John Carlos Baez on Nostr: - it's quite easy to check the counterexample using a computer: you take the function ...

- it's quite easy to check the counterexample using a computer: you take the function

𝐹(𝑥,𝑦,𝑧) =
(𝑦²(3𝑥𝑦+4)(𝑥𝑦+1)+𝑧(𝑥𝑦+1)³, 3𝑥𝑦²(3𝑥𝑦+4)+3𝑥𝑧(𝑥𝑦+1)²+𝑦,2𝑥−𝑥³𝑧−3𝑥²𝑦)

take its matrix of partial derivatives ∂𝐹ᵢ/∂𝑥ⱼ, show this has determinant of -2 everwhere, and check that 𝐹 is not one-to-one because

F(0, 0, -1/4) = F(1, -3/2, 13/2) = F(-1, 3/2, 13/2) = (-1/4, 0, 0).

A bunch of us have done it by now.

https://mathstodon.xyz/@gregeganSF/116950448993083825

quoting
note1hag…8dm2

The Jacobian conjecture is false!

The Jacobian conjecture proposed that if you have a map F:C^n→C^n with polynomials as the components, and the determinant of the matrix of partial derivatives of F is a non-zero constant, then F will have a polynomial inverse.

https://en.wikipedia.org/wiki/Jacobian_conjecture

But someone has now found a counterexample. This is easy to check, so there’s no doubt that it’s correct.

https://x.com/__alpoge__/status/2079028340955197566

𝐹(𝑥,𝑦,𝑧) =
(𝑦²(3𝑥𝑦+4)(𝑥𝑦+1)+𝑧(𝑥𝑦+1)³, 3𝑥𝑦²(3𝑥𝑦+4)+3𝑥𝑧(𝑥𝑦+1)²+𝑦,
2𝑥−𝑥³𝑧−3𝑥²𝑦)

F has a constant Jacobian determinant of -2, but it cannot have an inverse because it is not one-to-one:

F(0, 0, -1/4) = F(1, -3/2, 13/2) = F(-1, 3/2, 13/2) = (-1/4, 0, 0)