I have the following two parts question I would like to get a grasp on:
Given two non-zero polynomials $p,q$ over $\mathbb{C}$ we can consider the function:
$f(z):=\frac{p(z)}{q(z)}$
(i) Show that $f$ has a natural smooth extension to a map $g:\mathbb{C}P^1\rightarrow \mathbb{C}P^1$ (when viewing $\mathbb{C}P^1$ as a smooth manifold).
(ii) Compute the degree of $g$.
I know that there is a structure of a smooth manifold with the two coordinate charts $(U_1,\phi_1)$ and $(U_2,\phi_2)$ given by:
$U_1= \mathbb{C}P^1\setminus \{(0,1)\}\qquad$ and $\qquad\phi_1((z,w))=\frac{w}{z}$
and:
$U_2= \mathbb{C}P^1\setminus \{(1,0)\}\qquad$ and $\qquad\phi_2((z,w))=\frac{z}{w}$
Where $(0,1)\cong "\infty"$.
If I denote the roots of $q$ by $\zeta_1,...,\zeta_m\in \mathbb{C}$, a natural extension to me seems to be:
$g((z,w)):=\begin{cases} \phi_1 \circ f \circ\phi_1^{-1} & ;(z,w)\in U_1, \quad \text{and} \quad \phi_1([z:w])\neq \zeta_j \\ (0,1) &; (z,w)=(0,1)\quad \text{or} \quad \phi_1([z:w])=\zeta_j \end{cases}$
First question being is this truly a smooth extension?
Second question: is we know that a degree of a function is the same for two homotopic functions, and for this particular function it will be inconvenient calculating the degree. Is an easy homotopic function to calculate it's degree would be the extension of a rational function of the sort:
$z^{\deg(p)-\deg(q)}$ ?
P.S:
Given a smooth function $f:M\rightarrow N$ between smooth closed orient-able manifolds of the same dimension and a regular value $y\in N$, I define the degree of $f$ as the amount of times the differential at $f^{-1}[y]$ preserves orientation minus the amount of times at $f^{-1}[y]$ the differential reverses orientation.