Lifting sheaves from the special fibre to the generic fibre Let $R$ be a complete discrete valuation ring and let $S = \operatorname{Spec}(R)$.  Let $\mathcal{X}\to S$ be a complete, regular, flat, connected $S$-scheme of finite type whose fibres are smooth, projective, geometrically connected algebraic curves.  Let $s \in S$ be the closed point, let $\xi \in S$ be the generic point, let $\mathcal{X}_s$ be the special fibre of $\mathcal{X}$, and let $X = \mathcal{X}_\xi$ be the generic fibre of $\mathcal{X}$. 
If $\mathscr{L}$ is an invertible sheaf on $\mathcal{X}$, then there are maps $\Gamma(\mathcal{X}, \mathscr{L})\to \Gamma(X, \mathcal{L}_\xi)$ and $\Gamma(\mathcal{X}, \mathscr{L})\to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$.  In the right circumstances (I think when $\mathscr{L}$ is generated by global sections; maybe we need projective normality?), these two maps induce surjections $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(\xi) \to \Gamma(X, \mathcal{L}_\xi)$ and $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(s) \to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$.
My question involves going in the other direction:

Suppose we are given $\mathscr{L}_s$ on $\mathcal{X}_s$ which is generated by global sections. Under what conditions, and how, can one find an invertible sheaf $\mathscr{L}$ on $\mathcal{X}$ such that $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(\xi) \to \Gamma(X, \mathcal{L}_\xi)$ and $\Gamma(\mathcal{X}, \mathscr{L}) \otimes_R k(s) \to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$ are surjective.

ETA: Note that I am particularly interested in the "and how" part; which is to say that, given an (explicit) basis for $\Gamma(\mathcal{X}_s, \mathscr{L}_s)$, how do I go about finding an (explicit) basis for $\Gamma(\mathcal{X}, \mathscr{L})$ that satisfies the surjectivity properties above (assuming $\mathscr{L}$ exists)?
My idea is to take a basis of $\Gamma(\mathcal{X}_s, \mathscr{L}_s)$, lift it to an $R$-module $M$, and take the saturation $\operatorname{Sat}(M) = R^n \cap (M \otimes_R k(\xi))$ where $n = \dim\Gamma(\mathcal{X}_s, \mathscr{L}_s)$.  Some related "sub-questions" based on this idea:


*

*Suppose $\Gamma(\mathcal{X}_s, \mathscr{L}_s)$ is generated by global sections $\{\bar{s}_0, \ldots, \bar{s}_r\}$ and let $M$ be the $R$-module generated by lifts $\{s_0, \ldots, s_r\}$ of the $\bar{s}_i$.  Let $\mathscr{M} = \widetilde{M}$ be the sheaf associated to $M$.  Then $M = \Gamma(\mathcal{X}, \mathscr{M}) \to \Gamma(\mathcal{X}_s, \mathscr{L}_s)$ is surjective.  Under what conditions is $\mathscr{M}$ invertible?

*Is this $\mathscr{M}$ isomorphic to $\pi_*\mathscr{L}_s$, where $\pi\colon\mathcal{X}_s \to \mathcal{X}$ is the canonical "projection"?


N.B. The conditions I listed in the first paragraph describe the (rather restrictive) situation I'm interested in; I would certainly be interested in hearing answers valid in a more general context.  Feel free to strengthen any conditions as necessary; for example, I think projective normality  might play a role somewhere, but I'm not sure where exactly.
 A: The fact that you've assumed such niceness about the fibers and $\mathcal{X}$ should allow you to always do this regardless of whether or not $\mathcal{L}$ is globally generated or ample or anything.
The way I'm thinking of your setup is that you have some scheme $X_s$ over $R/m=k$, and $\mathcal{X}\to S$ is a deformation of it. Given $\mathcal{L}_s$ on the the special fiber a starting point is to ask whether or not this line bundle deforms with $X_s$ all the way to $X_\xi$.
Since the fibers are curves, and the obstruction to deforming a line bundle lies in $H^2(X_s, \mathcal{O})=0$ there is no obstruction and the line bundle deforms. Now you have some $\mathcal{L}_\xi$ on $X_\xi$ that when restricted to $X_s$ is $\mathcal{L}_s$. 
Since $\mathcal{X}$ is nice you should be able to think of $\mathcal{L}_\xi$ as a Weil divisor on $X_\xi$ and taking Zariski closure will give you a line bundle $\mathcal{L}$ on all of $\mathcal{X}$ that when restricted to the two fibers mentioned gives you the correct thing back (to do this all you needed was $\mathcal{X}$ to be locally factorial if I remember correctly).
Edit: I didn't interpret your maps properly. What we have now is a candidate with the property that $\mathcal{L}|_{X_s}\simeq \mathcal{L}_s$, which I thought would translate into your map upon taking sections, but it is still in theory possible that $\Gamma (\mathcal{X}, \mathcal{L})=0$ while $\Gamma (X_s, \mathcal{L}_s)\neq 0$ so we need to convert this to what you actually want somehow.
A: This is not a complete answer just a proposal: under the assumption you make the special fibre $\mathcal{X}_s$ is a regular curve, thus invertible sheaves correspond to Weil divisors. 
Assume that the sheaf $\mathcal{L}_s$ corresponds to a Weil prime divisor $\mathcal{P}$ on $\mathcal{X}_s$. There exists a Weil prime divisor $P$ on the generic fibre $X$ such that $\mathcal{P}$ is contained in the Zariski closure $\overline{P}$ of $P$ on $\mathcal{X}$. Since $R$ is henselian $\mathcal{P}$ is the only point of $\mathcal{X}_s$ lying in $\overline{P}$. Let $\mathcal{L}(P)$ be the $\mathcal{O}_\mathcal{X}$-ideal sheaf defining the reduced subscheme structure on $\overline{P}$. By construction this sheaf is reflexive and since $\mathcal{X}$ is regular it is thus invertible and should do the job.
If $\mathcal{L}_s$ corresponds to a Weil divisor $e_1\mathcal{P}_1+\ldots +e_r\mathcal{P}_r$ one can extend the construction linearly replacing  $\mathcal{L}(P)$ by the product (or tensor product) $\mathcal{L}(P_1)^{e_1}\cdot\ldots\cdot$ $\mathcal{L}(P_r)^{e_r}$.
