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Say we have a category $\mathcal C$, and for every category $\mathcal A$, we have a category $\mathcal D_{\mathcal A}$ and a functor $F_{\mathcal A} : \mathcal D_{\mathcal A} \to \mathcal C$, and, moreover, that this is 2-functorial in $\mathcal A$. Suppose further that $F_{\mathcal A}$ is a right adjoint for every $\mathcal A$, and call it's left adjoint $G_{\mathcal A}$. Then we have a function from categories $\mathcal A$ to functors $C \to \mathcal D_{\mathcal A}$, sending each category $\mathcal A$ to the functor $G_{\mathcal A}$.

Question: Is this function (2-)functorial in $\mathcal A$? (If yes, what is it's action on morphisms?; if no, is there an intuitive explanation of why construction of adjoints shouldn't be functorial?)

That is, given $\mathcal A$ and $\mathcal A'$ and a functor $f : \mathcal A \to \mathcal A'$ (which, by assumption, induces a functor $F_f : \mathcal D_{\mathcal A'} \to \mathcal D_{\mathcal A}$ and a natural transformation $T_f : F_{\mathcal A} \to F_{\mathcal A'}\circ F_f$), do we get a natural transformation between $F_f \circ G_{\mathcal A}$ and $G_{\mathcal A'}$?

I've been staring at the universal morphism characterization of adjunctions, and not seeing how it gives functoriality.

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    $\begingroup$ Uniqueness of adjoints gives pseudo-functoriality. $\endgroup$
    – Fernando Muro
    Commented Nov 26, 2014 at 23:19
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    $\begingroup$ What does "2-functorial in $\mathcal{A}$" mean? Do you mean a 2-functor $\mathfrak{Cat} \to \mathfrak{Cat}_{/ \mathcal{A}}$, where $\mathfrak{Cat}_{/ \mathcal{A}}$ is the strict slice 2-category? Or some other variation? $\endgroup$
    – Zhen Lin
    Commented Nov 26, 2014 at 23:48
  • $\begingroup$ It seems like what you're asking is the following: "Suppose I have a cartesian fibration which is locally cocartesian, is it also cocartesian?" The answer to that question is "yes", but I may be misunderstanding your question. $\endgroup$
    – Dylan Wilson
    Commented Nov 27, 2014 at 0:21

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Yes. The map $G_{A'} \to F_f \circ G_A$ is called the mate of $T_f : F_A \to F_{A'} \circ F_f$. It's the composite

$$ G_{A'} \to G_{A'} F_A G_A \to G_{A'} F_{A'} F_f G_A \to F_f G_A$$

of $T_f$ with the unit of the adjunction $G_A \dashv F_A$ and the counit of the adjunction $G_{A'}\dashv F_{A'}$.

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