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

- a functor $f^* : B\to [A°,Set]$ by $$(a,b)\mapsto B(fa,b)$$ and
- 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 $$ Nat(B(b,f\_\,), \hom(a,\_\,)) \cong B(fa,b). $$