# Fibers of the resolution of indeterminacy points of a rational map

I have a little problem to fully understand the next thing:

Let $F:\mathbb{P}^2 \dashrightarrow \mathbb{P}$ a rational map. Let $\sigma:\bar{\mathbb{P}}^2 \rightarrow \mathbb{P}^2$ be a composition of $\sigma$-processes resolving the indeterminacy points of the rational map $F$, so that $\bar{F} = F\circ \sigma$ is a morphism.

We denote by $\bar{Y}_z$ the fibers of the morphism $\bar{F}$. Let $\sigma^{-1}(L_{\infty}) = \bar{S}_{\infty} \cup \bar{Y_{\infty}} \cup(\bigcup L_i)$, where $\infty = \mathbb{P}^1 \setminus \mathbb{C}^1$, the curves $L_i$ are components of the fibers of $\bar{F}$, and the morphism $\bar{F}$ maps each of the irreducible components of the curve $\bar{S}_{\infty}$ onto $\mathbb{P}^1$.

My question is. What are exactly the sets $\bar{S}_{\infty}$, $\bar{Y_{\infty}}$ and $L_i$?.

[Edited] I forgot to define $L_{\infty} = \mathbb{P}^2 \setminus \mathbb{C}^2$

Thank you in advance.

-
Does not seem to be clear like this. Would be good to say what is $L_\infty$. –  Jérémy Blanc Aug 29 '13 at 8:38

## 1 Answer

It seems to me that the only way this can happen is if $L_\infty\subset \mathbb P^2$ maps to $\infty\in\mathbb P^1$.

So, $\overline Y_\infty$ consists of $\sigma^{-1}_*L_\infty$ (a.k.a., the strict transform of $L_\infty$) and possibly a few $\sigma$-exceptional curves that map to $L_\infty$ via $\sigma$.

The rest of the $\sigma$-exceptional curves that map to $L_\infty$ via $\sigma$ are contained in $\overline S_\infty$ and $\bigcup L_i$. The "essential" ones, that is, those who really cause $F$ to be not defined somewhere, i.e., those who map to the entire $\mathbb P^1$ are in $\overline S_\infty$, but these may lie over infinitely near points, so in order to "reach" them you might have to blow up points whose corresponding exceptional divisors get contracted by $\overline F$. Based on the little information you're giving, I don't think one can say more precisely which ones are in which group.

There is one more group of curves, the fibers of $F$ that intersect $L_\infty$ in a point (or points) where $F$ is not defined. The strict transforms of these also end up among the $L_i$.

-
Thanks for the answer. An additional condition is to assume that the generic fiber $Y_z$ is connected. –  Geri Aug 30 '13 at 3:19