Let $\pi: X \rightarrow C$ be a ruled surface with a section $\sigma : C \rightarrow X$ and let $D=\sigma(C)$ be a divisor on $X$ Let $\mathcal{F}= \pi_*\mathcal{L}(D)$ Then $\mathcal{F}$ is a locally free sheaf of rank $2$ and $\pi^*\mathcal{F} \rightarrow \mathcal{L}(D)$ is surjective. So there is a morphsim $g: X \rightarrow \mathbb{P}(\mathcal{F})$ over $C$ such that $g^*\mathcal{O}_{\mathbb{P}(\mathcal{F})}(1) \cong \mathcal{L}(D)$. Since $D$ on fibre has degree $1$ and any fibre is isomorphic to $\mathbb{P}^1$, $D$ is very ample on each fibre.
In proof of Harhshorne (V.2.2), By above facts, $g$ is an isomorphism on each fibre and so $g$ is isomorphism.. But I don't understand this...