Let $S$ be a surface given by $S:= \mathbb{P}^1 \times \mathbb{P}^1$ and we have the canonical morphism $pr_i: S \to \mathbb{P}^1$ for $i = 1,2$.
Define the invertible sheaf $\mathcal{L}:= pr_1 ^*(\mathcal{O}_{\mathbb{P}^1}(1))$ on $S$.
Let $a \in \mathbb{P}^1$ be a closed point, such that $pr_i^{-1}(a) \cong \mathbb{P}^1$.
My question is: How to see that the divisor $\{ a \} \times \mathbb{P}^1 \cong \mathbb{P}^1$ corresponds to $\mathcal{L}$? Or in other words, why in the exact sequence
$ 0 \to \mathcal{L}^{\vee} \to \mathcal{O}_{\mathbb{P}^2} \to \mathcal{O}_D \to 0$
$D$ coincides with $\mathbb{P}^1$ ?
Suppose that $\mathbb P^1=\mathbb PV$, where $V$ is a two-dimensional vector space with coordinate functions $x_1,x_2$. The sheaf $\mathcal O_{\mathbb P^1}(1)$ has a rational (in fact, global) section, corresponding to $x_1$. Then the sheaf $pr_1^*(\mathcal O_{\mathbb P^1}(1))$ has a section $pr_1^*(x_1)$. It is obvious that the divisor of zeros and poles of this section is the projective line.