Let $\pi:X \to S$ be a flat, projective morphism, $S$ irreducible. Suppose that for all $s \in S$, the fiber $X\_s$ satisfies $h^2(\mathcal{O}\_{X\_s})=0$. This means in particular that given an invertible sheaf $\mathcal{L}\_0$ on $X\_{s\_0}$, for some $s\_0 \in S$, there exists no obstruction to infinitesimal deformation of the invertible sheaf $\mathcal{L}\_0$ (see $\S 6$ of Hartshorne's "Deformation theory"). My question is: does there exist an invertible sheaf $\mathcal{L}$ on $X$ such that $\mathcal{L}|\_{X\_{s\_0}} \cong \mathcal{L}\_0$? Any reference will be most welcome (I am looking for techniques to go from local to global deformation of invertible sheaves).



Local to global deformation of invertible sheaves
The answer is no, even for locally constant families. Let $F$ be a smooth projective variety with an automorphism $\sigma$, and let $L$ be a line bundle on $F$ such that $\sigma^\* L$ is not isomorphic to $L$. For example, take $F=\mathbf{P}^1\times \mathbf{P}^1$ with $\sigma$ the coordinate-switching involution, and $L=\mathcal{O}(1,0)$. Take $S$ to be the nodal curve obtained by gluing $0$ to $\infty$ on $\mathbf{P}^1$, and let $X$ be obtained from $F\times \mathbf{P}^1$ by identifying $(f, 0)$ and $(\sigma(f), 1)$. The natural $X\to S$ is a locally trivial fibration with fiber $F$. Take $s$ to be the node and $L\_0$ to be $L$.


Think about it this way: in general, suppose that $X\to S$ is nice enough so that there exists a relative Picard scheme $P\to S$. The assumption that $H^2(X\_s, \mathcal{O})=0$ for all $s\in S$ implies that $P\to S$ is (formally) smooth, and so every section over a point (i.e. your $\mathcal{L}\_0$) extends to an infinitesimal/formal/etale neighborhood. But there is no reason to have a non-zero global section of $P\to S$ at all.


