Fibrations on blow-ups of $\mathbb{P}^2$ 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?
 A: 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 $f_4$.
A: As in Jason's answer take $n = 4$. The fibrations $\widetilde{f}_i$'s for $i=1,..,4$ are induced by the linear systems $\mathcal{L}_i$ of the lines through $p_i$ for $i=1,...,4$.
Now, take the linear system of $\mathcal{C}_{1,2,3,4}$ of conics through $p_1,...,p_4$. This is a pencil and therefore induces a rational map $g:\mathbb{P}^2\dashrightarrow\mathbb{P}^1$. Once you blow-up the four base points you get a morphism $\widetilde{g}:X_4\rightarrow\mathbb{P}^1$. Since $\widetilde{g}$ and the $\widetilde{f}_i$'s are induced by non-equivalent linear systems we have $\widetilde{g}\neq \widetilde{f}_i$ for any $i=1,...,4$.
