I have a question about the proof of proposition 2.3, Chap. V in Hartshorne's Algbraic Geometry:
Let $p: X \to C$ be a ruled surface, i.e. a (projective) surface birationally equivalent to $C \times \mathbb{P}^1$ for $C$ smooth base curve, let $C_0 \subset X$ be some fixed section and $F =p^{-1}(c)$ some fixed fiber.
Let $D' \in \text{Pic} X$ a divisor class with $D' \cdot F=0$; see $V.1$ how intersection numbers are defined. Lemma 2.1 shows that $p_* \mathcal{O}_X(D')$ is invertible.
Why $\mathcal{O}_X(D') $ is ismorphic to $
p^*p_*\mathcal{O}_X(D')$ as claimed in the proof? Note that we always have a map $
p^*p_*\mathcal{O}_X(D') \to \mathcal{O}_X(D') $ induced by adjunction $(f^*,f_*)$ called the "unit map" and since
$p_* \mathcal{O}_X(D')$ is invertible it is a map between
invertible sheaves, therefore injective since on a trivialization
this map is given by multiplication with a regular element.
Why is this map surjective?
Are there more general conditions known when this map turns out to be an isomorphism?