I'm currently struggling with the following exercise from Hatcher's Algebraic Topology.
2.2.8. A polynomial $f(z)$ with complex coefficients, viewed as a map $\mathbb{C}\to \mathbb{C}$, can always be extended to a continuous map of one-point compactifications $\hat{f}\colon S^2\to S^2$. Show that the degree of $\hat{f}$ equals the degree of $f$ as a polynomial. Show also that the local degree of $\hat{f}$ at a root of $f$ is the multiplicity of the root.
If we set $g_n(z)=z^n$ for each $n\geq 1$, then the induced map $\hat{g}_n\colon S^2\to S^2$ is the suspension of $g|_{S^1}\colon S^1\to S^1$ (where we think of $S^1$ as the unit circle in $\mathbb{C}$). It follows by Example 2.32 and Proposition 2.33 in the textbook that $\deg \hat{g}_n =n$. As $\hat{g}_n{}^{-1}(0)=\{0\}$, this also implies $\deg \widehat{g}_n|0 =n$.
Now suppose $f$ is an arbitrary polynomial with $\deg f>0$. Let $z_1,\ldots,z_k$ denote the distinct roots of $f$, with multiplicities $n_1,\ldots,n_k$, respectively. Then for each $j$ there exists an open neighborhood $U_j$ of $z_j$ in $\mathbb{C}$ together with a biholomorphism $\varphi_j\colon U_j\to \varphi(U_j)\subseteq \mathbb{C}$ such that $\varphi_j(z_j) =0$ and the following diagram commutes:
We thus obtain the following induced diagram on homology groups:
Since ${\varphi_j}_*$ is an isomorphism, this gives us $\deg \hat{f}|z_j = \pm n_j$, which almost solves the problem. How would I deduce from here that $\deg \hat{f}|z_j = n_j$? I think intuitively it's because $\varphi_j$ is conformal, so orientation preserving, but at this point Hatcher hasn't really introduced anything in terms of orientation/orientability. More generally, I am somewhat confused about how the sign of the local degree of a map $S^n\to S^n$ is defined, as discussed in this question.
