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Let $C$ be a base category, $F,G$ be two categories fibered over $C$ and $F \to G$ be a morphism. The following criterion is used very often: If all the fiber functors $F_U \to G_U$ ($U \in C$) are equivalences, then $F \to G$ is an equivalence. Although most books seem to use this as it is was trivial, a little work has to be done; see for example the notes by Angelo Vistoli on fibered categories, Section 3.5.

Question. Assume that all fiber functors $F_U \to G_U$ have a left adjoint $G_U \to F_U$. Is it possible to extend these to a left adjoint $G \to F$ of $F \to G$?

Somehow this feels like the assertion "adjunctions form a stack". I don't know if this is true at all. : Somehow we have to arrange that the units $1 \to F_U G_U$ and counits $G_U F_U \to 1$ are compatible in $U$, but is this possible? Any references (at last, this should be well-known) are welcome. Note the similarity to one of my former questions.

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Let $C$ be a base category, $F,G$ be two categories fibered over $C$ and $F \to G$ be a morphism. The following criterion is used very often: If all the fiber functors $F_U \to G_U$ ($U \in C$) are equivalences, then $F \to G$ is an equivalence. Although most books seem to use this as it is was trivial, a little work has to be done; see for example the notes by Angelo Vistoli on fibered categories, Section 3.5.
Question. Assume that all fiber functors $F_U \to G_U$ have a left adjoint $G_U \to F_U$. Is it possible to extend these to a left adjoint $G \to F$ of $F \to G$?