Connection between degree of dominant morphism and cardinality of fibres My exercise:
Let $f:X\rightarrow Y$ be a dominant morphism of curves.
For any dominant morphism, the degree of it is defined to be $[K(X):K(Y)]$ with $K(Y)$ identified with $f^*(K(Y))$.
Prove that the fibres of $f$ have at most $deg(f)$ points if $Y$ is non-singular. 
The notes we are working out skim over the whole degree/ramification story.  Other sources(i.e. Shafarevich Book 1) treat more material, but only for $f$ finite.
Could anyone elaborate(e.g. give some concrete examples) on what the connection is between the algebraic definition of degree and the behaviour of $f$ on the varieties and/or give a hint for the exercise so I can figure this connection out for myself?
 A: I assume by curves you mean integral projective curves over some field $k$. Then, maybe the question will be answered by the following fact:

Theorem: If $f:X\to Y$ is a map of curves over $k$. Then, either $f$ is constant, or it is surjective. If it is the latter, then it is finite.

The proof of this is not too difficult. Since $X$ and $Y$ are projective, by the Cancellation Theorem (see Vakil), $f$ is projective, and so closed. Thus, $f(X)$ is an irreducible closed subset of $Y$. Since $Y$ is a curve, this forces to $f(X)$ to be a point, or all of $Y$. 
Now, if $f$ is surjective, then it is necessarily quasi-finite. Indeed, it's clear that the only point which maps to the generic point is the generic point (because $f$ is closed). If $x\in Y$ is a closed point, then $f^{-1}(y)\subseteq X$ is closed. If it's not finite, then since $X$ carries the cofinite topology, it must be dense, which contradicts that $f$ isn't constant. But, $f$ is then quasifinite and projective, and so finite (by ZMT).
[There is an easier proof of this theorem, I believe in Hartshorne, under some regularity assumptions. I don't recall it though].
