Functors in Isbell duality exchange $f^*a$ and $f_*a$

Question

As you maybe remember, Isbell duality is an adjunction $$\mathcal O : [A°,Set] \leftrightarrows [A,Set]° : {\cal S}pec$$ as defined here; since every functor $f : A\to B$ defines both

1. a functor $f^* : B\to [A°,Set]$ by $$(a,b)\mapsto B(fa,b)$$ and
2. a functor $f_* : B\to [A,Set]°$ by $$(a,b)\mapsto B(b,fa)$$

then it is reasonable to ask

is it true that $\mathcal O \circ f^*\cong f_*$ and ${\cal S}pec \circ f_* \cong f^*$?

In other words the presheaf $B(f\_\,,b)$ is exchanged with $B(b, f\_\,)$ by $\mathcal O$, and similarly $f_*b$ becomes $f^*b$ when it is post-composed with the ${\cal S}pec$ functor. Both $\cal O$ and ${\cal S}pec$ admit very explicit descriptions as $$ \begin{gather*} {\cal O}(P)(A) = Nat(P, \hom(\_\,,A))\\ {\cal S}pec(Q)(A) = Nat(Q, \hom(A,\_\,)), \end{gather*} $$ and while it seems to me that $\cal O$ has this property, I see no way to prove that $$ Nat(B(b,f\_\,), \hom(a,\_\,)) \cong B(fa,b). $$

Answer 1

Let's give names to these two separate conditions: \begin{align} Nat(B(b,f-),hom(a,-)) &\cong B(fa,b) \tag{1} \\ Nat(B(f-,b),hom(-,a)) &\cong B(b,fa) \tag{2} \end{align} Neither of these must be true in general, though they hold under some natural conditions. I will treat (2) first, since (1) is completely dual.

As a counterexample, take $A = 1$ the terminal category, then $f$ corresponds to an object of $B$, and (2) reduces to a natural isomorphism $$ Set(B(f,b),1) \cong 1 \cong B(b,f) $$ which is false unless $f$ is a terminal object in $B$.

A sufficient condition for (2) is that $f : A \to B$ is both fully faithful and dense, meaning that there are natural isomorphisms \begin{align} A(a,a') &\cong B(fa,fa') \tag{$f$ fully faithful}\\ B(b,b') &\cong Nat(B(f-,b),B(f-,b')) \tag{$f$ dense} \end{align} We then derive (2) in two steps: $$ Nat(B(f-,b),hom(-,a)) \cong Nat(B(f-,b),B(f-,fa)) \cong B(b,fa) $$ (As an aside, it is perhaps worth mentioning that fully faithful and dense functors play a role in the theory of monads with arities.)

Essentially the same counterexample works with (1) (now taking $f : 1 \to B$ to be any non-initial object of $B$), and for a sufficient condition we can replace density with the codensity-like assumption that the functor $f_* : B \to [A,Set]^\circ$ is fully faithful, i.e., that there is an isomorphism $B(b,b') \cong Nat(B(b',f-),B(b,f-))$.

Update (14 June): Restoring original answer (with expanded commentary) after realizing the flaw in the OP's counterargument. The isomorphisms ${\mathcal O} \circ f^* \cong f_*$ and $Spec \circ f_* \cong f^*$ cannot follow by preservation of (left/right) Kan extensions along (left/right) adjoints, since $f^*$ is a left Kan extension while $f_*$ is a right Kan extension.

Answer 2

I think a formally similar result is true provided we redefine $f^*$ and $f_*$.

Covariant and contravariant Yoneda embeddings commute with Isbell functors in the sense that

$Spec \circ Z \cong Y$

$\mathcal{O} \circ Y \cong Z$

where $A \xrightarrow{Y(A)} [A^{op}, set]$ and $A \xrightarrow{Z(A)}[set, A]^{op}$.

The isomorphisms are just evaluations $\chi^{Y}$ and $\chi^{Z}$ (Street-Walters notation) which are iso since $Y$ and $Z$ are full and faithfull.

For $A \xrightarrow[]{f} B$ it follows directly that

$Spec \circ f^* \cong f_*$

$\mathcal{O} \circ f_* \cong f^*$

provided $f^* := Z(B) \circ f $ and $f_* := Y(B) \circ f $.

This means that $B(f(a), \_ )$ and $B(\_, f(a))$ are converted to each other by composition with Isbell functors.