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Let $\pi\colon Z \to \Delta$ be a smooth family of complex (projective) varieties, over a small disk in $\mathbb{C}$ such that $\pi^{-1}(0)$ is the only (normal-crossing) singular fiber, $\pi^{-1}(0)= X\cup_D Y$.

I have some questions (may be equal) about line bundle on the family vs individual fiber.

Under what condition (on $X$, $Y$, $D$), it is true that we extend a line bundle on the singular fiber to a line bundle on the whole family?

Under what condition (on $X$, $Y$, $D$), it is true that there is an isomorphism between the space of line bundles (Picard group) on the singular fiber and the space of line bundles on the smooth fiber?

I general I would like to know about the comparison between the Picard group of $Z$ and its fibers.

For $t\neq 0$, I expect Pic($Z_t$) to be bigger than Pic($Z_0$), can you provide an example of that?

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    $\begingroup$ I think that extension theorems of a line bundle to the whole family are only available in general in the situation where $\Delta$ is the spectrum of a complete local ring (in the algebraic setting), using Grothendieck's GAGA (see for instance the paragraph "Application to liftings problems" in Illusie's article in "FGA explained"). You may then reduce to strictly henselian local ring using Artin's approximation theorem. $\endgroup$
    – Damian Rössler
    Commented Sep 30, 2012 at 8:34
  • $\begingroup$ Concerning the Picard group, look at the example where $Z_t = \{xy-t^2=0\} \subset \mathbb P^2$. Then for $t\neq 0, Z_t \simeq \mathbb P^1$, but $Z_0$ is a degenerate conic whose Picard group surjects onto the Picard group of its normalization (which is 2 copies of $\mathbb P^1$. Therefore $\mathrm{Pic}(Z_0)$ is bigger than $\mathrm{Pic}(Z_t)$. $\endgroup$
    – Henri
    Commented Sep 30, 2012 at 12:36
  • $\begingroup$ Edit for the previous comment: of course $Z_t=\{xy-tz^2=0\} \subset \mathbb P^2$! $\endgroup$
    – Henri
    Commented Sep 30, 2012 at 13:37
  • $\begingroup$ @ Henri: In your example, line bundles on each fiber are extendable to whole family, is not it? $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Sep 30, 2012 at 14:30

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