If $M$ and $N$ are oriented $n$-manifolds and $f: M \to N$ then the degree of $f$ is given by
$$
\deg f = \sum_{p \in f^{-1}(q)} sign_p f
$$
where $q$ is a regular value and the sign is $+1$ if $f$ is orientation preserving at $p$ and $-1$ if is orientation reversing at $p$. If we identify $\mathbb S^2$ with the one point compactification of the complex plane, then any polynomial in $\mathbb C[X]$ is a smooth map from $\mathbb S^2$ to itself. Does the degree of a polynomial as a map between manifolds coincide the algebraic degree? Equal up to a sign? Easily related?
It's easy to see that they coincide for $f(z)=z^n$. The fundamental theorem of algebra assures that a regular point has exactly the algebraic degree number of preimages, but can the signs of the preimages lower the mapping degree?
Best Answer
Yes, they coincide. Any polynomial (or indeed any holomorphic function) on the complex plane is orientation-preserving, so all of the "signs" of the preimage are positive.
Note, however, that the maps $f(n)=z^{-n}$ do not have negative degree. These maps are also holomorphic on the sphere, so the degrees are positive (with $z^{-n}$ having degree $n$). To get negative degree, you have to compose with the complex conjugation map, which has degree $-1$.