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Let all varieties be projective and smooth over $\Bbb{C}$.

Let $f:S\to C$ be a surjective morphism from a surface onto a curve.

I'm trying to prove that any two fibers of $f$ are linearly equivalent.

Here's how far I could get:

Since $f$ is dominant, we can define a pullback $f^*:\text{Pic}(C)\to \text{Pic}(S)$. So if points $P,Q$ are linearly equivalent, $[P]=[Q]$ in $\text{Pic}(C)$, then $f^*[P]=f^*[Q]$ in $\text{Pic}(S)$, so the fibers are linearly equivalent.

But how do I know that any $P,Q$ are equivalent? I only know how to prove this when $C=\Bbb{P}^1$.

Maybe is it possible to prove that $[P-Q]\in\ker f^*$ somehow?

Thanks in advance!

rmdmc89
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    You are correct. Unless $C$ is the projective line, in general fibers are not linearly equivalent. – Mohan Jun 24 '21 at 18:31
  • @LazzaroCampeotti you're right, I totally misread Brenin's answer. Thanks for pointing that out! That was not pedantic at all. – rmdmc89 Jun 25 '21 at 00:46

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