$f:S\to X$ is generically finite of degree $d$, then $f_*f^*D=dD$ In Beuville's Complex Algebraic Surfaces, chapter 1, he consideres a map $f:S\to X$ between smooth projective varieties over $\Bbb{C}$ ($S$ is a surface, but I don't think this will be important) which is generically finite of degree $d$. If $C\subset S$ is an irreducible curve, he defines:
$$
f_*C:=\begin{cases}
0,\text{ if }f(C)\text{ is a point;}\\
r\Gamma,\text{ if }f(C)\text{ is a curve }\Gamma\text{ and }f\big|_C:C\to\Gamma\text{ is finite of degree }r.
\end{cases}$$
He also says that it follows from the definition that $f_*f^*D=dD$ for any divisor $D$ in $X$.
Here are my questions:

*

*What does he mean by generically finite? I know Hartshorne defines generically finite as a morphisms of schemes whose preimage of the generic point is finite. But here we have varieties, so how would that translate?


*I wish he would explain the map $f\big|_C:C\to\Gamma$ better. Is it necessarily finite? When could $r$ be different from $d$ ?


*Why do we get $d$ in $f_*f^*D=dD$?
 A: All you say makes sense only if both $S,X$ are smooth projective surfaces and $f$ generically finite. This means that for points in a dense Zariski open set of $X$, the inverse image is of some finite cardinality $d$.
Given such a situation and an irreducible curve $C\subset S$, either $f(C)$ is a point or a curve $\Gamma$ for dimension considerations. In the latter case, the restriction map $C\to\Gamma$ is onto and since $C$ is projective, this map is in fact finite of some degree $r\leq d$. A simple example (and there are plenty) to show that $r$ can be strictly smaller is as follows. Take a generic projection of $\mathbb{P}^1\times\mathbb{P}^1\subset \mathbb{P}^3\to\mathbb{P}^2$. You can easily check that its branch locus is a smooth conic $\Gamma$ in the plane and its inverse image is $2C$ for a curve $C$ and $C\to \Gamma$ is an isomorphism, so degree one while the map of surfaces has degree 2.
Finally, take an irreducible curve $D$ which intersects the open set of points of the plane where the fibers are $d$ points. Then, $f^*D$ is the inverse image of $D$ and thus, since $f(f^*D)=D$, and clearly has $d$ points in the inverse image of a general point of $D$, one sees that $f_*(f^*D)=dD$ by your definition.
