Let $f\colon \mathbb{P}^2\dashrightarrow\mathbb{P}^2$ be a rational map ($\mathrm{deg}(f)$ may be high), let $\Gamma\subset\mathbb{P}^2\times\mathbb{P}^2$ be the closure of the graph. Let $x\in\mathbb{P}^2$ be a point in the source, then the fiber $\Gamma|\_x$ can be viewed as a union of rational curves in the target $\mathbb{P}^2$. Do we know if $\Gamma|\_x\subset\mathbb{P}^2$ necessarily have a component of low degree (say, degree 1)?



Bounding degree of rational curves in the exceptional locus
The answer is no. Let us consider the case where $f$ is birational. In this case, for each point $x\in \mathbb{P}^2$ such that $f$ is not defined, the intersection of $\Gamma$ with $x\times \mathbb{P}^2$ contains all curves of $\mathbb{P}^2$ that are contracted by the inverse $f^{-1}$. The degree of these curves can be arbitrary large.


As an example, you can consider some so-called "Halphen map", that is a birational map of $\mathbb{P}^2$ that preserves a pencil of curves of genus $1$ (a simple case is given by a pencil given by two general cubics). You blow-up the $9$ points $p\_1,\ldots,p\_9$ of intersection and consider a translation given by $E\_1-E\_2$, where $E\_1$ and $E\_2$ are the exceptional divisors of $p\_1$ and $p\_2$ (which are two sections of the pencil). If you consider this map $f$ on $\mathbb{P}^2$, the degree of $f^n$ grows quadratically with $n$ and there are exactly $9$ points not defined for $f^n$, namely $p\_1,\ldots,p\_n$. There are exactly $9$ curves contracted by $f^{-n}$ and these are irreducible of degree that is also growing. For $n$ large, the degree of the curves contracted is then unbounded.


There is an explicit family given in Proposition 6.4 of <https://algebra.dmi.unibas.ch/blanc/articles/degencremona.pdf>
where the degree of $f^n$ is $36n^2+1$, and where $f$ contracts $7$ curves of degree $12n^2$, one of degree $12n^2-6n$ and one of degree $12n^2+6n$.


