The assertion is false. The Picard group of the <Hirzebruch surface> is freely generated by the negative section $S$ and a fiber $F=D_1$. Pullbacks from $\mathbb P^1$ form only the subgroup $\mathbb ZF$. For example, $\mathcal O(S)$ cannot be a pullback: its restriction to a fiber has degree $S\mathbin\cdot F=1$, whereas every pullback from the base restricts trivially to every fiber.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 152 / 3 / e / Solution

 Solution (source code)