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Let $f:X\to B$ and $g:Y\to B$ be smooth morphisms of complex projective varieties. Assume that for every closed point $b\in B$, the fibres $X_b=X\times \kappa(b)$ and $Y_b$ are derived equivalent. Assume furthermore that the Fourier-Mukai transforms are induced by complexes of sheaves $F_b \in D^b(X_b \times Y_b)$ that glue to a sheaf $F\in D^b(X \times_B Y)$. Does it follow that the total spaces $X$ and $Y$ are derived equivalent?

Edit: As pointed out by pro, the answer to the original question (without the italic part) is negative. An explicit example is provided by taking $P^1\times P^1$ and any Hirzebruch surface $F_n$. Both are $P^1$ bundles over $P^1$, so the fibres are even isomorphic, but since $P^1\times P^1$ has an antiample canonical bundle both varieties could only be derived equivalent if they were isomorphic, which they are not.

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    $\begingroup$ I can't imagine this could ever be true, there's no gluing data. Take X and Y to be the total spaces of projectivizations of vector bundles on B. I would bet you could already find a counterexample there (perhaps with total spaces being Fano so you can use Bondal-Orlov to reduce to the case of vector bundles with non-isomorphic projectivizations). $\endgroup$
    – pro
    Dec 22, 2015 at 0:19

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With the added assumptions the answer is yes. In other words, one can check equivalences by looking at fibres.

This is Proposition 2.15 of http://arxiv.org/pdf/math/0610319.pdf (it might be that when everything is smooth this was already known).

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