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Isbell Duality

$\newcommand{\IsbellSpec}{\mathsf{Spec}}\newcommand{\IsbellO}{\mathsf{O}}\newcommand{\Sets}{\mathsf{Sets}}\newcommand{\rmL}{\mathrm{L}}\newcommand{\rmR}{\mathrm{R}}\newcommand{\B}{\mathrm{B}}\newcommand{\Nat}{\mathrm{Nat}}$The Yoneda lemma states that, given a presheaf $\mathcal{F}$ on a category $\mathcal{C}$ and a copresheaf $F$ on $\mathcal{C}$, we have \begin{align*} \mathcal{F}(A) &\cong \Nat(h_{A},\mathcal{F}),\\ F(A) &\cong \Nat(h^{A},F), \end{align*} naturally in $A\in\mathrm{Obj}(\mathcal{C})$.

Isbell duality then arises by considering instead the assignments $A\mapsto\Nat(\mathcal{F},h_{A})$ and $A\mapsto\Nat(F,h^{A})$, which define functors

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leading to the Isbell conjugacy adjunction

$$(\IsbellO\dashv\IsbellSpec)\colon\mkern9mu\mathsf{PSh}(\mathcal{C})\rightleftarrows\mathsf{CoPSh}(\mathcal{C})^\mathsf{op}.$$

This adjunction comes with a unit and counit having components of the form \begin{align*} \eta_{\mathcal{F}}&\colon\mathcal{F}\to[\IsbellSpec\circ\IsbellO](\mathcal{F}),\\ \epsilon_{F}&\colon[\IsbellO\circ\IsbellSpec](F)\to F, \end{align*} and a presheaf $\mathcal{F}$ is called Isbell self-dual if $\eta_\mathcal{F}$ is an isomorphism, with a corresponding definition for copresheaves $F$.

The case of monoids and groups

Given any monoid or group $A$, we have a category $\B{A}$, called the delooping of $A$, consisting of a single object $\bullet$ with $\mathrm{Hom}_{\mathcal{C}}(\bullet,\bullet)=A$, the unit and composition maps coming from the monoid maps of $A$.

Taking $\mathcal{C}=\B{A}$ in the context of Isbell duality discussed above, we obtain an adjunction between left and right $A$-sets: $$(\mathsf{O}\dashv\mathsf{Spec})\colon\mkern9mu\Sets^\rmL_A\rightleftarrows\Sets^{\rmR,\mathsf{op}}_A,$$ where:

  1. The functor $\mathsf{O}$ sends a left $A$-set $(X,\lambda)$ to the set $\Sets^{\rmL}_{A}(X,A)$ of maps of left $A$-sets from $(X,\lambda)$ to the left regular representation $(A,\cdot_A)$ of $A$.
  2. Similarly, the functor $\mathsf{Spec}$ sends a right $A$-set $(X,\rho)$ to $\Sets^{\rmR}_{A}(X,A)$.
  3. The unit of this adjunction has components of the form $$\eta_X\colon X\to\Sets^{\rmR}_{A}(\Sets^{\rmL}_{A}(X,A),A)$$ and is given by evaluation, i.e. by $x\mapsto[f\mapsto f(x)]$.
  4. Similarly, the counit has components $$\epsilon_X\colon X\to\Sets^{\rmL}_{A}(\Sets^{\rmR}_{A}(X,A),A)$$ also given by evaluation, i.e. by $x\mapsto[f\mapsto f(x)]$.

Question

In general, what is known about Isbell duality in this setting? For instance:

  1. Can we deduce any interesting properties of $A$ based on properties of $\IsbellO$, $\IsbellSpec$, the unit and counit of the Isbell adjunction, etc.?

    In particular, what would be some references on semigroups or group theory developing some of the theory of "dual $A$-sets" $\mathrm{Sets}^{\mathrm{L}}_A(X,A)$ and $\mathrm{Sets}^{\mathrm{R}}_A(X,A)$, similarly to dual modules?

  2. What is known about Isbell self-dual objects in this setting, i.e. those $A$-sets $X$ for which the map $$\mathrm{ev}_{(-)}\colon X\to\Sets^{\rmR}_{A}(\Sets^{\rmL}_{A}(X,A),A)$$ is an isomorphism? What about those $A$-sets for which this map is injective (cf. torsionless modules) or surjective?

    For instance, when $A$ is a group, there is a complete characterisation of the Isbell self-dual $A$-sets, given by Avery–Leinster in Example 6.5 of Isbell conjugacy and the reflexive completion.

  3. The algebras for the Isbell duality adjunction consist of $A$-sets $X$ with a map $$\phi\colon\Sets^{\rmR}_{A}(\Sets^{\rmL}_{A}(X,A),A)\to X$$ satisfying $\phi(\mathrm{ev}_x)=x$ for each $x\in X$ and a more complicated condition relating applying $\phi$ twice with evaluation.

    Is there a nice description of them?

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    $\begingroup$ Sets of the form Ae with e idempotent have this property. I’m not sure what else would. Would it not depend very much on the monoid? $\endgroup$ Commented Apr 14 at 11:47
  • $\begingroup$ @BenjaminSteinberg Perhaps! I've refactored the question to account for this (and also try to improve it in general) $\endgroup$
    – Emily
    Commented Apr 14 at 17:17
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    $\begingroup$ For groups this should be easily calculable. Every (either left or right) $A$-set is a coproduct of orbits, each of the form $A/A'$, resp. $A'\backslash A$, for some subgroup $A'$ of $A$, and there is no morphism from such an orbit to $A$ unless $A'$ is trivial. So values of the adjoints are the free left, resp. right $A$-sets $\endgroup$ Commented Apr 14 at 18:47
  • $\begingroup$ @მამუკაჯიბლაძე Yep! Thanks, მამუკა. Incidentally, Example 6.5 of Avery–Leinster's Isbell conjugacy and the reflexive completion is a reference for this; since posting the original question I've found out about their paper and have edited the question accordingly $\endgroup$
    – Emily
    Commented Apr 19 at 21:44

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