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Stream: learning: questions

Topic: definition of an enriched natural transformation

Bruno Gavranović (Nov 25 2020 at 12:18):

Hi all, I've recently posted a question on twitter asking why the definition of an enriched natural transformation is the way it is: https://twitter.com/bgavran3/status/1331229889756848132

Interestingly enough, exactly a year later I ended up thinking about the same question. @PaoloPMath's perspective here https://twitter.com/PaoloPMath/status/1194605406951227393) helped quite a bit, and after some discussions with @mattecapu some more exciting questions came up! 1/n https://twitter.com/bgavran3/status/1193951310283976706

- Bruno Gavranović (@bgavran3)

I'm only now realizing that perhaps I should've been using this chat all along for all my CT queries :grinning_face_with_smiling_eyes:
The question is pretty well laid out on the link, but the main idea is: why does an enriched natural transformation use maps from the monoidal unit $I$, as opposed from some object $P : \mathcal{V}$?

Using objects that are different from the monoidal unit seems to be in line what enriched CT is all about, and the whole reasoning of using weighted limits. But Kelly's enriched CT book (and all other sources I could find) define it as a map from the monoidal unit.

Matteo Capucci (he/him) (Nov 25 2020 at 12:54):

I want to point out that using $I$ is not necessary to express naturality. Well, it is if you do it the way Kelly does, using the unitors of $\mathcal V$, but it is not if you just write down the equation expressing naturality.
What I mean here is that given two $\mathcal V$-functors $F, G : \mathbf C \to \mathbf D$, one can define a $\mathcal V$-natural transformation $\alpha:F \to G$ to be an $ob \mathbf C$-indexed collection of morphisms $\alpha_X : P \to \mathbf D(FX, GX)$.
Then you ask naturality in this way: given $f : Q \to \mathbf C(X,Y)$, you want

$\alpha_X \otimes (f;G_{X,Y}) ; \circ_D = ((f;F_{X,Y} \otimes \alpha_Y) ; \circ_D$

Matteo Capucci (he/him) (Nov 25 2020 at 13:04):

Fixing $P$ for all the $\alpha_X$ seems unavoidable, since there would be no way to express naturality then (basically, the left and right hand of the above equation would have a different source!)
To relax this, one could go galaxy brain and use the end definition of natural transformations:
$[F,G] = \int_{X \in \mathbf C} \mathbf D(FX,GX)$
Then one would add a weight $W : \mathcal C \to \mathcal V$ and define $W$-weighted natural transformations to be
$[F,G]^W = \int_{X \in \mathbf C} [WX, \mathbf D(FX,GX)]$

Matteo Capucci (he/him) (Nov 25 2020 at 13:07):

Indeed, this is what one does to define co/limits in the enriched settings! So why not here?

Matteo Capucci (he/him) (Nov 25 2020 at 13:14):

Matteo Capucci said:

[D]efine $W$-weighted natural transformations to be
$[F,G]^W = \int_{X \in \mathbf C} [WX, \mathbf D(FX,GX)]$

Also, let me precise that one would then consider natural transformations $F \to G$ to be natural transformations with any weight, so terms of $\int^W [F,G]^W$ or simply $\coprod_W [F,G]^W$.
I'm fairly sure putting all of them togheter saves their compositional structure: if you have natural transformations $\alpha : [F,G]$ (weighted, say, by $W$) and $\beta : [G,H]$ (weighted by $T$), you should be able to compose them by tensoring, obtaining $\beta\alpha : [F,H]$ weighted by $W \otimes T$.
The identity natural transformation is $I \overset{1_X}\to \mathbf C(X,X) \overset{F}\to \mathbf D(FX,FX)$, and indeed it is the unit of the composition defined above

Reid Barton (Nov 25 2020 at 15:18):

I guess a more down-to-earth viewpoint is what's implicitly in the first few messages of the twitter thread: $\mathcal{V}$-functors between two $\mathcal{V}$-categories form a $\mathcal{V}$-category, and the morphisms of the underlying category are the enriched natural transformations.

Reid Barton (Nov 25 2020 at 15:21):

Whether looking at the underlying category of a $\mathcal{V}$-category is useful or not probably depends quite a lot on $\mathcal{V}$. For example, for $\mathcal{V}$ = abelian groups it's quite a sensible thing to do.

Bruno Gavranović (Nov 25 2020 at 22:19):

Reid Barton said:

Whether looking at the underlying category of a $\mathcal{V}$-category is useful or not probably depends quite a lot on $\mathcal{V}$. For example, for $\mathcal{V}$ = abelian groups it's quite a sensible thing to do.

Right, but the idea is to set up the definitions so that they work for all choices of $\mathcal{V}$. It can be argued that for Lawvere metric spaces the definition of an enriched natural transformation doesn't quite work the way it's set up, or at the very least, misses a lot of interesting structure.

Reid Barton (Nov 25 2020 at 22:29):

Well don't pass to the underlying category if your $\mathcal{V}$ is not the right sort to do it

Bruno Gavranović (Nov 26 2020 at 22:07):

I'm not suggesting to pass to the ordinary category. I want to stay in the enriched setting, but the widely used definition seems to suggest that we have to pass back, if we want to consider 2-cells.

Reid Barton (Nov 26 2020 at 22:28):

Well that's what a 2-cell is--a morphism of an ordinary category

Nathanael Arkor (Nov 26 2020 at 22:44):

Perhaps the relevant distinction is that it is only sensible to consider individual enriched natural transformations in certain situations, whereas considering the collection of enriched natural transformations between two enriched functors is always sensible (for exactly the same reason that passing to the underlying category is only sensible in some settings).

Nathanael Arkor (Nov 26 2020 at 22:45):

And, in your instance, considering individual natural transformations is not something you want to do.

Joshua Meyers (Dec 01 2020 at 15:13):

Matteo Capucci said:

$\alpha_X \otimes (f;G_{X,Y}) ; \circ_D = ((f;F_{X,Y} \otimes \alpha_Y) ; \circ_D$

The left and right hand sides seem to have different domains! The left hand side has domain $P\otimes Q$ and the right hand side has domain $Q\otimes P$. Is there a way to fix this?

Emily (Mar 09 2021 at 19:24):

Matteo Capucci (he/him) said:

Matteo Capucci (he/him) said:

[D]efine $W$-weighted natural transformations to be
$[F,G]^W = \int_{X \in \mathbf C} [WX, \mathbf D(FX,GX)]$

Also, let me precise that one would then consider natural transformations $F \to G$ to be natural transformations with any weight, so terms of $\int^W [F,G]^W$ or simply $\coprod_W [F,G]^W$.
I'm fairly sure putting all of them togheter saves their compositional structure: if you have natural transformations $\alpha : [F,G]$ (weighted, say, by $W$) and $\beta : [G,H]$ (weighted by $T$), you should be able to compose them by tensoring, obtaining $\beta\alpha : [F,H]$ weighted by $W \otimes T$.
The identity natural transformation is $I \overset{1_X}\to \mathbf C(X,X) \overset{F}\to \mathbf D(FX,FX)$, and indeed it is the unit of the composition defined above

Me an Fosco have been working on a project involving precisely this notion! At least for the case $\mathcal{V}=\mathsf{Sets}$, this indeed works (where you can more generally take $W$ to be of the form $W\colon\mathcal{C}^\mathsf{op}\times\mathcal{C}\longrightarrow\mathsf{Sets}$), giving a kind of "graded Godement calculus" of natural transformations and weighted natural transformations. The only problem we are having with the general case is that it seems it's quite hard (if possible at all) to define the end

$$\int_{X\in\mathcal{C}}[W^{X}_{X},\textbf{Hom}_{\mathcal{D}}(F(X),G(X))]_{\mathcal{V}}$$

without requiring $\mathcal{V}$ to have diagonals: how does one define a $\mathcal{V}$-functor $(X,Y)\mapsto[W^{X}_{Y},F(X),G(Y)]_{\mathcal{V}}$ without them?

Matteo Capucci (he/him) (Mar 09 2021 at 19:37):

It's nice to hear this is interesting you as well!

Matteo Capucci (he/him) (Mar 09 2021 at 19:42):

Something like

$$\int_{X \in \mathcal C} \mathcal C(X, Y) \pitchfork_{\mathcal V} \int_{Y \in \mathcal C} [W^X_Y, \mathcal D(FX, GY)]_{\mathcal V}$$

Matteo Capucci (he/him) (Mar 09 2021 at 19:43):

which I guess can be Fubinized to

$$\iint_{X,Y \in \mathcal C} \mathcal C(X,Y) \pitchfork_{\mathcal V} [W^X_Y, \mathcal D(FX, GY)]_{\mathcal V}$$

Matteo Capucci (he/him) (Mar 09 2021 at 19:45):

I don't know if I'm fooling myself here though

Matteo Capucci (he/him) (Mar 09 2021 at 19:48):

The idea is to work with $(X,Y,X',Y') \mapsto [W^X_Y, \mathcal D(FX', GY')]_{\mathcal V}$ so we don't need to duplicate variables. Then we collapse the duplicate variables by imposing equality

Emily (Mar 09 2021 at 21:05):

Matteo Capucci (he/him) said:

The idea is to work with $(X,Y,X',Y') \mapsto [W^X_Y, \mathcal D(FX', GY')]_{\mathcal V}$ so we don't need to duplicate variables. Then we collapse the duplicate variables by imposing equality

This is a nice idea! I think we're still trapped though: the assignment

$$(A,B) \mapsto \int_{Y}[W^{B}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}}$$

defines a $\mathcal{V}$-bifunctor $\int_{Y}[W^{-_{2}}_{Y},\mathcal{D}(F(-_{1}),G(Y))]_{\mathcal{V}}\colon\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}\longrightarrow\mathcal{V}$ which when combined with¹ $\mathcal{C}(-,K)\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathcal{V}$ would give a $\mathcal{V}$-functor

$$\mathcal{C}(-_{3},K)\pitchfork_\mathcal{V}\int_{Y}[W^{-_{2}}_{Y},\mathcal{D}(F(-_{1}),G(Y))]_{\mathcal{V}}$$

from $\mathcal{C}\boxtimes_{\mathcal{V}}\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}$ to $\mathcal{V}$. On the other hand, to get a functor $\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}\longrightarrow\mathcal{V}$, we'd need to set up a morphism

$$\mathcal{C}(A,A')\otimes_{\mathcal{V}}\mathcal{C}(B,B') \longrightarrow \left[ \mathcal{C}(B,K)\pitchfork_\mathcal{V}\int_{Y}[W^{B}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}} , \mathcal{C}(B',K)\pitchfork_\mathcal{V}\int_{Y}[W^{B'}_{Y},\mathcal{D}(F(A'),G(Y))]_{\mathcal{V}} \right]_{\mathcal{V}}$$

which would in particular require having a map

$$\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'}) \otimes_{\mathcal{V}} \left( \mathcal{C}(\textcolor{red}{B},K)\pitchfork_\mathcal{V}\int_{Y}[W^{\textcolor{red}{B}}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}} \right) \longrightarrow \mathcal{C}(\textcolor{blue}{B'},K)\pitchfork_\mathcal{V}\int_{Y}[W^{\textcolor{blue}{B'}}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}}$$

(is using colour okay for you, Matteo?)

so we still need some kind of duplication :/

¹We'll set $K=Y$ in the end, but let's stick with $K$ for the moment to differentiate between "$Y$ the object" and "$Y$ the integration variable"

Emily (Mar 09 2021 at 21:07):

Also, this is completely unrelated to the above, but here's something nice: a main example of a $W$-weighted natural transformation is for $W=\mathrm{Hom}_{\mathcal{C}}(-_{1},-_{2})$: in this case, a $\mathrm{Hom}$-weighted natural transformation is precisely a dinatural one =)

Nathanael Arkor (Mar 09 2021 at 21:14):

On the nLab page, it states that enriched dinatural transformations only make sense when the enriching category is cartesian. Presumably this is related to the fact that you don't seem to be able to avoid requiring diagonals?

Matteo Capucci (he/him) (Mar 09 2021 at 21:33):

Théo said:

(is using colour okay for you, Matteo?)

Absolutely :) thanks for asking

Matteo Capucci (he/him) (Mar 09 2021 at 21:34):

Théo said:

Also, this is completely unrelated to the above, but here's something nice: a main example of a $W$-weighted natural transformation is for $W=\mathrm{Hom}_{\mathcal{C}}(-_{1},-_{2})$: in this case, a $\mathrm{Hom}$-weighted natural transformation is precisely a dinatural one =)

This is very cool

Matteo Capucci (he/him) (Mar 09 2021 at 21:36):

Théo said:

Matteo Capucci (he/him) said:

The idea is to work with $(X,Y,X',Y') \mapsto [W^X_Y, \mathcal D(FX', GY')]_{\mathcal V}$ so we don't need to duplicate variables. Then we collapse the duplicate variables by imposing equality

This is a nice idea! I think we're still trapped though: the assignment

$$(A,B) \mapsto \int_{Y}[W^{B}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}}$$

defines a $\mathcal{V}$-bifunctor $\int_{Y}[W^{-_{2}}_{Y},\mathcal{D}(F(-_{1}),G(Y))]_{\mathcal{V}}\colon\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}\longrightarrow\mathcal{V}$ which when combined with¹ $\mathcal{C}(-,K)\colon\mathcal{C}^\mathsf{op}\longrightarrow\mathcal{V}$ would give a $\mathcal{V}$-functor

$$\mathcal{C}(-_{3},K)\pitchfork_\mathcal{V}\int_{Y}[W^{-_{2}}_{Y},\mathcal{D}(F(-_{1}),G(Y))]_{\mathcal{V}}$$

from $\mathcal{C}\boxtimes_{\mathcal{V}}\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}$ to $\mathcal{V}$. On the other hand, to get a functor $\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}\longrightarrow\mathcal{V}$, we'd need to set up a morphism

$$\mathcal{C}(A,A')\otimes_{\mathcal{V}}\mathcal{C}(B,B') \longrightarrow \left[ \mathcal{C}(B,K)\pitchfork_\mathcal{V}\int_{Y}[W^{B}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}} , \mathcal{C}(B',K)\pitchfork_\mathcal{V}\int_{Y}[W^{B'}_{Y},\mathcal{D}(F(A'),G(Y))]_{\mathcal{V}} \right]_{\mathcal{V}}$$

which would in particular require having a map

$$\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'}) \otimes_{\mathcal{V}} \left( \mathcal{C}(\textcolor{red}{B},K)\pitchfork_\mathcal{V}\int_{Y}[W^{\textcolor{red}{B}}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}} \right) \longrightarrow \mathcal{C}(\textcolor{blue}{B'},K)\pitchfork_\mathcal{V}\int_{Y}[W^{\textcolor{blue}{B'}}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}}$$

(is using colour okay for you, Matteo?)

so we still need some kind of duplication :/

¹We'll set $K=Y$ in the end, but let's stick with $K$ for the moment to differentiate between "$Y$ the object" and "$Y$ the integration variable"

So if I understand what you're saying, we still get into trouble when we try to define the inner end as a universal wedge?

Emily (Mar 09 2021 at 22:36):

Nathanael Arkor said:

On the nLab page, it states that enriched dinatural transformations only make sense when the enriching category is cartesian. Presumably this is related to the fact that you don't seem to be able to avoid requiring diagonals?

It is! I think this statement in the nLab comes from Remark 4.3.8 of version 5 of Fosco's book, who first worked this out. The point there was precisely that we need diagonals to define $\mathcal{V}$-dinaturality. Requiring $\mathcal{V}$ to be Cartesian works, but the weaker condition of having diagonals suffices =)

Emily (Mar 09 2021 at 22:58):

Matteo Capucci (he/him) said:

So if I understand what you're saying, we still get into trouble when we try to define the inner end as a universal wedge?

The inner end works, i think the problem is to combine it with $\mathcal{C}(A,B)$: the cotensor lets us take two $\mathcal{V}$-functors $F,G\colon\mathcal{C}\rightrightarrows\mathcal{V}$ and form another $\mathcal{V}$-functor $F\pitchfork G\colon\mathcal{C}^\mathsf{op}\boxtimes_{\mathcal{V}}\mathcal{C}\longrightarrow\mathcal{V}$. However, doing this for $F=\mathcal{C}(-_{1},K)$ and $G=\int_{Y}[W^{-_{2}}_{Y},\mathcal{D}(F(-_{1}),G(Y))]_{\mathcal{V}}$ doesn't give us the correct $\mathcal{V}$-functor, as it goes from $\mathcal{C}^{\mathsf{op}}\boxtimes_\mathcal{V}\mathcal{C}\boxtimes_\mathcal{V}\mathcal{C}$ to $\mathcal{V}$, rather than from $\mathcal{C}^{\mathsf{op}}\boxtimes_\mathcal{V}\mathcal{C}$ to $\mathcal{V}$: cotensoring together the codomains of $F$ and $G$ doesn't give us something we can take a coend of :/

So we might then try to define it by hand. However, for its components on the Hom-objects, you find that $\mathcal{C}(B,B')$ has to act on two variables at once: the components of the $\mathcal{V}$-functors above go

$$\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'}) \otimes_{\mathcal{V}}\mathcal{C}(\textcolor{red}{B},K) \longrightarrow \mathcal{C}(\textcolor{blue}{B'},K)$$

and

$$\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'}) \otimes_{\mathcal{V}}\int_{Y}[W^{\textcolor{red}{B}}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}} \longrightarrow \int_{Y}[W^{\textcolor{blue}{B'}}_{Y},\mathcal{D}(F(A),G(Y))]_{\mathcal{V}}$$

So since we "use up" one $\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'})$ on each action, to combine them both into a cotensor would require more $\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'})$'s than we have, and a solution to this would be to have a diagonal map $\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'})\longrightarrow\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'})\otimes_{\mathcal{V}}\mathcal{C}(\textcolor{red}{B},\textcolor{blue}{B'})$

Emily (Mar 09 2021 at 23:01):

(By the way, defining $W$-weighted natural transformations as above and defining $\mathcal{V}$-dinaturality via a "formula" like

$$\mathrm{DiNat}_{\mathcal{V}}(F,G) \text{``$=$''} \int_{X}\mathcal{C}(F^{X}_{X},G^{X}_{X})$$

turn out to be equivalent! if we can do one of them, we can do both =)

Matteo Capucci (he/him) (Mar 10 2021 at 08:45):

Ok so maybe there's no hope, by the remark in Fosco's book you linked above

Matteo Capucci (he/him) (Mar 10 2021 at 08:51):

Matteo Capucci (he/him) said:

Something like

$$\int_{X \in \mathcal C} \mathcal C(X, Y) \pitchfork_{\mathcal V} \int_{Y \in \mathcal C} [W^X_Y, \mathcal D(FX, GY)]_{\mathcal V}$$

Also, now that I take a fresh look at it, this doesn't make sense: I use $Y$ out of scope. Also my attempt is doomed by the start: in the formula $\varphi(x,y) \land x = y$, you see both $x$ and $y$ are still duplicated, so I don't get rid of the need of diagonals...

Matteo Capucci (he/him) (Mar 10 2021 at 08:52):

It's so frustrating though

Emily (Mar 10 2021 at 20:42):

It is!!

Emily (Mar 10 2021 at 20:42):

I've tried a lot of different strategies, but they all failed :/

Emily (Mar 10 2021 at 20:44):

I wonder if maybe "$\mathbf{DiNat}_\mathcal{V}(F,G)$ exists"¹ implies "$\mathcal{V}$ has diagonals"

¹in some sense, e.g. "for each pair $(F,G)$ of parallel functors $\mathcal{C}^\mathsf{op}\boxtimes_\mathcal{V}\mathcal{C}\rightrightarrows\mathcal{D}$, there exists an object $\mathbf{DiNat}_\mathcal{V}(F,G)$ of $\mathcal{V}$ satisfying the following properties"

Emily (Mar 10 2021 at 20:47):

Actually maybe this might be a good idea? Can one axiomatise dinatural transformations in terms of their properties?

Matteo Capucci (he/him) (Mar 10 2021 at 21:05):

Théo said:

I wonder if maybe "$\mathbf{DiNat}_\mathcal{V}(F,G)$ exists"¹ implies "$\mathcal{V}$ has diagonals"

I think so... Indeed, to 'weigh' a functor one needs to use up a variable twice: one to compute the weight and one to compute the functor. So it's not a 'linear' operation.

Matteo Capucci (he/him) (Mar 10 2021 at 21:06):

Théo said:

Actually maybe this might be a good idea? Can one axiomatise dinatural transformations in terms of their properties?

:thinking: maybe?

Emily (Mar 10 2021 at 21:46):

I'll think more about this; it seems quite promising!

Emily (Mar 10 2021 at 21:47):

Thanks so much for this conversation!! :smiley:

Matteo Capucci (he/him) (Mar 10 2021 at 21:56):

:blush: Thanks to you for starting it!

Emily (Jun 13 2021 at 21:30):

@Matteo Capucci (he/him) and @Nathanael Arkor: I finally figured out how this works! I think we indeed can't define $\mathbf{DiNat}_{\mathcal{V}}(F,G)$ in general: there seems to be a kind of microcosm principle going on here, and the answer is to enrich in $\mathsf{CoMon}(\mathcal{V})$ instead:

• First, the reason this is not an issue in ordinary category theory is because $\mathsf{CoMon}(\mathcal{V})\cong\mathcal{V}$ for $\mathcal{V}$ Cartesian monoidal, and hence $\mathsf{Cats}_{\mathsf{CoMon}(\mathcal{V})}\cong\mathsf{Cats}_{\mathcal{V}}$ also, so $\mathsf{CoMon}(\mathsf{Sets})$-enriched categories are just plain categories.
• Second, the comultiplication maps on the Homs: they are exactly what you need to define $\mathbf{DiNat}_{\mathcal{V}}(F,G)$ (the comonoid axioms are used to show that it is functorial, for instance);

• Second+1/2: There's a kind of "loss of $\mathbb{E}_{k}$-ness" going on for dinaturality (like taking categories of monoids of ordinary, braided, or symmetric monoidal categories gives a category, a monoidal category, or a symmetric monoidal category respectively):
  - For $\mathcal{V}$-enrichment, $\mathbf{DiNat}_{\mathcal{V}}(F,G)$ can't be defined;
  - For $\mathsf{CoMon}(\mathcal{V})$-enrichment, $\mathbf{DiNat}_{\mathcal{V}}(F,G)$ is an object of $\mathcal{V}$;
  - For $\mathsf{CCoMon}(\mathcal{V})$-enrichment, $\mathbf{DiNat}_{\mathcal{V}}(F,G)$ is a comonoid in $\mathcal{V}$ (and happens to be also a cocommutative one for enriched categories, but this won't happen at this level of cocommutativity for enriched $\infty$-categories).
  So $F,G\mapsto\mathbf{DiNat}_{\mathcal{V}}(F,G)$ gives a functor from pairs of functors enriched in $\mathbb{E}_k$-coalgebras in $\mathcal{V}$ to $\mathbb{E}_{k-1}$-coalgebras in $\mathcal{V}$, where "$\mathbb{E}_{0}$" corresponds to just an object in $\mathcal{V}$, $\mathbb{E}_{1}$ to a comonoid, and $\mathbb{E}_{2}=\mathbb{E}_{3}=\cdots$ to a cocommutative comonoid.

• Third: the counit maps turn out to be exactly what you need to define constant $\mathcal{V}$-functors!

Nathanael Arkor (Jun 13 2021 at 23:05):

That's really fascinating! It is very surprising to me that there's a connection between dinaturality (which to me seems a way to capture mixed variance) and (higher) comonoidal structure. I'd love to understand what's going on more conceptually. Does this mean you can carry out the development of weighted natural transformations as discussed above?

Mike Shulman (Jun 14 2021 at 04:05):

Note that the monoidal structure of $\mathsf{CCoMon}(\mathcal{V})$ is cartesian. So passing to more-commutative comonoids is a way of making the monoidal structure "closer to cartesian".

Matteo Capucci (he/him) (Jun 14 2021 at 09:01):

@Théo wow, this is beautiful

Emily (Jun 15 2021 at 01:24):

@Nathanael Arkor @Matteo Capucci (he/him) I'm very happy that you liked this idea! :smiley:

Emily (Jun 15 2021 at 01:25):

@Mike Shulman I had no idea these were the products in $\mathsf{CCoMon}(\mathcal{V})$, amazing!

Emily (Jun 15 2021 at 01:26):

Nathanael Arkor said:

That's really fascinating! It is very surprising to me that there's a connection between dinaturality (which to me seems a way to capture mixed variance) and (higher) comonoidal structure. I'd love to understand what's going on more conceptually. Does this mean you can carry out the development of weighted natural transformations as discussed above?

The reason the "reduction of cocommutativity" appears is because of Eckmann–Hilton: given $\mathcal{V}$-functors $F,G\colon\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}\rightrightarrows\mathcal{D}$, we'd like to define $\mathbf{DiNat}_{\mathcal{V}}(F,G)$ as the $\mathcal{V}$-end of the assignment

$$(A,B)\mapsto\mathbf{Hom}_{\mathcal{D}}(F^{B}_{A},G^{A}_{B}).$$

However, to make this $\mathcal{V}$-functorial, we need a "comultiplication" $\Delta_{\mathcal{C}}\colon\mathcal{C}\to\mathcal{C}\boxtimes_{\mathcal{V}}\mathcal{C}$, as $\mathbf{Hom}_{\mathcal{D}}(F^{-_{2}}_{-_{1}},G^{-_{1}}_{-_{2}})$ is then given by the composition

image.png

It is here that the reduction of cocommutativity appears: for $\Delta_{\mathcal{C}}$ to exist, we need an enrichment in $\mathsf{CoMon}(\mathcal{V})$, and for it to be a $\mathsf{CoMon}(\mathcal{V})$-functor, we need an enrichment in $\mathsf{CCoMon}(\mathcal{V})$: being $\mathsf{CoMon}(\mathcal{V})$-functorial means in particular that the comultiplication morphisms $\mathbf{Hom}_{\mathcal{C}}(A,B)\to\mathbf{Hom}_{\mathcal{C}}(A,B)\otimes_{\mathcal{V}}\mathbf{Hom}_{\mathcal{C}}(A,B)$ are morphisms of comonoids, and this is is equivalent to the $\mathbf{Hom}$- objects being cocommutative, by Eckmann–Hilton.

Emily (Jun 15 2021 at 01:27):

(This is essentially the many-object version of the isomorphism $\mathsf{CCoMon}(\mathcal{V})\cong\mathsf{CoMon}(\mathsf{CoMon}(\mathcal{V}))$, as we have $\mathsf{Cats}_{\mathsf{CoMon}(\mathcal{V})}\cong\mathsf{CoMon}(\mathsf{Cats}_{\mathcal{V}})$, and then $\mathsf{Cats}_{\mathsf{CCoMon}(\mathcal{V})}\cong\mathsf{CoMon}(\mathsf{Cats}_{\mathsf{CoMon}(\mathcal{V})})\cong\mathsf{CCoMon}(\mathsf{Cats}_{\mathcal{V}})$!)

Emily (Jun 15 2021 at 01:28):

Nathanael Arkor said:

Does this mean you can carry out the development of weighted natural transformations as discussed above?

I think this definitely should work for defining weighted natural transformations, but there seems to be a new complication now: for an enrichment in $\mathsf{CoMon}(\mathcal{V})$, DiNat exists and things go normally, but for an enrichment in $\mathsf{CCoMon}(\mathcal{V})$, it actually exists too much! We have both

• A $\mathcal{V}$-functor $\mathbf{Hom}_{\mathcal{D}}(F^{-_{2}}_{-_{1}},G^{-_{1}}_{-_{2}})\colon\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathcal{V}}\mathcal{C}\longrightarrow\mathcal{V}$,
• A $\mathsf{CoMon}(\mathcal{V})$-functor $\mathbf{Hom}_{\mathcal{D}}(F^{-_{2}}_{-_{1}},G^{-_{1}}_{-_{2}})\colon\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathsf{CoMon(\mathcal{V})}}\mathcal{C}\longrightarrow\mathsf{CoMon}(\mathcal{V})$, and
• A $\mathsf{CCoMon}(\mathcal{V})$-functor $\mathbf{Hom}_{\mathcal{D}}(F^{-_{2}}_{-_{1}},G^{-_{1}}_{-_{2}})\colon\mathcal{C}^{\mathsf{op}}\boxtimes_{\mathsf{CCoMon(\mathcal{V})}}\mathcal{C}\longrightarrow\mathsf{CCoMon}(\mathcal{V})$;
  and taking $\mathcal{V}$-, $\mathsf{CoMon}(\mathcal{V})$-, or $\mathsf{CCoMon}(\mathcal{V})$-ends of these functors will give different universal properties!

(It also doesn't help that it's somewhat hard to find interesting examples of categories enriched in (cocommutative) comonoids for non-Cartesian $\mathcal{V}$ :sweat_smile:)

(Edit: actually I'm a bit confused now: if say Hom(F,G) is a CoMon(V)-functor, then we may take its CoMon(V)-end, but the universal property of ends gives a comonoid structure on its V-end as well; are these two comonoids isomorphic?)

Emily (Jun 15 2021 at 01:30):

But still, I'm pretty sure this should work, even if we end up having three different $\mathbf{Nat}^{[W]}(F,G)$'s, with three levels of "strength" for the associated universal properties...

Emily (Jun 15 2021 at 01:30):

(In any case, I think this should definitely be worth the trouble: weighted natural transformations pave the way to a lot of very nice constructions: they allow you to define weighted co/ends, weighted Kan extensions, weighted adjunctions, weighted co/monads, "coend-y" versions of adjunctions, monads, and Kan extensions (i.e. which are to co/ends what ordinary adjunctions, monads, and Kan extensions are to co/limits), a way to (vertically) compose dinatural transformations into "higher" dinatural ones (i.e. natural transformations weighted by $\mathrm{Hom}^{2}$, $\mathrm{Hom}^{3}$, and so on, which turn out to be related to conjugacy classes), notions of weighted co/limits which come with universal weighted co/cones (i.e. universal elements of $\mathrm{Nat}^{[W]}(\Delta_{\mathrm{lim}^{[W]}(D)},D)$), conical co/limits defined through constant enriched functors, and probably other nice things!)

Nathanael Arkor (Jun 15 2021 at 12:31):

This does sound very interesting, but my intuition is definitely struggling now! I think I will need to wait for some examples to understand how to think about these variations. I was also wondering about some examples of $\mathsf{CoMon(\mathcal V)}$-enriched categories… Looking forward to seeing where this goes next!

Matteo Capucci (he/him) (Nov 08 2024 at 14:16):

@Emily (she/her) did you (and maybe @fosco ) ever continue working on this?
I might need a refreshment of all the above, but today I was once again pondering 'weighted natural transformations' of enriched functors and remembered of this thread.

Emily (she/her) (Nov 08 2024 at 17:50):

Matteo Capucci (he/him) said:

Emily (she/her) did you (and maybe fosco ) ever continue working on this?
I might need a refreshment of all the above, but today I was once again pondering 'weighted natural transformations' of enriched functors and remembered of this thread.

Hey Matteo! This is definitely something I want to finish working on :) (I've sent you a DM with more details)