The fibers of a map to $\mathbb{P}^1$ Suppose $X$ is a surface and $D$ is a prime divisor such that $|D|$ gives rise to a surjective morphism $f:X\to \mathbb{P}^1$. Is every fiber linearly equivalent to $D$? My thought process was that we can find one point $p\in \mathbb{P}^1$ whose fiber is $D$ and since every other point in $\mathbb{P}^1$ is linearly equivalent to $p$, they would pull back to linearly equivalent divisors. But I'm not sure if it's true or how to formalize it.
 A: Yes, every fiber is linearly equivalent to $D$. What you have is a pencil of curves inside $X$, and the divisors (curves in this case) in the pencil are exactly the fibers of $f$. More generally:
Fact 1. In $\mathbb P^n$, two divisors are linearly equivalent if and only if they have the same degree. For instance, two hyperplanes are always linearly equivalent.
Fact 2. The pull-back map of a morphism of varieties $f:X\to Y$ (where $X$ is normal and $Y$ is locally factorial) preserves linear equivalence. (Useless comment: this is nice because it means that linear equivalence descends to give a morphism Pic $Y\to$ Pic $X$ between the Picard groups.)
Fact 3. Let $X$ be a normal variety (say, projective over a field $k=\overline k$). To give a morphism $f:X\to\mathbb P^n$ is the same as to give a line bundle $L=O_X(D)$ on $X$, plus $n+1$ global sections generating $L$. If you wish, this is the same as to give the (base point free) linear system $|D|$. 
Using the above, we come to the

Key point: the divisors appearing in the linear system $|D|$ (i.e. the
  divisors which are linearly equivalent to $D$) are exactly those of
  the form $f^{-1}(H)$, where $H$ is a hyperplane in $\mathbb P^n$.

I hope this helps.
