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Stream: learning: questions

Topic: Representable Isbell Dual + (conditions) => Representable

Ruby Khondaker (she/her) (Jan 28 2026 at 16:07):

Hi, there's a calculation I want to check whether I've gotten correct, as well as if this falls into some larger story.

Let $\mathcal{C}$ be a small category and $P : \mathcal{C}^\text{op} \to \mathbf{Set}$ a presheaf. We can use Isbell conjugation to obtain a functor $P_* : \mathcal{C} \to \mathbf{Set}$ defined by $P_*(c) := \text{Nat}(P, \text{Hom}(-, c))$.

Suppose $P_*$ is representable by $r \in \text{Ob}(\mathcal{C})$ with universal element $\alpha : P \Rightarrow \text{Hom}(-, r)$. There are two main claims I want to check.

Claim 1: Suppose there is a $u \in P(r)$ such that the composite natural transformation $P \Rightarrow \text{Hom}(-, r) \Rightarrow P$ is the identity. Then $u$ is a universal element for $P$, i.e. $P$ itself is represented by $r$.

Proof: It suffices to check that the composite $\text{Hom}(-, r) \Rightarrow P \Rightarrow \text{Hom}(-, r)$ is the identity. But since this is a map $r \to r$, using the universal property it suffices to check the composite $P \Rightarrow \text{Hom}(-, r) \Rightarrow P \Rightarrow \text{Hom}(-, r)$ is equal to $\alpha$. (So $\alpha$ is "epimorphic" but only for maps $\text{Hom}(-, r) \to \text{Hom}(-, s)$, I guess?). And this follows by rebracketing.

Claim 2: Suppose $P$ sends colimits to limits. Then $P$ is representable.

Proof: By unfolding definitions, we see that $r$ represents the coend $\int^c c \times P(c)$, which may be realised as a colimit over a suitable category of elements. Thus $P$ sends this coend to an end. This means the maps $P(\alpha_s(x)) : P(r) \to P(s)$ for $s \in \text{Ob}(\mathcal{C}), x \in P(s)$ form a universal wedge. By considering the singleton set, we deduce the existence of a $u \in P(r)$ such that $P(\alpha_s(x))(u) = x$ for all $s \in \text{Ob}(\mathcal{C}), x \in P(s)$. But that's precisely the statement that $P \Rightarrow \text{Hom}(-, r) \Rightarrow P$ is the identity! Thus, by claim 1, $P$ is representable.

Ruby Khondaker (she/her) (Jan 28 2026 at 16:08):

The reason I ask is that something akin to this seems to come up in the kan extension formulae for representing objects (of which the formulae for adjoints fall out as a special case).

Ruby Khondaker (she/her) (Jan 28 2026 at 23:25):

Another thing that I think is always true is that $P_*$ preserves limits, so is much more “likely” to be representable than $P$, which may or may not send colimits to limits. It seems like the strategy here to represent P is to find a representation of $P_*$ (relatively straightforward), and then convert this to a representation of $P$ itself (harder).

Morgan Rogers (he/him) (Jan 29 2026 at 11:44):

The last thing ("$P^*$ preserves limits") is true, but I'm not sure that the "likelihood" consequence follows. It really depends if you have some other presentation of $P^*$ around that you can exploit to show representability!

Ruby Khondaker (she/her) (Jan 29 2026 at 11:57):

Morgan Rogers (he/him) said:

The last thing ("$P^*$ preserves limits") is true, but I'm not sure that the "likelihood" consequence follows. It really depends if you have some other presentation of $P^*$ around that you can exploit to show representability!

I guess what I was thinking is $P_*(x) = \int_c \text{Hom}(P(c), \text{Hom}(c, x)) \cong \int_c \text{Hom}(P(c) \times c, x) \cong \text{Hom}(\int^c Pc \times c, x)$. So, as long as $\mathcal{C}$ has the coend $\int^c c \times P(c)$, $P_*$ is representable.

Ruby Khondaker (she/her) (Jan 29 2026 at 12:02):

If it helps, I wrote up some thoughts on this in more detail in my most recent article - https://pseudonium.github.io/2026/01/27/Baby_Yoneda_3_Know_Your_Limits.html, under the section "Representing the Dual". The case I discuss there is for preorders, which avoids some of the subtler points regarding naturality and size issues for arbtirary categories.

Ruby Khondaker (she/her) (Jan 29 2026 at 12:11):

Incidentally, that coend is precisely the weighted colimit you use in computing a left kan extension of the yoneda embedding along the identity C -> C

Ruby Khondaker (she/her) (Jan 29 2026 at 12:47):

Maybe one sense in which “more likely” can be made precise is that $P$ representable always implies $P_*$ representable, but the reverse isn’t necessarily true.

Ruby Khondaker (she/her) (Jan 29 2026 at 14:07):

Actually, thinking on this a little more - since $P_*$ necessarily preserves limits, we have that $P_* \text{ representable } \iff (P_*)^* \text{ representable }$. So you never need to apply this trick more than once!

Morgan Rogers (he/him) (Jan 29 2026 at 17:36):

I see, more likely in that there are potentially more presheaves P with P^* representable that there are representable presheaves! Okay, I can buy that. The connection with ends and Kan extensions is nice, I hope people absorb that.
(Regarding veracity: I haven't spotted any errors but I also haven't made much effort to verify things :) )

Ruby Khondaker (she/her) (Jan 29 2026 at 17:42):

As an example, presheaves of the form $P(-) = \text{Hom}(-, A) \coprod \text{Hom}(-, B)$ are almost never representable - the reason is that the universal element has to either be a morphism into $A$ or a morphism into $B$, but this is then preserved by precomposition. So e.g. if the universal element is a morphism into $A$, you're stuck with morphisms into $A$ and can't reach any morphisms into $B$.

However, in this case $P_*(-) = \text{Hom}(A, -) \times \text{Hom}(B, -)$, which is representable whenever $A$ and $B$ have a coproduct! So this gives a large class of non-representable presheaves for which $P_*$ is representable.

Ruby Khondaker (she/her) (Jan 29 2026 at 17:44):

Thinking about it more carefully, it might actually be impossible for $\text{Hom}(-, A) \coprod \text{Hom}(-, B)$ to be representable.

Josselin Poiret (Jan 29 2026 at 17:48):

Ruby Khondaker (she/her) said:

Suppose $P_*$ is representable by $r \in \text{Ob}(\mathcal{C})$ with universal element $\alpha : P \Rightarrow \text{Hom}(-, r)$. There are two main claims I want to check.

Am I wrong to assume that a representable for $P^* : \mathcal{C} \to \mathbf{Set}$ is given by a natural transformation $\mathrm{Hom}(r, -) \Rightarrow P^*$ instead?

Ruby Khondaker (she/her) (Jan 29 2026 at 17:49):

Yes, but by yoneda that’s an element of $P_*(r)$, which by definition is a natural transformation $P \Rightarrow \text{Hom}(-, r)$.

Josselin Poiret (Jan 29 2026 at 17:51):

right, i didn't unfold at all

Ruby Khondaker (she/her) (Jan 31 2026 at 09:15):

Perhaps related - I recently learned about the definition of a [[total category]], which appears to be a category for which $P_*$ is always representable.

Ruby Khondaker (she/her) (Feb 03 2026 at 08:19):

It appears that exercise 2.17.2 in Borceux Vol 1 is a corollary of this result. The task is to show that a functor F has a limit if the forgetful functor from the category of cones on F has a colimit. The observation which connects these two is that a cocone on the forgetful functor is precisely a natural transformation P -> Hom(-, r), with P(c) the set of cones on F with tip c, viewed as a presheaf. P automatically sends colimits to limits, so it is representable iff P_* is representable, which implies the exercise.

Morgan Rogers (he/him) (Feb 03 2026 at 08:25):

I've really enjoyed this thread. It almost certainly has interesting ramifications in an enriched setting (e.g. in an abelian setting where more limits and colimits coincide, and the absolute limits/colimits are similarly more interesting).

Ruby Khondaker (she/her) (Feb 03 2026 at 08:32):

Thank you! It’s definitely given me an interesting alternative perspective on computing representations. Results like the adjoint functor theorem seem to be about providing sufficient conditions to compute $\int^c c \times P(c)$. I suppose this has emphasised the special role that limits and colimits play among all representable functors?