I'm working on a problem from Hatcher's Algebraic Topology, and I want to show that if we take polynomial $f(x)$ definded on the Riemann sphere mapping to the Riemann sphere, the local degree of the map (when viewed as a map $S^2 \to S^2$) at a root is equal to the multiplicity of the root.
Intuitively this seems obvious because if we look at a neighborhood of some root with multiplicity $n$ and translate the function to the origin, we get a map that kind of looks like the map $z \to z^n$, given a sufficiently small neighborhood, which has degree $n$ as desired. My problem is making my claim rigorous because the polynomial does not map perfectly from $z \to z^n$ in the neighborhood. I'm not sure how to compute the local degree directly. I know it means looking at the induced map $H_2(U,U-x_0) \to H_2(f(U), f(U)-0)$ where $x_0$ is a root and $U$ is a neighborhood of $x_0$.
Best Answer
I believe I figured it out.
We can use the argument principle, which states the number of zeros minus the number of poles around a small contour around the root $x$ is the change in the argument of $f(z)$ as $z$ travels around the contour, divided by $2 \pi$. This means the multiplicity of $x$ is number of times $f(z)$ travels around $0$ when $z$ travels around a sufficiently small contour surrounding $x$. In other words, this means the local degree of $f$ at the root is the multiplicity of the root as desired.