Is manifold mapping degree equal to algebraic degree for polynomials?

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?

Solution 1:

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 $-1$.