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.



Fibers of the resolution of indeterminacy points of a rational map
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 $L\_i$.


