Let $X\_n = Bl\_{p\_1,...,p\_n}\mathbb{P}^2$ be the blow-up of $\mathbb{P}^2$ in $n$ **general points** $p\_1,...,p\_n\in\mathbb{P}^2$.


Let $f\_i:\mathbb{P}^{2}\dashrightarrow\mathbb{P}^1$ be the linear projection from the point $p\_i\in\mathbb{P}^2$. Then $f\_i$ lifts to a fibration $\widetilde{f}\_i:X\_n\rightarrow\mathbb{P}^1$.


So we get $n$ fibrations $\widetilde{f}\_i:X\_n\rightarrow\mathbb{P}^1$ for $i=1,...,n$. Now let $g:X\_n\rightarrow\mathbb{P}^1$ be a fibration of $X\_n$ on $\mathbb{P}^1$. Is it true that $g$ must be one of the $\widetilde{f}\_i$'s?



Fibrations on blow-ups of $\mathbb{P}^2$
As Mohan indicated, this is not always going to be true. Let $n$ be $4$. The group $\text{Aut}(\mathbb{P}^2)$ acts transitively on the set of ordered $4$-tuples $(p\_0,p\_1,p\_2,p\_3)$ such that no $3$ of these points are collinear. Thus, up to projective linear transformation, assume the points are $p\_0=[1,1,1]$, $p\_1=[1,0,0]$, $p\_2=[0,1,0]$, and $p\_3=[0,0,1]$. 


Denote the homogeneous coordinates on $\mathbb{P}^2$ by $[X,Y,Z]$. Then consider the rational function $$g = X(Z-Y)/Z(Y-X).$$ This gives a rational map $\mathbb{P}^2\dashrightarrow \mathbb{P}^1$ that extends to a regular, projective, flat morphism $$g:X\_4\to \mathbb{P}^1.$$ None of the $4$ exceptional divisors is contained in a fiber of $g$, thus it equals none of $f\_1$, $f\_2$, $f\_3$, nor $f\_4$.


