Rational map on $\mathbb P^1$ and its fibers

Question

Consider a non-singular complex projective surface $S$ and a rational map $\psi:S\longrightarrow \mathbb P^1$; moreover suppose that $\psi$ is not defined on $\Delta=\{x_1,\ldots,x_m\}\subset S$. Now I have the following question:

Let $y\in\mathbb P^1$ be a closed point and consider the fiber $\psi^{-1}(y)$. Is it true that $\psi^{-1}(y)\cup\Delta$ is the closure in $S$ of $\psi^{-1}(y)$? (Certainly it is a closed subset containing $\psi^{-1}(y)$).

Practically in my mind I imagine a fiber of $\psi$ as a curve on $S$ minus $m=|\Delta|$ points. Can $\psi^{-1}(x)$ be a curve of $S$ minus $m-k$ points (with $k=1,\ldots,m$)?

Many thanks in advance.

Answer 1

Yes, this is true. Let me give an answer here, and see if you are happy with it; I will give extra details if necessary.

Fix two points in $\mathbf P^1$, which we'll call $0$ and $\infty$. Let $D_0$ and $D_\infty$ be the closures in $S$ of the fibres over these points. So $D_0$ and $D_\infty$ are divisors on $S$. There is a rational function on $\mathbf P^1$ whose divisor is $(0)-(\infty)$; pulling this rational function back via $\psi$, we get a rational function on $S$ whose divisor is $D_0-D_\infty$. (It might seem like we only proved this on the locus of definition of $\psi$, but since $S$ is smooth this extends to all of $S$.)

So $D_0$ and $D_\infty$ are linearly equivalent divisors on $S$. In the language of line bundles, there is a line bundle $L$ with two global sections $s_0, s_\infty$ whose zero loci are precisely the two divisors. Now consider sections of $L$ of the form $\lambda s_0 + \mu s_\infty$: the zero loci of these sections are exactly the closures of the fibres of $\psi$! Explicitly, $D_{[\lambda:\mu]} = \{ \lambda s_0 + \mu s_\infty =0\}$ is the closure of the fibre over the point $[\lambda : \mu] \in \mathbf P^1$, where $[1 : 0] = 0$ and $[0:1]=\infty$. (If this seems sketchy: note that each $D_{[\lambda:\mu]}$ is a subvariety of $S$ of pure codimension 1, and when we remove finitely many points from it we get the fibre over $[\lambda:\mu]$. )

Now for any given point $p \in S$, the condition $\{ \lambda s_0(p) + \mu s_\infty(p) =0\}$ is a homogeneous linear condition on $\lambda$ and $\mu$, and it is satisfied precisely when $p \in D_{[\lambda:\mu]}$. So there are 3 possibilities:

1. Only solution is $\lambda=\mu=0$, so that $p$ is not in the closure of any fibre. But this case is impossible: one equation in two variables must have at least one non-trivial solution.

2. 1-dimensional space of solutions. Then there is a unique $[\lambda: \mu] \in \mathbf P^1$ such that $p \in D_{[\lambda:\mu]}$. In this case the map $\psi$ is well-defined at $p$, and sends it to $[\lambda: \mu]$. (If this seems a bit sketchy: trivialise $L$ near $p$ and use this to write $s_0$ and $s_\infty$ as regular functions near $p$. Then check that the formula $x \mapsto [s_0(x):s_1(x)]$ extends $\psi$ over $p$.)

3. 2-dimensional space of solutions. In this case $p \in D_{[\lambda:\mu]}$ for all $[\lambda: \mu] \in \mathbf P^1$ — in other words, $p$ is in the closure of every fibre. By the previous cases, this is what must happen if $\psi$ is not defined at $p$.