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Let $\mathcal{C}$ be a (sufficiently complete and cocomplete) closed monoidal category with internal hom $[-,-]$. Let $F : \mathcal{A} \to \mathcal{C}$ be a functor obtained as the left Kan extension of $F' : \mathcal{B} \to \mathcal{C}$ along a fully faithful functor $i : \mathcal{B} \to \mathcal{A}$. If $G : \mathcal{A} \to \mathcal{C}$ is another functor I can construct the composite $$ [F, G] : \mathcal{A}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}^\mathrm{op} \times \mathcal{C} \to \mathcal{C}. $$ Since the value of $F$ on $\mathcal{A}$ depends only on the value of $F'$ on $\mathcal{B}$, it is reasonable to assume that the value of $[F, G]$ on $\mathcal{A}^\mathrm{op} \times \mathcal{A}$ depends only on the value of $[F', G]$ on $\mathcal{B}^\mathrm{op} \times \mathcal{A}$. In fact, thanks to the formula for left Kan extensions, $$ [F(a_1), G(a_2)] \simeq \left[ \operatorname*{colim}_{(b \to a_1) \in i/a_1} F'(b), G(a_2) \right] \simeq \operatorname*{lim}_{(b \to a_1) \in i/a_1} [F'(b), G(a_2)]. $$ This exhibits $[F, G] : \mathcal{A}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}$ as a right Kan extension of $[F', G] : \mathcal{B}^\mathrm{op} \times \mathcal{A} \to \mathcal{C}$ along $\mathcal{B}^\mathrm{op} \times \mathcal{A} \to \mathcal{A}^\mathrm{op} \times \mathcal{A}$.

Is it possible to take this a step further and compute $[F, G]$ in terms of $[F', G \circ i]$? That is, how closely can I mimic the formula $\mathrm{Nat}(F, G) \simeq \mathrm{Nat}(F', G \circ i)$ (where $\mathrm{Nat}$ denotes natural transformations of functors) in my category $\mathcal{C}$ by using the internal hom instead of the hom of functors?

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The analogous adjunction for the enriched category of functors can be deduced from the adjunction for the ordinary category of functors using the Yoneda lemma. Indeed, to establish a natural isomorphism $$[F,G]≅[F',G∘i]$$ it suffices (by the Yoneda lemma) to establish an isomorphism $$\def\cC{{\cal C}}\cC(V,[F,G])≅\cC(V,[F',G∘i]),$$ natural in $V∈\cC$. Commuting the functor $\cC(V,-)$ with the limit that computes $[-,-]$, we transform the above relation to $$\def\Nat{\mathop{\sf Nat}}\Nat(F,G^V)≅\Nat(F',(G∘i)^V).$$ Since the powering $(-)^V$ is computed objectwise, we have $(G∘i)^V=G^V∘i$, so the above isomorphism follows from the ordinary adjunction for Kan extensions applied to $F'$ and $G^V$: $$\Nat(F,G^V)≅\Nat(F',G^V∘i).$$

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    $\begingroup$ Note also that this is true for every adjunction, not only for left extensions and precomposition! $\endgroup$
    – fosco
    Commented Jul 30, 2023 at 7:52
  • $\begingroup$ Sorry, I am confused by your first equation. I am thinking of $[F, G]$ and $[F', G \circ i]$ as functors with different domains (and so I wouldn't know how to interpret an isomorphism between them); are you implicitly taking a limit? $\endgroup$
    – Lorenzo Riva
    Commented Jul 30, 2023 at 15:28
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    $\begingroup$ @LorenzoRiva: I was answering your question precisely as it was formulated in the last sentence (“That is, how closely…”), using $[-,-]$ as one of the standard notations for the enriched hom, i.e., taking limits of the functors you previously described. Of course, if you are interested in individual values of these functors instead, you can repeat the same Yoneda trick, deriving the enriched case from the nonenriched statement. $\endgroup$
    – Dmitri Pavlov
    Commented Jul 30, 2023 at 17:03
  • $\begingroup$ I understand now, thank you! $\endgroup$
    – Lorenzo Riva
    Commented Jul 30, 2023 at 18:13

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