TITLE: Topological degree of polynomial maps.
QUESTION [8 upvotes]: 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).

REPLY [3 votes]: 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. 
That's projective solutions. In generic cases, you would expect no solutions at infinity. Unfortunately as David points out below, you may have an odd number of solutions at infinity.
So in the case when there's nothing at infinity, we have solutions over the complexes, but non-real solutions will come in complex conjugate pairs, so in that nice case the degree mod 2  is given by de.