> Jacobian conjecture [...] states that if a polynomial function from an n-dimensional space to itself has a Jacobian determinant which is a non-zero constant, then the function has a polynomial inverse.

> ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0)

But 1 != -1 and -3/2 != 3/2 . So it's not its own inverse. Is the conjecture that it is its own inverse or that is has an inverse?

Edit: it was worded a bit strangely, but it is saying that [ (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) ].map(F) all produce (-1/4, 0, 0). Thus it has no inverse and indeed disproves the Jacobian conjecture.

	▲	krackers a day ago \| parent \| next [-]
		I think the map sends (1, -3/2, 13/2), -> (-1/4, 0, 0) and also (-1, 3/2, 13/2) -> (-1/4, 0, 0) so it's not invertable which disprove the jacobian conjecture that polynomial maps over complex numbers with a jacobian that's non-zero are globally invertible. (Just as a note for myself, I had to think of why the fact that such jacobians are constant is a byproduct, I guess it's because of lioville's theorem implying that any polynomial over C that never hits 0 must be a constant [because the reciprocal is bounded and thus must also be a constant])
	▲	jameshart a day ago \| parent \| prev [-]
		That it has an inverse. If f(a) = f(b) for a≠b then f can't have an inverse. Suppose f has inverse g; then g(f(x)) must = x for all x. But then g(f(a)) would have to equal a, and g(f(b)) would have to equal b. But they can't, because f(a)=f(b)