(Homework) Homological degree of a complex rational polynomial Let $P, Q \in \mathbb{C}[Z]$ be polynomials with $Q \ne 0$. Complex Analysis gives that we get a continuous map 
$$
f: \mathbb{C} \cup \{\infty\} \longrightarrow \mathbb{C} \cup \{\infty\} \\
f(z) = \frac{P(z)}{Q(z)}\,,\quad z \in \mathbb{X}\\
$$
and using $\mathbb{C} \cup \{\infty\}\cong S^2 $, we can assign a degree $f \mapsto d(f)$.
How do I go about this?
I think I will have to use that
$$
d(f) = \sum_{x \in f^-1\{z_0\}}d(f,\,x)
$$ 
where $d(f,\,x)$ denotes the local degree, and $z_0$ is one of the uncountably many nonzeroes/regular values of $df/dz$, and also use the role of the determinant in computing local degrees.
However, I don't really know how to go about this (admittedly, I am also very tired and somewhat worn down from working through 'sketchy' proofs).
Since, for $Q(z) \ne 0\,$, $\left(\frac{P(z)}{Q(z)} = \zeta \iff P(z) - Q(z)\zeta = 0\right)$, the above formula can be applied for $z_0$ as described above (i.e., we have a finite fiber): Simply choose $z_0$, such that $$Q(z_0) \ne 0\, \land\, df/dz(z_0) \ne 0$$
which is possible, because $df/dz$ has infinitely many nonzeroes, and $Q$ has only finitely any zeroes.
I'm not even sure how I would compute the degree for a complex polynomial - I think I might have to know how the homological degree interacts with multiplication and addition of functions (if there are simple rules, how do I prove them?).
 A: The degree of a map $f \colon S^2 \rightarrow S^2$ is by definition
$$ \deg(f) = \sum_{f(z) = w} \pm 1 $$
where $w \in S^2$ is a regular value of $f$ and each point $z$ in the preimage $f^{-1}(w)$ is counted as $+1$ or $-1$ according to whether $f$ is orientation preserving or orientation reversing at $z$. 
A basic fact about holomorphic functions is that they are orientation preserving so in your case, you actually have $\deg(f) = |f^{-1}(w)|$ for any regular value $w$. Assume that $\operatorname{gcd}(P,Q) = 1$ (so $P,Q$ have no common factors). For a generic $w$, the number of distinct solutions to $f(z) = w$ is $\max(\deg P, \deg Q)$ because
$$ f(z) = w \iff P(z) = Q(z)w \iff P(z) - Q(z)w = 0 $$
is a polynomial equation in $z$ of degree $\max(\deg P, \deg Q)$ so it should have $\max(\deg P, \deg Q)$ distinct solutions. 
Formally, let $w \neq 0, \infty, f(\infty)$ be a regular value of $f$. Then all the preimages of $w$ are in $\mathbb{C}$ and $$f(z) = w \iff P(z) = Q(z) w.$$
If $f(z) = w$ then $f'(z) \neq 0$ which shows that 
$$P'(z)Q(z) - Q'(z)P(z) = P'(z)Q(z) - Q'(z)Q(z)w = Q(z)(P'(z) - Q'(z)w) \neq 0$$
so $z$ is not a repeated root of $P(z) = Q(z)w$.
