$ relative to that of the line $L_\rho$. $\kappa$ is the value of the cross ratio of 4 points on the line $\ell_{h(x),Q}$, two of the points being $Q$ and $h(x)$ and the remaining points a pair of auxiliary points determined by the geometry (read the caption for more details and visualizations).
From the perspective of elementary projective geometry, the precise construction of $(\sigma,\kappa)$ is unimportant. Recall that $\mathbf A^n$ is affine $n$-space (which you can think of as $\mathbf C^n$ if you are not an algebraic geometer). We write $\mathbf P^1$ for the projective line (the usual line with an additional point at infinity), and $\mathbf P^2$ is the projective plane (the usual plane with an extra point at infinity for every direction points can go off to infinity to). The basic point is that for fixed $\pi(x,Q)$, the set of valid points $(x,Q)$ in $U$ with the given value of $\pi$ is (since $Q$ determines $x$, except when $Q=\rho$) the blow-up of the projective plane $\mathbf P^2$ at $\rho$, minus the regions where $Q$ is on $R$ or where $Q$ is on $L_\rho$. Removing $L_\rho$ turns this blow-up into $\mathbf P^1\times\mathbf A^1$ (here $\mathbf A^1$ is the algebro-geometric name for $\mathbf C$) and removing $R$ is removing a section of the latter over $\mathbf A^1$ (since every line through $\rho$ intersects $R$ in exactly one other point), making the remaining points a copy of $\mathbf A^1\times\mathbf A^1=\mathbf A^2$. Thus $U$ is a family of $\mathbf A^2$s parameterized by points $\pi$ of $\mathbf A^1$, which has to be a copy of $\mathbf A^1\times\mathbf A^2$. The particular choice of $\kappa$ makes this explicit (read the “$(\sigma,\kappa)$ chart” caption for more details).
Move input coordinates to compute output coordinates, or vice versa. Caution: for some special values of the output coordinates there will be no input coordinates. If an issue arises, click “Reset defaults”. The input coordinates show $(\pi,\sigma,\kappa)$. These determine $(x,Q)$; these determine the outputs as follows. In “The divisor”, the plot shows three points in the complex plane. The point $x$ is $x$ as earlier, forced to lie on the real axis in this visualization since $x$ is always a real number. The points $(q_1,q_2)$ are the pair of complex roots of the real polynomial $z^2-Sz+P$, where $(S,P)$ are the coordinates of $Q$. The coefficients of the polynomial $(z-x)(z-q_1)(z-q_2)$ are $e_1,e_2,e_3$ (up to sign). In general, all quantities should be allowed to be complex numbers, but this is unhelpful for visualization.
Write a degree-two divisor $y+z$ on the affine line $\mathbf A=\mathbf P^1\setminus\{\infty\}$ in its symmetric coordinates $S=y+z$, $P=yz$. This identifies $\operatorname{Sym}^2(\mathbf A)$ with $\mathbf A^2_{S,P}$, and it is the plane you are looking at. The Veronese conic $\Gamma=\{2x\}$ becomes the parabola $$\Gamma:\ P=\tfrac{S^2}{4},\qquad \Gamma(x)=g(x)=(2x,\,x^2).$$ The $S$-axis is the locus $P=0$, so the point $(x,0)$ is the divisor $0+x$; that is where the draggable $x$ sits. It is also the foot of the tangent, since setting $P=0$ in $D_x:\ P=xS-x^2$ returns $S=x$. So $D_x$ runs from $x$ on the axis through $\Gamma(x)$, and the doubling $x\mapsto 2x$ is visible along it. The point $p_\infty=g(\infty)$ is not in this chart: it is the point at infinity in the vertical direction, which is why every line of the pencil $E_x$ appears vertical.
The ramification divisor $D$ meets the fiber over $x$ in the tangent line to $\Gamma$, and the hyperplane pullback $E$ meets it in the zero-sum locus: $$D_x:\ P=xS-x^2,\qquad E_x:\ S=-x.$$ So the $D_x$ sweep out all tangents to $\Gamma$ while the $E_x$ are the complete pencil of lines through $p_\infty$. They meet at $$h(x)=x+\alpha(x)=(-x,\,-2x^2),\qquad \alpha(x)=-2x,$$ and as $x$ varies this traces the auxiliary conic $R:\ P=-2S^2$. The arrow $\mu$ in the plot is the map $g(x)\mapsto h(x)$, namely $(S,P)\mapsto(-S/2,\,-2P)$.
Given $(x,Q)\in U$ we have $Q\neq h(x)$, so the line $\ell_{h(x),Q}=\overline{h(x)Q}$ is defined. It meets $R$ at $h(x)$ and at one residual point $\rho(x,Q)$; projecting that from $p_\infty$ — which in this chart is simply reading off its $S$-coordinate — gives $$\pi(x,Q)=S_{\rho}\in\mathbf A^1.$$ That projection is the dashed segment in the plot, dropping from $\rho$ straight down to the $S$-axis. Concretely, if $\ell_{h(x),Q}$ has slope $m$ then the two intersections with $R$ have $S$-coordinates summing to $-m/2$, so $S_\rho=-\tfrac m2+x$.
Point. You choose $x$ and $Q$; the applet computes $\rho$ and $\pi$. This is the construction as written.
Fiber. You fix the value of $\pi$ and move $Q$; the applet inverts the construction to recover $x$. The configurations you can reach this way are precisely the fiber $\pi^{-1}(\rho)$, which the paper identifies with $\mathbf A^2$: it is $\operatorname{Bl}_\rho\mathbf P^2$ with the proper transform of $R$ and one ruling deleted. Switching modes leaves the picture unchanged, since each mode hands the other the quantity it derives.