I am reading Hartshorne's book chapter 5 (on surface) and I have a question:
on page 371, proposition 2.3, it says:
Let $X$ be surface, $C$ curve, $\pi:X\to C$ ruled surface. $f$ be a fiber, $\sigma$ be a section of $\pi$, $C_0=\sigma(C)$. Let $D$ be a divisor on $X$, $D.f$=n, and set $D'=D-nC_0$. Then he claims:
$L(D')\cong \pi^*\pi_* L(D')$
I don't know how this comes from? Can someone helps me? Thanks in advance.
By Lemma 2.1 of the same section, $\pi_*(L(D'))$ is a locally free sheaf of rank $D'.f +1= D.f-nC_0.f +1= n-n+1=1$. By the counit of the adjunction $f^*\leftrightarrows f_*$, we have a map $\pi^*\pi_*L(D') \to L(D')$ - note that both of these object are line bundles on $X$. By the adjunction formula, we have that $\operatorname{Hom}_{\mathcal{O}_X}(\pi^*\pi_*L(D'),L(D'))\cong \operatorname{Hom}_{\mathcal{O}_C}(\pi_*L(D'),\pi_*L(D'))$. Since $\pi_*L(D')$ is a line bundle on $C$, our map has to be a global section of $\mathcal{O}_C$. But $C$ is a projective curve, so therefore the map is either $0$ or an isomorphism. It is clear the map is not $0$, so it is therefore an isomorphism.