A Counterexample to the Jacobian Conjecture

Some conjectures are believed because no one can find a counterexample. The Jacobian conjecture was believed because it felt like there could not be one. If a polynomial map has a Jacobian determinant that never vanishes, the inverse function theorem hands you a local inverse at every single point, and polynomials are too rigid to hide the kind of infinite folding that lets a smooth map wrap the plane onto itself. For 85 years that intuition held, with no proof and no counterexample. In July 2026 a candidate counterexample appeared: an explicit polynomial map of $\mathbb{C}^3$ that is locally invertible everywhere and yet sends different points to the same place.¹

The announcement came on July 19, 2026, from Levent Alpöge, a mathematician at Anthropic, who posted the map in a few lines and said he had found it with the AI model Claude Fable 5.² Because the claim is one explicit map, checking it takes no new theory, only substitution and a determinant. Abhishek Saha of Queen Mary University of London noted that the short post was easily verifiable, and by the next day it had been confirmed by many mathematicians.³ Formal journal peer review is a separate and slower process, still pending, but the arithmetic itself, which is all this page really shows, is not in question.

This page is not a proof sketch or a claim of priority, just a close look at the map. Its Jacobian determinant is the constant $-2$, and three different inputs share one output. Both facts are exact and were checked symbolically before anything here was drawn, so the picture illustrates the arithmetic rather than standing in for it.

What the conjecture claims

Ott-Heinrich Keller asked the question in 1939.⁴ In modern form:

Every word earns its place. Polynomial means each output coordinate is built from the inputs with addition and multiplication only, no division and no transcendental functions. The domain and codomain share the dimension $n$, so that an inverse can even be asked for and the Jacobian matrix $JF$ is square, with a determinant. And that determinant is required to be a constant, not merely nonzero: a uniform, rigid condition on the map everywhere at once.

The real version of the statement, with $\mathbb{R}^n$ in place of $\mathbb{C}^n$, is already known to be false. Pinchuk constructed a polynomial map of $\mathbb{R}^2$ with nowhere-vanishing Jacobian that is not injective, of total degree 35.⁵ The complex conjecture is the one that stayed open, and the one that carried the folklore, because over $\mathbb{C}$ the extra structure was supposed to rule such maps out.

Why it looked true

The inverse function theorem says a map is locally invertible at any point where its Jacobian determinant is nonzero. If that determinant is a nonzero constant, the map is locally invertible at every point, with no bad spots anywhere. It is tempting to conclude the map is therefore globally invertible, and that temptation is exactly the trap. As John Cook put it, being locally invertible everywhere does not imply being invertible.⁶ A map can be a perfect local bijection near each point and still fold distant regions of space onto each other, so that two far-apart inputs land on one output. Polynomials were believed too finite to do this while keeping the determinant constant. The map below does it anyway.

The map

Writing $u = 1 + xy$ to keep it readable, the counterexample is

with coordinate degrees 7, 6, and 4. Differentiate and simplify, and the $3\times 3$ Jacobian determinant collapses to

a nonzero constant, so $F$ satisfies the hypothesis of the conjecture exactly. Now take the two points

which differ already in their first coordinate, and push both through $F$. They land on the same point,

Every coordinate of $A$, $B$, and $P$ here is a real number, even though the conjecture lives over $\mathbb{C}$. That is a lucky break for drawing it: $\mathbb{C}^3$ is really six real dimensions and cannot be pictured whole, but because these points are real we can plot the honest real slice of the map in ordinary 3D. On the left is the input space with $A$, $B$, and several paths joining them. On the right is the output space: each path's image, computed by evaluating $F$ along it, is a loop that leaves $P$ and comes back to $P$.

The asymmetry in the animation is the whole argument. The input markers converge on two distinct spots, $A$ at $t=0$ and $B$ at $t=1$. Their images converge on one spot, $P$, at both ends. A map that collapses two inputs to a single output cannot be undone: from $P$ alone there is no way to know whether you started at $A$ or at $B$, so no inverse function can exist. And $F$ is worse than two-to-one. Solving $F(x,y,z) = P$ exactly returns three preimages, not two: a third point $C_0 = \left(0, 0, -\tfrac14\right)$ also lands on $P$, and generically every point in the image has three preimages.⁷

One more thing to notice right at $P$: most loops meet it at a sharp corner rather than smoothly. A loop leaves $P$ along $JF(A)$ applied to the path's starting velocity and returns along $JF(B)$ applied to its ending velocity, and those two directions generally disagree, folding the loop by as much as a right angle for the sharpest paths. (The lone straight path is the exception: there they line up, and its image glides through $P$.) It is a reminder that the hypothesis pins only the determinant of $JF$ to the constant $-2$, never the matrix $JF$ itself, which keeps turning from point to point.

This also settles a natural worry. As you add more paths, their image loops cross one another all over the output space, not only at $P$. Each such crossing is a point with two different preimages, so it is one more collision, one more small counterexample. That is not a problem, it is the rule: $F$ is three-to-one almost everywhere, so shared outputs are generic, and $A$, $B$, $P$ are special only in having tidy real coordinates worth plotting. The paths in the input space may cross too, but a crossing there means nothing: $F$ is a function, so a shared input still has exactly one image. Only coincidences in the output carry information, and those are everywhere.

When you rotate the output view, notice that the loops fan out almost within a plane. That is real: the third output coordinate $c = 2x - 3x^2 y - x^3 z$ stays small along these paths, so the flower is genuinely thin in one direction and can look like a sheet edge-on. The same sensitivity in $x$ forces the input family to be thin too, always near the $x$ direction: the shape and tilt controls fill it out as roundly as that allows and swing the thin direction around the horizontal plane, but they cannot remove it, because paths that wander far in $x$ send their images off to infinity.

Checking it yourself

None of this rests on the picture. The map is a short polynomial, and the claims are things you can verify by hand or with a few lines of exact (rational) arithmetic. Substituting the three points costs nothing:

At $A$, for instance, $xy = -\tfrac32$ so $u = 1 + xy = -\tfrac12$ and $4 + 3xy = -\tfrac12$; the first coordinate is $\left(-\tfrac12\right)^3\!\left(\tfrac{13}{2}\right) + \left(-\tfrac32\right)^2\!\left(-\tfrac12\right)\!\left(-\tfrac12\right) = -\tfrac{13}{16} + \tfrac{9}{16} = -\tfrac14$, and the other two vanish. The same happens at $B$ and at $C_0$. The determinant being the constant $-2$ is one more expansion. The demo does exactly this: the curves on the right are not drawn to a script, they are $F$ evaluated at runtime along each path, and they meet at $P$ because the arithmetic makes them.

What it would mean

Injectivity is the crux. A map with a two-sided inverse must be injective, so a polynomial map that is not injective cannot be invertible, whatever its Jacobian does. This $F$ has a constant nonzero Jacobian and is not injective. If it holds up, the Jacobian conjecture is false over $\mathbb{C}$ for $n \ge 3$. The planar case $n = 2$, the one people most wanted, remains open.⁶

It is worth sitting with how ordinary the object is. Not a degree-35 monster like the real counterexample, but a degree-7 map with small integer coefficients, the kind of thing that could have been written down decades ago. The conjecture survived 85 years not because such maps are hard to build once you know the shape to aim for, but because almost no one believed the shape existed. The lesson is the old one the inverse function theorem always warned about, now with a witness you can rotate in your hands: locally invertible everywhere is not the same as invertible.