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Given adjoint functors $F: A \to B$, $G: B \to A$, if you then take their pullback functors $F^* : Set^B \to Set^A$ and $G^* : Set^A \to Set^B$ given by pre-composition, are these two also adjoint (assuming conditions for their adjoints to exist)? According to wikipedia http://en.wikipedia.org/wiki/Functor_category, this is true for post-composition, is it true for pre-composition? I realise these functors have Kan extensions as adjoints but can find nothing in the literature about taking the Kan extension along the adjoint of a functor.

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If $F \dashv G$, then $G^\ast \dashv F^\ast$. Is that what you're asking? –  Todd Trimble Aug 14 at 13:37
However, for the precompositions $F^\ast: Set^{B^{op}} \to Set^{A^{op}}$ and $G^\ast: Set^{A^{op}} \to Set^{B^{op}}$, we do have $F^\ast \dashv G^\ast$ if $F \dashv G$. –  Todd Trimble Aug 14 at 13:45
Yes, that is what I meant :) Ah ok, it's true for $Set^{A^{op}}$? Do you know of where I could find a proof? Because surely then I could just apply this to the opposite functors $F^{op}: A^{op} \to B^{op}$ etc to get the result I need –  Charles Craven Aug 14 at 13:48
It's an easy exercise if you use the definition of adjunction with the triangle identities. –  Zhen Lin Aug 14 at 14:23
If I recall correctly you can find this statement in SGA 4, in the exposé about functoriality of presheaf categories. –  Adeel Aug 14 at 14:46

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This is best understood via the bicategorical characterization of adjunctions in terms of unit and counit 2-morphisms. It is then clear that any bifunctor preserves adjunctions in the obvious sense. And taking presheaf categories and pullback functors is clearly a bifunctor from (small categories)$^{op}$ to categories.

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