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It is well known that under the Isbell duality $\text{Spec}\dashv {\cal O} : {\cal V}^{A^°} \leftrightarrows \big({\cal V}^A\big)^°$ representable functors are self-dual, i.e. fixed by the unit and counit of the adjunction: $$\text{Spec}({\cal O}(\hom(-,x)))\cong \hom(-,x)\qquad\qquad {\cal O}(\text{Spec}(\hom(y,-)))\cong \hom(y,-)$$

No mention is made about representables being the unique self-dual objects. Is it true?

If not, how to characterize a bigger class of Isbell self-dual objects? My initial guess was that at least a class of colimits of representables are Isbell-self-dual. They are not, even for coproducts, if this computation is correct:

  1. ${\cal O}(\hom(-,x)\amalg \hom(-,y)) = \hom(x,-)\times \hom(y,-)$;

  2. $\text{Spec}$ of this evaluated in $a$ now equals $$ Nat(\hom(-,x)\times \hom(y,-), \hom(a,-)) $$ that doesn't seem related to the initial presheaf. If you assume that $A$ has coproducts, in fact, it is true that this is $\hom(a,x\amalg y)$, so that $\text{Spec}({\cal O}(h_x\amalg h_y)) \cong h_{x\amalg y}$, different from $h_y\amalg h_y$ in general.

This question has already been asked on math.SE, and it was me that pointed the OP to another MO thread discussing the topic, but I'm seeking some explicit answer now.

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    $\begingroup$ One very obvious type of counterexample is where $A$ is not Cauchy complete, since the presheaf and copresheaf categories for $A$ and its Cauchy completion coincide. A keyphrase btw is "Isbell envelope" aka "Isbell completion". $\endgroup$ – Todd Trimble Jan 23 '17 at 14:28
  • $\begingroup$ So in such a case there are non-representable self-dual objects? And what are they? $\endgroup$ – Fosco Loregian Jan 23 '17 at 14:43
  • $\begingroup$ Maybe I see your point, looking at the $n$Lab page of Cauchy completion: if $A$ is not Cauchy complete there are profunctors $* \to A$ that admit a right adjoint in $\bf Prof$ without being functors. This, however, does not answer to the question how to find one of such profunctor and how/why it is Isbell self-dual. But maybe I'm not seeing the obvious? Thanks! $\endgroup$ – Fosco Loregian Jan 23 '17 at 14:51
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    $\begingroup$ Fosco, all I was saying in the first sentence is that if $\bar{A}$ denotes the Cauchy completion and $a \in Ob(\bar{A})$, then the representable $\bar{A}(-, a)$ is a fixed point of the Isbell conjugation between $\mathcal{V}^{\bar{A}}$ and $\mathcal{V}^{\bar{A}^{op}}$, which coincides with the Isbell conjugation between $\mathcal{V}^A$ and $\mathcal{V}^{A^{op}}$. If $a$ is not already an object of $A$, then this gives an example of a non-$A$ representable that is a fixed point. $\endgroup$ – Todd Trimble Jan 23 '17 at 14:57
  • $\begingroup$ So, are Isbell self-dual objects for $A$ all and the only representables of the Cauchy completion $\bar A$? $\endgroup$ – Fosco Loregian Jan 23 '17 at 15:07
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A quick answer for now, which I might add more to later. Recall that for every adjunction $F \dashv G: C \to D$ there is the notion of "fixed point" of the adjunction which has two faces: either it is an object $c$ of $C$ for which the counit $\epsilon_c: FGc \to c$ is an isomorphism, or it is an object $d$ of $D$ for which the unit $\eta_d: d \to GFd$. The adjunction $F \dashv G$ then induces an adjoint equivalence between the full subcategories $\text{Fix}_{FG}(C)$ and $\text{Fix}_{GF}(D)$, and thus we identify these categories. In the case of the Isbell conjugation adjunction for a small category $X$, the category of fixed points is often called the Isbell envelope or Isbell completion; let's denote it $I(C)$.

In the special case where $C$ is a preorder, i.e., a $\mathbf{2}$-enriched category, $I(C)$ is a preorder which goes by a more famous name: the Dedekind-MacNeille completion. You can find some discussion of this in the MO-answer here. In this case $I(C)$ is both complete and cocomplete (i.e. admits small limits and colimits).

In other cases, the Isbell completion need not be $\mathcal{V}$-complete/cocomplete, as you correctly surmise (see comments below the MO-answer I just mentioned), but it can be quite a bit larger (and certainly larger than the Cauchy completion). It's maybe best to point to some examples. For the case where $\mathcal{V}$ is the monoidal closed category $([0, \infty), \geq, +)$, i.e., where $\mathcal{V}$-categories are metric spaces in the sense of Lawvere, Simon Willerton has an interesting article which connects the Isbell completion to the tight span of metric spaces; you can find a Café discussion here where some examples are computed.

Isbell completions also figure in "comparative smootheology", but I'd need to study and think more before commenting on that. You can also find more references to the Isbell completion in the nLab; as I said, I may come back and add more.

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  • $\begingroup$ "I may come back and add more" you already helped a lot, but please come back! $\endgroup$ – Fosco Loregian Jan 23 '17 at 15:46
  • $\begingroup$ I have an additional question, is $A\mapsto \text{Spec}_A \dashv {\cal O}_A$ a functor from $Cat$ to the category $Adj$ of adjunctions? $\endgroup$ – Fosco Loregian Jan 23 '17 at 20:09
  • $\begingroup$ Maybe it depends what is meant by $Adj$ (a bicategory), and it may depend on which functor you mean. If morphisms in $Adj$ are pairs of functors which commute with both the left and right adjoint parts, then one thing to look at is a functor $Cat^{op} \to Adj$ that takes $f: A \to B$ to the pair $(V^{f^{op}}, (V^f)^{op})$, but a calculation shows this doesn't work. I didn't check, but I'm skeptical that other possibilities involving left or right adjoints to $V^{f^{op}}$ and $(V^f)^{op}$ work either. $\endgroup$ – Todd Trimble Jan 24 '17 at 14:56
  • $\begingroup$ By the way, there's one fact of life which is either disconcerting or interesting depending on your point of view, given by the saying "Injective Hulls are not natural", a title of a paper by Adamek, Herrlich, and Rosicky. An example is the MacNeille completion of a poset, which is a special case of the Isbell completion/envelope. So if you're hoping that Isbell completions are cleanly functorial, you're probably in for a disappointment. :-( I've been burned by this type of hope in the past. $\endgroup$ – Todd Trimble Jan 24 '17 at 15:02
  • $\begingroup$ Argh! So $\lnot$"everything is a functor"? That's something I must remember next time I brag about the good old saying. Let's say that I want to know if, in any sense whatsoever, sending a category to "its" Isbell adjunction is functorial. $\endgroup$ – Fosco Loregian Jan 24 '17 at 15:59

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