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Stream: learning: questions

Topic: When are exponential objects = limits?

John Onstead (Apr 30 2025 at 11:24):

The category $\mathrm{FinSet}$ has an interesting property in that all exponential objects can be rewritten as limits: the exponential set $X^Y$ can be equivalently expressed as the limit of the constant functor $X: Y \to \mathrm{Set}$ where $Y$ is the discrete category for the set $Y$. In other words, this is a categorification of how exponentiation is repeated multiplication.

John Onstead (Apr 30 2025 at 11:25):

My question is: What are some other categories where a similar property- the ability to express all exponential objects as some kind of limit (perhaps not just products)- holds? Does this property have a well defined name? What are some other consequences of this property, and does any of this have any bearing on Cat and the concept of functor categories?

Morgan Rogers (he/him) (Apr 30 2025 at 13:56):

It's true internally in any CCC, in the sense that we can interpret $X^Y$ as the limit of a $Y$-indexed diagram constant at $X$ (where $Y$-indexed diagram literally means a morphism $Z \to Y$). The colimit of such a diagram always exists (it's $Z$) and the constant diagram is the projection $X \times Y \to Y$. The limit, if it exists, is a right adjoint to the constant diagram functor.

Morgan Rogers (he/him) (Apr 30 2025 at 14:37):

The existence of this limit functor is a property between cartesian closed and locally cartesian closed.

John Onstead (Apr 30 2025 at 19:22):

Ok, let me see if I'm understanding... so as I mentioned above, in $\mathrm{Set}$, $X^Y$ is the limit of a diagram constant at $X$ given by $X: Y \to \mathrm{Set}$. By the wording above, this would count as a "$Y$ indexed" diagram constant at $X$. Then I think I can see where $X \times Y \to Y$ comes in, at least for sets. If you think of the constant functor from $Y$ as an indexed family of $X$ sets, indexed by elements of $Y$, then you can recontextualize it as a bundle where every element of $Y$ has an $X$ set "over" it.

John Onstead (Apr 30 2025 at 19:22):

If this is true, I'm still not sure where the limit comes in. $X \times Y \to Y$ might encode the same data as a functor from $Y$ into $\mathrm{Set}$ constant at $X$, but you can only take limits of functors, not bundles.

John Onstead (Apr 30 2025 at 19:35):

I think I can see the reasoning though. If there exist some category where morphisms and projections in some category are the objects, then there might be a functor from the original category to it sending an object $X$ to the projection map $X \times Y \to Y$, and it might have an adjoint. This sets up a situation formally analogous to a literal category of diagrams $[Y, C]$ where the limit is adjoint to the literal constant diagram functor. So is this truly a "limit" in the strict sense, or just something formally analogous to one?

Nathanael Arkor (Apr 30 2025 at 19:35):

Every category with finite products is powered over $\text{FinSet}$, where the power of a finite set $X$ with an object $C$ is given by $\prod_{x \in X} C$. In particular, in $\text{FinSet}$ itself, exponentials are powers, and so exponentials may be expressed in terms of products. So I would say that powers precisely capture the intuition you describe. (Dually, the fact that products of sets can be expressed as repeated sums is captured by the notion of [[copowers]].)

John Onstead (Apr 30 2025 at 19:39):

Nathanael Arkor said:

So I would say that powers precisely capture the intuition you describe. (Dually, the fact that products of sets can be expressed as repeated sums is captured by the notion of [[copowers]].)

Hmm that sounds interesting! I'm not too familiar yet with powers/copowers so I'll certainly take a look into it.

John Baez (Apr 30 2025 at 20:15):

By the way, I always found the term "copowers to be very cryptic until I heard a synonym for it: "tensors".

Sometimes we have a symmetric monoidal closed category $V$ and a $V$-enriched category $A$ and we can tensor any object $a \in A$ with any object $v \in V$ and get an object $v \otimes a \in A$ obeying a nice universal property. Then we can say $A$ is tensored over $V$ - but if we want to impress and confuse our friends, we can say $A$ is copowered over $A$.

There's also a dual concept of $A$ being powered over $V$, which means we can form objects like $a^v$ obeying a nice universal property. But if we want to impress and confuse our friends, we can say $A$ is cotensored over $V$.

(All the definitions can be found at the link.)

So: completely nice people say "tensored" and "powered" - while completely nasty ones instead say "copowered" and "cotensored". There are also logical but not very friendly people who say "tensored" and "cotensored" or "copowered" and "powered."

Nathanael Arkor (Apr 30 2025 at 20:18):

I think copowered is much better terminology, because copowers are colimits and powers are limits. Also, "tensor" is so overloaded in category theory that it's confusing to use it for yet another concept.

Nathanael Arkor (Apr 30 2025 at 20:18):

Moreover, I think it's confusing to have dual concepts and use completely different words for them.

Kevin Carlson (Apr 30 2025 at 20:38):

Interestingly, though, "tensored" in particular and especially "tensored over" are pretty unambiguous compared to "tensor".

John Onstead (Apr 30 2025 at 21:17):

I've learned more about powers and indeed this seems like something I was looking for. It says that for $V$ any symmetric monoidal closed category, it is powered over itself, where the power and internal hom converge. Since the power is expressed as a weighted limit, I guess this can be taken to mean that any internal hom of any symmetric monoidal closed category can be re-expressed as a weighted limit. That's quite surprising, especially since it seems the category does not have to even necessarily be Cartesian (exponential objects are the internal homs specifically of Cartesian closed categories)

John Onstead (Apr 30 2025 at 21:21):

John Onstead said:

So is this truly a "limit" in the strict sense, or just something formally analogous to one?

I also think I figured this out too. It seems this refers to the concept of an "internal limit". I had no idea about these, and to be honest I thought an "internal limit" was an equipment theoretic notion when it comes to trying to define the internal notion of the universal property of limit to some equipment or virtual equipment. But it seems to instead refer to what was discussed above, quite fascinating!

John Onstead (Apr 30 2025 at 22:11):

I also wanted to bring up my initial motivation for this question, which was about limits in $\mathrm{Cat}$. I wondered if any functor category can be expressed as some kind of limit (I guess the answer to that is yes by the above reasoning). But now I want to see it in action and actually try to characterize a functor category explicitly by a limit.

John Onstead (Apr 30 2025 at 22:12):

For instance, a good first nontrivial example would be $[2, C]$ where $2$ is the walking morphism/arrow, hence this functor category is $\mathrm{Arr}(C)$. By the definition of powering, this is the weighted limit of the diagram $* \to \mathrm{Cat}$ picking out $C$, weighted by a similar functor picking out $2$. But I'm not sure how to show this truly is the arrow category, since I have no experience with weighted limits. Conical limits are easier since you can break them apart into products and equalizers, but that clearly won't work here.

John Onstead (Apr 30 2025 at 22:12):

So my question is: how would you compute this weighted limit, and is there any useful tricks (like the limit = product + equalizer) to help in the computation? (IE, maybe there's a way to convert these kinds of weighted limits in Cat into something like a conical limit, and then do something analogous to product + equalizer)

Kevin Carlson (Apr 30 2025 at 22:17):

The most relevant "trick" is essentially that weighted limit = cotensor+equalizer, or product+cotensor+equalizer, depending on the details of the setting and how you think about the situation. So this particular case is rather atomic.

Kevin Carlson (Apr 30 2025 at 22:20):

If we denote by $(*\mapsto 2)$ the weight $*\to \mathbf{Cat}$ picking out the arrow, then by definition the weighted limit $\{2,C\}$ has the universal property that $\mathbf{Cat}(D, \{2,C\})$ (assuming $D,C$ are categories; they could equally well be objects in any 2-category) is isomorphic to the category of 2-natural transformations $(*\mapsto 2) \to (*\mapsto \mathbf{Cat}(D,C)).$ It's pretty quick to confirm that this category is the same as the functor category $\mathbf{Cat}(2,\mathbf{Cat}(D,C)),$ which is the desired result.

John Onstead (Apr 30 2025 at 22:52):

Kevin Carlson said:

The most relevant "trick" is essentially that weighted limit = cotensor+equalizer, or product+cotensor+equalizer, depending on the details of the setting and how you think about the situation. So this particular case is rather atomic.

I see; it's quite ironic that the evaluation trick involves converting the limit right back into a cotensor/power, when the goal was to define the power in terms of the limit!

Kevin Carlson said:

then by definition the weighted limit $\{2,C\}$ has the universal property that $\mathbf{Cat}(D, \{2,C\})$ (assuming $D,C$ are categories; they could equally well be objects in any 2-category) is isomorphic to the category of 2-natural transformations $(*\mapsto 2) \to (*\mapsto \mathbf{Cat}(D,C)).$

That seems related to the idea that a natural transformation can be given as a functor into an arrow category!

John Onstead (Apr 30 2025 at 22:56):

Maybe a more interesting example would be trying to find the category of spans $a \leftarrow c \to b$. It seems you could potentially start with the arrow category and take the product $\mathrm{Arr}(C) \times \mathrm{Arr}(C)$ to get the category of pairs of morphisms in $C$. Then you can take the projections maps to each individual instance of the arrow category, compose them with the domain functor, and equalize to get the category of pairs of morphisms that share a domain. That sounds like a good candidate for a category of spans! Maybe more complex functor categories can be constructed in a similar way?

Kevin Carlson (Apr 30 2025 at 23:01):

In terms of weighted limits, you could construct the category of spans as the cotensor $C^{\bullet\leftarrow \bullet\to \bullet}.$ It's true that you could break this down as an equalizer in the way you say, for instance because $\bullet\leftarrow \bullet\to \bullet$ is itself a coequalizer of two maps $\bullet \rightrightarrows (\bullet\to\bullet + \bullet\to\bullet)$ and cotensoring maps colimits in the exponent to limits.

John Onstead (May 01 2025 at 02:09):

Kevin Carlson said:

It's true that you could break this down as an equalizer in the way you say, for instance because $\bullet\leftarrow \bullet\to \bullet$ is itself a coequalizer of two maps $\bullet \rightrightarrows (\bullet\to\bullet + \bullet\to\bullet)$ and cotensoring maps colimits in the exponent to limits.

That's quite interesting! In some sense it reminds me of the situation with sheaves- that's another place where taking a hom turns a coequalizer (IE, the expression of a sieve as a coequalizer) into an equalizer (the sheaf condition)!

John Onstead (May 01 2025 at 02:11):

This was a very interesting discussion, I think I'm satisfied with the result. Thanks to all for your help!

Mike Shulman (May 01 2025 at 05:57):

Nathanael Arkor said:

Moreover, I think it's confusing to have dual concepts and use completely different words for them.

Well, people who call copowers "tensors" usually call powers "cotensors". But I also prefer "power" and "copower" due to the alignment with "limit" and "colimit".

My understanding is that until about 15 years ago "tensors" and "cotensors" were the standard terms for enriched categories, with "power" and "copower" reserved for Set-enrichment. I think the switch began with a message by Jeff Egger to the categories email list in September 2008 in which he wrote

In "basic concepts of enriched category theory",
Kelly writes:

Since the cone-type limits have no special position of
dominance in the general case, we go so far as to call
weighted limits simply "limits", where confusion
seems unlikely.

My question is this: why does he not apply the same
principle to the concept of powers? Instead, he
introduces the word "cotensor", apparently in order
to reserve the word "power" for that special case
which could sensibly be called "discrete power".
[This leads to the unfortunate scenario that a
"cotensor" is a sort of limit, while dually a
"tensor" is a sort of colimit.] Is there perhaps
some genuinely mathematical objection to calling
cotensors powers (and tensors copowers) which I may
have overlooked?

to which Steve Lack replied

I had a chat about this with a couple of other long-time users of the
terms tensor and cotensor (Ross Street and Dominic Verity). We
all think that, given the current overburdening of the word tensor,
this would be a sensible change.

Nathanael Arkor (May 01 2025 at 07:11):

Mike Shulman said:

Nathanael Arkor said:

Moreover, I think it's confusing to have dual concepts and use completely different words for them.

Well, people who call copowers "tensors" usually call powers "cotensors". But I also prefer "power" and "copower" due to the alignment with "limit" and "colimit".

(My comment was directed at John's use of "tensored" in combination with "powered".)

Nathanael Arkor (May 01 2025 at 07:13):

I hadn't heard about the emails; that's interesting!

John Onstead (May 01 2025 at 11:19):

There's actually a new question I had about powers that I was struck with a bit ago. The philosophical hand-wavy explanation is that a power is like an "object of morphisms" from some set, if that set were an object of the category. But usually explanations like this aren't meant to be taken seriously or literally. In any case, it did get me wondering about if it were actually possible for there to be a setting where a power is a literal exponential or hom.

John Onstead (May 01 2025 at 11:20):

The best place I could think of is none other than the presheaf category $[C^{op}, \mathrm{Set}]$, since both $C$ and $\mathrm{Set}$ embed. In such a category, there's a notion of a cartesian closed structure with internal hom $\mathrm{HOM}$. This makes me wonder what the object $\mathrm{HOM}(S, X)$ for some set $S$ and some representable $y(X)$ is like and does. If anything is like an "object of morphisms" from some set into an object of a category, it would be whatever presheaf corresponds to that!

John Onstead (May 01 2025 at 11:21):

So that's my question: what is $\mathrm{HOM}(S, X)$ like, is it ever representable, and is there any connection to the powering (either in the original or presheaf category since Yoneda preserves limits) of $X$ by $S$?

Mike Shulman (May 01 2025 at 13:44):

People do/did say "cotensored" as well. But I get that your point was that John didn't.

John Onstead (May 01 2025 at 19:12):

According to this nlab article, there is a relation $\mathrm{HOM}(X, Y)(c) = \mathrm{Hom}_{[C^{op}, \mathrm{Set}]} (y(c) \times X, Y)$. Substituting in everything we get $\mathrm{HOM}(S, y(X))(c) = \mathrm{Hom}_{[C^{op}, \mathrm{Set}]} (\mathrm{Hom}(-, c) \times S, \mathrm{Hom}(-, X))$. But I don't how how to evaluate this term. Normally, you can use Yoneda lemma to reduce expressions involving sets of natural transformations in presheaf categories, but this requires the domain be representable, which $\mathrm{Hom}(-, c) \times S$ is not always. So how would you evaluate this expression (if it can be evaluated)? Does it require end calculus?

John Onstead (May 02 2025 at 05:40):

For completeness here is a proof via end calculus I found online. I rewritten it up here:

(end formula for natural transformations)

$$\mathrm{Hom}_{[\mathcal{C}^{\mathrm{op}}, \mathrm{Set}]}\big(\mathrm{Hom}(d, c) \times S,\ \mathrm{Hom}(d, X)\big) \cong \int_{d} \mathrm{Set}(\mathrm{Hom}(d, c) \times S,\ \mathrm{Hom}(d, X))$$

(currying)

$$\mathrm{Set}(\mathrm{Hom}(d, c) \times S,\ \mathrm{Hom}(d, X)) \cong \mathrm{Set}\big(S,\ \mathrm{Set}(\mathrm{Hom}(d, c),\ \mathrm{Hom}(d, X))$$

(hom preserves limits)

$$\int_{d} \mathrm{Set}(S,\ \mathrm{Set}(\mathrm{Hom}(d, c),\ \mathrm{Hom}(d, X))) \cong \mathrm{Set}\left(S,\ \int_{d} \mathrm{Set}(\mathrm{Hom}(d, c),\ \mathrm{Hom}(d, X))\right)$$

John Onstead (May 02 2025 at 05:40):

(Yoneda lemma)

$$\int_{d} \mathrm{Set}(\mathrm{Hom}(d, c),\ \mathrm{Hom}(d, X)) \cong \mathrm{Hom}_{[\mathcal{C}^{\mathrm{op}}, \mathrm{Set}]}(y(c), y(X)) \cong y(X)(c) = \mathrm{Hom}(c, X)$$

(Definition of powering)

$$\mathrm{Set}(S, \mathrm{Hom}(-, X)) \cong \mathrm{Hom}(-, \mathrm{Pow}(S, X))$$

So, $\mathrm{HOM}(S, y(X))(-) \cong \mathrm{Set}(S, \mathrm{Hom}(-, X)) \cong y(\mathrm{Pow}(S, X))$. Thus proving that a power can indeed be seen as a literal hom between a set and an object in the presheaf category!

John Onstead (May 02 2025 at 05:44):

To be honest, I'm a little disappointed in myself that I've been studying category theory for over 2 years now and still couldn't solve this pretty standard CT problem without internet help. I guess I got to keep practicing- hopefully I'll get there one day!

Kevin Carlson (May 02 2025 at 17:50):

IIUC you haven't been doing your CT studying in a formal setting. There are very few self-taught folks who've gotten as far as you, regardless of time invested. And there's always much more to learn. Don't be discouraged!

John Baez (May 02 2025 at 18:20):

I've been studying category theory for 30 years and still couldn't do that coend calculation. That either means you shouldn't be discouraged, or you should be very discouraged. :upside_down:

I think it's fine for people to learn about the topics they care about, knowing that it's never more than an infinitesimal fraction of all there is to learn.

John Onstead (May 02 2025 at 19:13):

Kevin Carlson said:

IIUC you haven't been doing your CT studying in a formal setting. There are very few self-taught folks who've gotten as far as you, regardless of time invested. And there's always much more to learn. Don't be discouraged!

John Baez said:

I've been studying category theory for 30 years and still couldn't do that coend calculation. That either means you shouldn't be discouraged, or you should be very discouraged. :upside_down:

I think it's fine for people to learn about the topics they care about, knowing that it's never more than an infinitesimal fraction of all there is to learn.

Thanks, I will try my best!