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

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 $$Nat(B(b,f\_\,), \hom(a,\_\,)) \cong B(fa,b).$$

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.

• And yet it seems to me that ${\cal O}$ must send $Lan_fy=f^*$ to $Lan_fy^\sharp=f_*$ since it is a left adjoint: $${\cal O} \circ B(f,1) = {\cal O}\circ Lan_f y_A \cong Lan_f ({\cal O} \circ y_A) \cong Lan_f y_A^\sharp = B(1,f)$$ where $y^\sharp : A \to [A,Set]°$... – Fosco Loregian Jun 13 '18 at 9:18
• I must be missing something. When $A = 1$, the Isbell adjunction reduces to the contravariant adjunction on Set induced by exponentiation into the singleton set. I don't see how composing with $1^{(-)}$ has a hope of transporting $f^*$ to $f_*$, or vice versa, unless $f : 1 \to B$ is an initial/terminal object. – Noam Zeilberger Jun 13 '18 at 10:11
• @FoscoLoregian Okay, I've thought about this again (sleep is good!), and I'm pretty sure my original answer and counterexample were correct. The reason why your argument above doesn't work is simply that $f_*$ is not the left Kan extension of $y^\sharp$ along $F$, it is the right Kan extension. – Noam Zeilberger Jun 14 '18 at 7:26
• Isbell adjunction is contravariant, I thought this took care of the fact you point out. But maybe I've been tricked into believing this by an incorrect number of dualizations? I agree with your counterexample, but I need to understand more what's going on. So: I'm surely accepting the answer, but can you please elaborate on $B(1,f)$ being a right and non-left extension? (perhaps you prefer to discuss the topic in chat?) – Fosco Loregian Jun 14 '18 at 8:32
• I was confused as well, but since $[A,Set]^\circ = [A^\circ,Set^\circ]$, the point is that a natural transformation $G \Rightarrow H : B \to [A,Set]^\circ$ corresponds to a family of maps $G(b)(a) \to H(b)(a)$ in $Set^\circ$, hence a family of functions $H(b)(a) \to G(b)(a)$. Then $B(1,f)$ is a right Kan extension rather than a left Kan extension, with the counit $\epsilon : B(1,f) \circ f \Rightarrow y^\sharp$ corresponding to the family of functions $A(a',a) \to B(fa',fa)$. – Noam Zeilberger Jun 14 '18 at 9:08

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.

• Yes, that is precisely what's written in the "Roma lecture notes"! In fact one has simply to distinguish (what you call) $B\langle f,1\rangle$ from (what you call) $B(f,1)$. These functors correspond each other under op-ing and un-currying. – Fosco Loregian Jun 14 '18 at 14:07
• This is solved and I'm happy with it, but do you have any clue about how to prove that $B\langle f,1\rangle$ and $B(f,1)$ correspond each other under the adjunction $P^\sharp\dashv P$ recalled here? I expect it to be true. – Fosco Loregian Jun 14 '18 at 14:18
• What exactly do you mean by that question? Not sure if this gets at what you are asking, but working in terms of distributors, $B(f,1) : B \to A$ can be defined as the left Kan extension of the identity $A \to A$ along $B\langle f,1\rangle : A \to B$, so automatically $B\langle f,1\rangle \dashv B(f,1)$. – Noam Zeilberger Jun 14 '18 at 15:12
• The question means the following: given $f :A\to B$, does the functor $B(f-_2, -_1) : B\to PA$ correspond to $B(f-_1,-_2) : A\to P^\sharp B$ under the adjunction isomorphism $[B, PA]\cong [A, P^\sharp B]$. I have a proof for this, with a certain amount of pain to write down unit and counit for $P^\sharp\dashv P$. – Fosco Loregian Jun 15 '18 at 11:23