# Topological degree of polynomial maps.

The $\mathbb{Z}_2$ topological degree of a (non-constant) polynomial in one variable, clearly, coincides with its degree as a polynomial, $\mod 2$.

Consider further a polynomial self-mapping $F$ on $\mathbb{R}^2$, and assume it is a proper map (in case, even more generally a map in higher dimension)

Is there a short way to decide what's the parity of the topological degree of $F$?

E.g. it's odd if $F$ is an odd map, or more generally, if $F$ can be transformed into an odd map by a proper homotopy. Actually: is there a short way to understand if a polynomial map is a proper map? What about the case of a gradient map (I mean, the gradient of a polynomial $f\in\mathbb{R}[x,y]$)?

I'm somehow confident that there may be a simple criterion known, at least in $\mathbb{R}^2.$ After all, what is required is just a one-bit information (well this argument doesn't convince me either).

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(whether $\gamma^\pi$ is rational or not is also a one-bit answer) – Pietro Majer Sep 14 '10 at 18:45
Indeed, the size of the output can only give a lower bound on the complexity of the question... – Thierry Zell Sep 14 '10 at 19:23
...unless the speaker is a politician ;-) – Pietro Majer Sep 15 '10 at 15:58
To compute whether a polynomial map $F:R^2\to R^2$ is onto, I think you have to consider the closure of $Graph(F)\subset P^2\times P^2$, and project to the second factor (the map $F$ doesn't necessarily extend to $P^2$: consider $F(x,y)=(x,xy)$). If the image of $\overline(Graph(F))-Graph(F)$ in $P^2$ lies in $P^1$ at infinity, then the map should be proper. But I don't know enough algebraic geometry to make this precise (e.g. Harris only considers algebraically closed fields) books.google.com/… – Ian Agol Sep 16 '10 at 3:46

Here is an idea, I'm not 100% confident that it makes sense in all cases, but I'll try it anyway. Assuming that $f(\mathbb{R}^2)$ is 2-dimensional, the degree mod 2 of your map is the cardinality of the preimage of a generic point. If your components have degree respectively d and e, then Bezout gives you a preimage size of de.
This is a very good answer which isn't 100% right. The trouble is that Bezout gives you $de$ roots in the projective plane, so you might not get the right count at $\infty$. For example, consider $(x,y) \mapsto (x^3+y^3+x, x^3+y^3+y)$. Bezout gives $9$ complex roots, but $3$ of them are on the line at $\infty$. Still, this approach will work in a lot of cases. – David Speyer Sep 14 '10 at 19:45
Thierry -- in the affine case B\'ezout gives an inequality only: think of a map of the form $(x,y)\mapsto (x,y+f(x))$ where $deg(f)>1$. – algori Sep 14 '10 at 20:01
Thierry -- I started writing down basically the same answer when I thought about that example. It doesn't look like this has an easy soltion in terms of the degrees of the components. E.g. if the components are $(x,y^2+x)$ the map is even, but if the components are $(x,y+x^2)$, it is odd. – algori Sep 14 '10 at 20:33
Correction to my above: the three roots on the line at infinity are double roots. If you look at $(x,y) \mapsto (x^3+y^3+p(x,y), x^3+y^3+q(x,y))$ for $p$ and $q$ generic quadrics, then there should only three roots at $\infty$. – David Speyer Sep 14 '10 at 23:13