$f(z) = z^3 + 2z + 7$. Calculate $f_* : H_2\to H_2$. Let $f(z) = z^3 + 2z + 7$. $f: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$, with $f(\infty) = \infty$. Calculate $f_* : H_2(\hat{\mathbb{C}}, \mathbb{Z}) \to H_2(\hat{\mathbb{C}}, \mathbb{Z})$.
What I have done: I think that it is just asking where the generator of $H_2(\hat{\mathbb{C}}, \mathbb{Z}) \simeq \mathbb{Z}$ is send by $f_*$, ie how many times the whole image is hitted by $f$. Since $f$ is of degree $3$, for every $c \in \mathbb{C}$ $z^3 + 2z + 7 - c = 0$ has 3 solutions (no matter if repeated). Also $\infty$ is a pole of degree 3, so $f_*(n) = 3n$. But I can't formalize.
 A: Proposition
Given a non constant morphism of compact Riemann surfaces $f:X\to Y$,  the algebraic degree $\deg f$ of $f$ coincides with the topological degree $\operatorname {degtop}f$ of $f$ : $$\operatorname {deg}_{\operatorname{alg}}f=  \operatorname {deg}_{\operatorname{top}}f$$
Remark
The topological degree is the one about which you are asking. In your case the common value of these degrees is $3$.
Proof
The map $f$ is surjective and, except for finitely many exceptions, all points of $Y$ are regular values for $f$.  Consider one such regular value y. Its inverse image is a finite discrete subset $$f^{-1}(y)=\{x_1,\dots,x_n\}\subset X$$ $\bullet$ By definition (Forster, page 29) the algebraic degree of $f$ is then $\operatorname {deg}_{\operatorname{alg}}f=n$.
$\bullet \bullet$ The topological degree is also $n$.
Indeed it is the sum of the local degrees at the $x_i$'s (Madsen-Tornehave, Theorem 11.9 ), and these local degrees are of absolute value $1$, since $f$ is a local diffeomorphism, and of positive sign because holomorphic maps preserve orientation.
Hence the topological degree of $f$ is $\operatorname {deg}_{\operatorname{top}}f=n$.   
Supplementary Bibliography
There is an interesting discussion of the various visions of degree (topological, algebro-geometrical, differential-geometric) in Griffiths-Harris, starting on page 216.
A: See here for many interpretations of degree: https://en.wikipedia.org/wiki/Degree_of_a_continuous_mapping
When $f^{-1}(a)$ is a finite set, the degree can be computed using the induced maps on local homology (see wikipedia for definition, but roughly speaking it measures the local twisting of some continuous map at a point). In your case, when there are exactly 3 roots of $f(x) = a$, the function restricted to a neighborhood around each those roots has a local inverse, so there the induced map on local homology has degree 1. So the total degree is three.
Another way to see this is to explicitly homotope $z^3$ to your function, then try to compute the degree of $z^3$ by reducing to the case of maps from the circle to itself using some theorems you may know about suspensions (the degree of a suspension is the same as the degree of the original map). (Then you can explicitly write down the homology class representing 1 and its image under $z^n$ for the circle, maybe using what you know about $H_1$ and $\pi_1$, or some other techniques.)
Hatcher does all of this systematically in his Algebraic Topology book.
