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). 



Topological degree of polynomial maps.
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*.


