Is there an alternative approach to proving that the degree of induced map $\hat{f}$ and the degree of a polynomial $f$ on Riemann surface coincide?

Question

Since I'm not too familiar with complex analysis, I've been wondering whether it would be possible to solve the problem below (from Hatcher's Algebraic Topology) using some alternative approach to this one (which relies heavily on complex analysis), such as by induction and/or simplifying down the the irreducible factors.

The question is:

A polynomial $f(z)$ with complex coefficients, viewed as a map $C → C$, can always be extended to a continuous map of one-point compactifications $\hat{f}: S^2 → S^2$. Show that the degree of $\hat{f}$ equals the degree of $f$ as a polynomial. Show also that the local degree at $\hat{f}$ at a root of $f$ is the multiplicity of the root.

I thought of writing out the polynomial as $f(z)=(z-c_1)(z-c_2)\dots (z-c_n)$ and then claiming something in the lines of:

• Showing that the extension of $z-c_i$ has degree $1$, and since it's a polynomial of degree $1$ the claim follows.
• [This step is clearly flawed since $h$ returns a new polynomial rather than a value in $\mathbb{C}$, but even so it conveys the general degree-related tools I was hoping to be able to use] denoting $g(z)=(z-c_2)\dots (z-c_n)$ and $h(m)=(z-c_1)m$ we obtain that $f(z)=(hg)(z)$ and then we can use the fact that the degree of a composition is the product of the degrees.

Is there anything along similar lines, or that does not heavily rely on complexity analysis, that can be used to solve this problem?

Answer 1

For any $a \in \mathbf{C}$, $f + a$ is homotopic to $f$ (both as a map $\mathbf{C} \to \mathbf{C}$ and $S^2 \to S^2$) so has the same degree. If we choose $a$ such that $f(\alpha) + a \ne 0$ for each root $\alpha$ of $f'$ then $f + a$ has distinct roots. Therefore we can assume $c_i$ are distinct.

Now we can calculate the local degree at $c_i \mapsto 0$. In a sufficiently small disk neighborhood $D_i$ of $c_i$, $f_i(z) = f(z)/(z- c_i)$ takes values away from $0$, so the map $z \mapsto f(z) = f_i(z)(z - c_i)$ is homotopic^* to $f_i(c_i)(z - c_i)$ as a map $(D_i, \partial D_i) \to (B_\epsilon(0), \partial B_\epsilon(0))$. But the latter map is an oriented homeomorphism on the quotient $S^2 \to S^2$, so has degree $1$.

(*) It's easy to write down a homotopy assuming that $f_i(z)$ does not vanish on $D_i$, but maybe not so obvious why that's a necessary condition.