Two-variable adjunction

Question

In Riehl's Category Theory in Context, the following result has proof left as an exercise:

Proposition 4.3.6. Suppose that $F: \mathsf{A} \times \mathsf{B} \rightarrow \mathsf{C}$ is a bifunctor so that for each object $a \in \mathsf{A}$, the induced functor $F(a,-): \mathsf{B} \rightarrow \mathsf{C}$ admits a right adjoint $G_a: \mathsf{C} \rightarrow \mathsf{B}$. Then:

(i) These right adjoints assemble into a unique bifunctor $G: \mathsf{A}^{\mathrm{op}} \times \mathsf{C} \rightarrow \mathsf{B}$, defined so that $G(a, c)=G_a(c)$ and so that the isomorphisms $$ \mathsf{C}(F(a, b), c) \cong \mathsf{B}(b, G(a, c)) $$ are natural in all three variables.

If furthermore for each $b \in \mathsf{B}$, the induced functor $F(-, b): \mathsf{A} \rightarrow \mathsf{C}$ admits a right adjoint $H_b: \mathsf{C} \rightarrow \mathsf{A}$, then:

(ii) There is a unique bifunctor $H: \mathsf{B}^{\mathrm{op}} \times \mathsf{C} \rightarrow \mathsf{A}$ defined so that $H(b, c)=H_b(c)$ and the isomorphisms $$ \mathsf{C}(F(a, b), c) \cong \mathsf{B}(b, G(a, c)) \cong \mathsf{A}(a, H(b, c)) $$ are natural in all three variables.

(iii) In this case, for each $c \in \mathsf{C}$, the functors $G(-, c): \mathsf{A}^{\mathrm{op}} \rightarrow \mathsf{B}$ and $H(-, c): \mathsf{B}^{\mathrm{op}} \rightarrow \mathsf{A}$ are mutual right adjoints.

(We say that a pair of contravariant functors $F:\mathsf{C}^\mathrm{op}\to\mathsf{D}$ and $G:\mathsf{D}^\mathrm{op}\to\mathsf{C}$ are mutually right adjoint if there exists a natural isomorphism $\mathsf{D}(d,Fc)\cong\mathsf{C}(c,Gd)$; on this case (iii) follows by definition and by (ii).)

Note that (ii) follows from (i) plus the isomorphism $\mathsf{A}\times\mathsf{B}\cong\mathsf{B}\times\mathsf{A}$. But how do we show (i)?

Answer 1

For each $c\in\mathsf{C}$, we define a functor $G^c:\mathsf{A}^\mathrm{op}\to\mathsf{C}$ in the following way: on objects, we have $G^ca=G_ac$. To define the action on a morphism $f:a\to a'\in\mathsf{A}$, we first consider the composite natural transformation $$ \mathsf{B}(-,G_ac) \cong\mathsf{C}(F(a,-),c) \stackrel{F(f,-)^*}{\Longrightarrow}\mathsf{C}(F(a',-),c) \cong\mathsf{B}(-,G_{a'}c). $$ By the Yoneda lemma, this natural transformation comes from a morphism $G^cf:G^ca\to G^ca'$. Now, it is easy to see that $G^c$ is a functor. To get a functor $G:\mathsf{A}^\mathrm{op}\times\mathsf{C}\to\mathsf{B}$, we need to verify that $$ \tag{1}\label{cond} G_{a'}g\cdot G^cf=G^{c'}f\cdot G_ag $$ for any $f:a'\to a$ in $\mathsf{A}$ and $g:c\to c'$ in $\mathsf{C}$ (see Corollary in [ref]). The left and right members of last identity correspond to the outer paths in this rectangle:

Note that the left and right squares commute since $F(a,-)\dashv G_a$, whereas the middle one commute since post-composition and pre-composition of morphisms commute. Hence \eqref{cond} holds.

Thus, we have an isomorphism $\mathsf{C}(F(a,b),c)\cong\mathsf{B}(b,G(a,c))$ natural in $b$ and $c$, by hypothesis. Naturality in $a$ is the definition of $G^cf$.