Let $f:X\to Y$ be an algebraic fibre space, and $\dim Y=1$. What can I say about the numerical relation of the fibres?



When are the fibres numerically equivalent?
They are algebraically equivalent, which is stronger. Consider the subvariety on $X \times Y$ which is the inverse image of the diagonal on $Y \times Y$ under $(f,id)$. This is flat, because it's a hypersurface in a flat morphism, so all its fibers, which are the fibers of $f$, are algebraically equivalent.


This is only true as long as none of the fibers contain entire irreducible components $X$, which seems to be allowed in the definition of algebraic fiber bundle that I saw, but is obviously a pretty pathological scenario.


