The Jacobian counterexample, explained

1 · The Jacobian conjecture

The inverse function theorem is local: if a smooth map F has Jacobian determinant ≠ 0 at a point, then near that point F is invertible. The theorem says nothing by itself about the whole space: locally invertible maps can overlap themselves globally. The exponential-style map (x,y) ↦ (e^xcos y, e^xsin y) has nonzero Jacobian everywhere and hits every target infinitely often.

The Jacobian conjecture asserted that polynomial maps cannot exhibit this behavior. Precisely: if F : ℂⁿ → ℂⁿ has polynomial coordinates and det JF is a nonzero constant, then F has a global (polynomial!) inverse. In one variable it is trivially true: a polynomial derivative with no zeros is a nonzero constant (fundamental theorem of algebra), so f is linear. The conjecture claimed that the same local-to-global rigidity survives in every dimension.

About the field. The symbol ℂ means that coordinates may be complex numbers. The collision below uses only real numbers, so exact substitution certifies noninjectivity over both ℝ and ℂ.

shear s area scale = 1, always

a 3-to-1 local covering of a punctured region

Left. The polynomial shear (x,y) ↦ (x + s·y², y) has det J = 1 for every s. It bends the grid while remaining globally invertible, with polynomial inverse (x − s·y², y). Right. For the complex cube map w ↦ w³ on a ring, each of the three colored sectors maps onto the whole target ring, and the moving target point (violet) always has three preimages. No point is a fold — the map is locally 1-to-1 everywhere. But this map needs a hole in the middle. The conjecture said a polynomial map of the whole space, with constant det J, can never behave like this.

Terminology. Over a target region where every point has three preimages, the source separates locally into three layers, each mapping one-to-one onto that region. We call each layer a sheet. The three sectors in the right-hand figure are three such sheets.

Context. The conjecture is conventionally associated with Keller’s 1939 paper. Alpöge posted the explicit three-variable map below on 19 July 2026. Because the public record is new, this page calls it an announced counterexample and independently checks the two decisive identities.

2 · The counterexample

Here is the map, exactly as announced. Write u = 1 + xy:

F(x, y, z) = ( u³z + y²u(4+3xy), y + 3xu²z + 3xy²(4+3xy), 2x − 3x²y − x³z )

These are three integer-coefficient polynomials with ordinary total degrees (7, 6, 4). The disproof is just two finite certificates:

1 · Constant Jacobian

det JF(x,y,z) = −2 for every input.

The inverse function theorem therefore supplies a local inverse around every finite point. Nothing is crushed or folded infinitesimally.

2 · Explicit noninjectivity

F(0,0,−¼) = F(1,−3/2,13/2)
= F(−1,3/2,13/2) = (−¼,0,0)

Three distinct inputs share one output. A function with a global inverse cannot do that.

These two identities disprove the conjecture: the first satisfies its hypothesis and the second contradicts the existence of a global inverse. There are no limits or numerical approximations in the certificate. Appending untouched coordinates extends the example to every dimension n ≥ 3. No claim is made here about separate downstream conjectures.

Check the certificate yourself in 30 seconds

Any computer algebra system will do; this exact SymPy check uses rational arithmetic:

import sympy as sp

x,y,z = sp.symbols('x y z'); u = 1+x*y

F = sp.Matrix([u**3*z + y**2*u*(4+3*x*y), y + 3*x*u**2*z + 3*x*y**2*(4+3*x*y), 2*x - 3*x**2*y - x**3*z])

print(sp.factor(F.jacobian((x,y,z)).det()))  # -2

print(F.subs({x:0,y:0,z:-sp.Rational(1,4)}).T, F.subs({x:1,y:-sp.Rational(3,2),z:sp.Rational(13,2)}).T)

3 · How fiber size can change

Suppose a map like this is generically 3-to-1, and imagine sliding the target point around. The three preimages slide too. Now ask: how could the number of preimages ever change?

Finite merging would create a critical point. A fiber can instead lose points through a nonproper sequence whose inputs diverge while their images remain bounded.

Merging is forbidden. If two distinct preimages collided at a finite point p, then arbitrarily close to p the map would identify pairs of points — contradicting the inverse function theorem at p, where det JF = −2 ≠ 0. So the sheets of this map can never touch, anywhere.

Escaping is the only exit. The number of (real) preimages can change only when a preimage runs away to infinity while its target stays bounded. Maps with that behavior are called non-proper. This is the loophole: local invertibility controls every finite neighborhood, but it does not control what happens along a sequence that leaves every bounded region. The announced map uses exactly this failure of properness; §§5–6 identify it in the fiber equation.

4 · The fiber over C=0

Where do the three colliding points actually sit? Look at the third coordinate of the map, C = 2x − 3x²y − x³z = x·(2 − 3xy − x²z). The target plane C = 0 receives exactly two pieces of source space:

the flat plane x = 0, which maps by (0,y,z) ↦ (z+4y², y, 0) — a bijection onto the plane C = 0; and
the curved surface x²z = 2−3xy, which maps by ((v+1)(v+2)/x², 2(2v+3)/x, 0) with v = xy — covering the plane C = 0 twice more.

flat sheet x = 0 curved sheet x²z = 2−3xy the three preimages of (−¼, 0, 0) drag or use arrow keys

The two source sheets over the target plane C = 0 (vertical axis compressed 3×; the curved sheet is clipped where it climbs past |z| = 9). The point (0, 0, −¼) sits on the flat sheet; (±1, ∓3/2, 13/2) sit on the two lobes of the curved sheet — all three map to (−¼, 0, 0). As x → 0 the curved sheet diverges like 2/x². The two sheets never meet at a finite point; their limiting relation occurs at infinity, as required by §3.

The collision is therefore part of an entire double layer over the plane C=0, separated from the flat layer by the 1/x² divergence. The next section treats fibers off that plane.

5 · Reduction to one variable

The announced formula has a weighted scaling symmetry: rescale (x, y, z) ↦ (λx, y/λ, z/λ²) and the outputs simply rescale as (A, B, C) ↦ (A/λ², B/λ, λC). The combinations v = xy and t = x²z do not feel λ at all. They record the underlying shape while x records scale.

In these weighted coordinates, the coefficients are generated by the single quadratic p(w)=2w−3w².

Show the exact weighted identity

v = xy, t = x²z, u = 1+v, γ = 1 − (3/2)v − (1/2)t, w = uγ,
p(w) = 2w − 3w², q(w) = w² − 2w³,
α = u + q(w)/γ², β = 1 + p(w)/γ.

Then exact polynomial identity

F(x,y,z) = ( α/x², 2β/x, 2xγ ).

The apparent divisions cancel after substitution; they are bookkeeping for the weights, not poles of the polynomial map.

The inverse problem reduces in three steps:

1 · Compress 3 → 2

From a target (A,B,C), form P = BC/4 and Q = AC²/4. The scale coordinate x disappears.

2 · Compress 2 → 1

Eliminate γ to obtain Φ(w) = wP − Q, where Φ(w) = w² − w³.

3 · Reconstruct

Each root w determines γ, then x, y, and z—one complete preimage of the original map.

Show the reconstruction formulas

γ = P − p(w), u = w/γ, v = u−1,
x = C/(2γ), y = v/x, t = 2(1−(3/2)v−γ), z = t/x².

Thus finding all preimages of a target point means solving w² − w³ = Pw − Q — intersecting one fixed cubic curve with one straight line. The line's slope is P and its height is −Q, both read off the target. Generically, the resulting cubic equation has three complex roots. Each root reconstructs one input, so the map has generic fiber degree three.

6 · The cubic fiber equation

Below, the fixed cubic y = Φ(w) = w² − w³ and the target's line y = Pw − Q. The slider moves the target along the path (A, 0, 1) in target space (so the line is at height −A/4; the tilt slider adds a B-component, tilting the line to slope B/4). Every intersection is converted — by the exact Step-3 formulas — into a genuine point of ℝ³, whose image under the original degree-7 map F is computed live and compared against the target.

For −16/27 < A < 0, the horizontal line meets the cubic three times and reconstructs three real preimages; at A=−8/27 the table gives an exact three-point fiber over (−8/27,0,1). As A approaches either endpoint, two reconstructed preimages diverge in opposite directions. At tangency, γ=P−p(w)=0, so the reconstruction formula x=C/(2γ) places the repeated cubic root at infinity rather than at a critical point of F. The finite fiber at each endpoint contains exactly one point.

Show that exact three-point fiber

(−3/2, 4/3, 104/27),
(3/4, 2(−1+√3)/3, 8(13−9√3)/27),
(3/4, −2(1+√3)/3, 8(13+9√3)/27).

Substitution sends all three points exactly to (−8/27, 0, 1).

7 · The weighted-lift construction

Everything in §5 used only three properties of the little polynomial p(w) = 2w − 3w². Reversing that derivation gives the following general construction, obtained in this workspace while analyzing the announced map (see the derivation note). Take any polynomial p and any constant c ≠ 0 with:

Scope. The algebra in this section is exact and machine-checked. Whether this construction already appears in the literature has not yet been reviewed, so no historical priority is claimed.

seed degree:

① p(0) = 0 ② p(1) = −c ③ ∫₀¹ p(w) dw = 0

Degree count. If the seed p has degree d, the inverse equation has degree d+1; on a nonempty Zariski-open set its roots are distinct and reconstruct d+1 distinct inputs.

Show the complete weighted-lift recipe

Set q by q′ = w·p′/c, q(0) = 0, let κ = p′(1)/c with κ ≠ −2, put a = −(1+κ)/(2+κ), and choose b ≠ 0. With γ = 1 + av + bt and w = uγ, define

F_p(x,y,z) = ( α/x², β/x, xγ ),
α = u + q(w)/γ², β = c + p(w)/γ.

The endpoint conditions make the apparent divisions cancel, so F_p is polynomial. Direct calculation gives det JF_p = bc, while its fibers satisfy

∫₀ʷ p(s)ds = wP − cQ.

If p has degree d, this is generically a degree-(d+1) equation. Concrete degrees 3 through 7 and an independently built degree-5 instance are checked by the repository scripts.

The announced map is the minimal member of this family

A degree-1 seed cannot exist (p = aw has ∫₀¹p = a/2 ≠ 0), and in degree 2 the three conditions force p = 2w − 3w² up to one overall scale. By the exact identity of §5, Alpöge's map is this minimal seed (post-scaled by diag(1, 2, 2)). verified Thus, within this weighted-lift family, the announced map is the first nontrivial member. That is a structural statement about this family—not a claim that no other notion of “simplest counterexample” exists.

8 · An explicit degree-four map

Choose the degree-3 seed p(w) = w − 2w³ and normalize the resulting lift to integer coefficients. The construction produces a polynomial map G with

det JG = −6, generic fiber degree = 4, component degrees = (12,11,4).

Show the integer-coefficient map G

Put u = 1 + 3xy and γ = 1 − 4xy − x²z. Then

G(x,y,z) = ( (2u + u² − 3u⁴γ²)/x²,
(1 + u − 2u³γ²)/x, xγ ).

Both apparent quotients cancel exactly; expansion gives three polynomials with integer coefficients.

It also has the following short collision certificate:

G(1, 0, 0) = G(−1, 0, 2) = (0, 0, 1)

Its fiber equation is the quartic Φ₄(w) = (w² − w⁴)/2 = wP − Q with P = BC, Q = AC²/2 — a line meeting a two-humped quartic, so up to four sheets. The target (1, 0, 1) has four preimages (two real, two complex; the quartic is w⁴ − w² − 1), and the target (−1/8, 0, 1) has four distinct real preimages.

The graph of Φ₄ has two maxima of height 1/8. At A=−1/4, the horizontal line is tangent at both maxima, and two pairs of reconstructed real preimages diverge simultaneously; the real fiber count changes from four to zero. For A<−1/4 this real slice is empty, while the generic complex fiber still has four points.

9 · Why degree four is genuinely different

The generic fiber degree—the number of preimages above a typical output—is unchanged by invertible polynomial coordinate changes. It therefore distinguishes a new map up to polynomial coordinate equivalence from a more complicated presentation of the same map.

Operation starting from F	Generic fiber degree	What it means
Polynomial coordinate changes before or after F	3	Different formulas, but coordinate-equivalent.
Iterate or sandwich copies of F	3^m	An automatic hierarchy: 3, 9, 27, …
Vary the one-variable seed p	every n ≥ 3	Produces 4, 5, 6, … as well as 3.

Because 4 is neither 3 nor a power of 3, G is not the announced map in disguised coordinates and is not an iteration of it. That is the precise inequivalence established here. The seed family also realizes every generic fiber degree at least three.

Show one seed for every generic degree n ≥ 3

p_d(w) = 2w − 3w² + w(1−w)(w^d−2 − 6/[d(d+1)]),
d = n−1 ≥ 2.

For d=2 the added term vanishes. For d≥3 this has degree d, satisfies the three endpoint and integral conditions, and has p′_d(1)=−5+6/[d(d+1)]≠−2. Taking b=c=1 in §7 therefore gives det JF_n=1 and generic fiber degree n.

For a deliberately redundant finite check, the exact counterexample atlas instantiates degrees 3 through 100. It lists a rational target and all n rational preimages for each of 98 maps, together with a standard-library exact verifier.

Generic sheet counts realized by the constructions checked in this workspace. Coordinate changes keep the count at 3; composition gives powers of 3; changing the seed fills in every integer 3, 4, 5, 6, … .

Result established here. The every-n family and the degree-four inequivalence are the main follow-on findings of this work. This is a mathematical claim, not a claim of historical priority; comparison with the literature remains incomplete.

10 · Verification & provenance

Who did what

This page names both the human responsibility and the AI assistance. AI output is part of the production history; it is not treated as mathematical evidence.

Alexis Gallagher

Prompted the follow-on analysis and exposition, and is publishing this page.

GPT‑5.6‑sol

Did substantial work on structural reverse-engineering, the follow-on construction, writing, visual design, implementation, and automated checks.

Levent Alpöge + Fable

Alpöge is credited with the counterexample; his public announcement says it was found through work with Fable. This page does not claim that its later weighted-lift analysis was their derivation.

Independent checks

Exact symbolic scripts verify the displayed identities, determinants, collisions, and extensions. Those reproducible calculations—not the identity of any human or model—support the mathematical claims.

This page's figures recompute the mathematics live: the intersection points you see are converted to points of ℝ³ and pushed through the original polynomial maps in your browser. This numerical runtime check is compared with exact precomputed fibers: running…

Announced externally

The explicit map F and its attribution to Levent Alpöge on 19 July 2026.
The primary public source is the linked announcement post.
This page says “announced” because the public record is still new.

Established exactly here

det JF = −2 and both displayed three-point fibers.
The weighted identity, cubic reconstruction, two walls, and escape mechanism.
Map G: det −6, degrees (12,11,4), and exact four-point fibers.
A weighted lift for every generic fiber degree n ≥ 3.

Not established here

Literature priority for the weighted-lift interpretation or seed family.
Whether the discoverer's derivation used this mechanism.
A classification of all maps with this scaling symmetry.
The complete nonproper-value set, including every boundary piece at C = 0.

Reproduce the exact symbolic checks

The canonical derivation is RESEARCH.md. The main checker and page-fact audit use exact rational and radical arithmetic, not floating point:

uv run --with sympy python verify.py
uv run --with sympy python verify_page_facts.py

They check the two Jacobian determinants, all displayed collisions and fibers, the seed identity, reconstruction formulas, wall fibers, and concrete higher-degree members. See the repository root for the complete supporting material.