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. 



Does derived equivalence of the fibres imply derived equivalence of the total spaces?
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).


