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Stream: practice: terminology & notation

Topic: Counit of an adjunction is "separating"

Patrick Nicodemus (May 22 2023 at 01:10):

let $f, g : X\to Y$, $h : Y\to Z$. We say that the relationship $R$ holds between $f,g,h$ if for all $s : W\to X$, $fs=gs$ iff $hfs=gfs$. Thus if the category had equalizers, $hf$ and $hg$ have the same equalizer as $f$ and $g$.

What is a good name for this property?

I am looking at adjunctions $F : \mathcal{C} \to \mathcal{D}, G: \mathcal{D}\to\mathcal{C}$ where $F\dashv G$ and for $X$ in $\mathcal{C}$, $Y$ in $\mathcal{D}$ and $f,g$ are parallel morphisms $X\to GY$, we have $R(Ff,Fg,\varepsilon_Y)$, where $\varepsilon$ is the counit of the adjunction.

Patrick Nicodemus (May 22 2023 at 01:38):

That is, I am considering adjunctions where, for all $f, g : X\to G(Y)$ and $s : W\to F(X)$, we have $F(f)\circ s = F(g)\circ s$ iff $f^{\sharp}\circ s = g^\sharp\circ s$ (i.e., if $\varepsilon_Y\circ F(f)\circ s= \varepsilon_Y\circ F(g)\circ s$, where $\varepsilon$ is the counit of the adjunction.

Jean-Baptiste Vienney (May 22 2023 at 01:41):

Could you provide a concrete example of such an adjunction or more generally some triplet of morphisms $f,g,h$ such that $R(f,g,h)$?

Jean-Baptiste Vienney (May 22 2023 at 01:43):

Other question: what is $f^{\sharp}$?

Patrick Nicodemus (May 22 2023 at 02:11):

Let $\mathcal{C}$ be a category with products. Let $Y$ be an arbitrary object and write $G$ for the comonad $-\times Y$. I am looking at the co-Kleisli category of co-free $G$-algebras. The objects of this category are "trivial bundles" over $Y$, i.e., projections of the form $\pi_Y : A\times Y\to Y$ for arbitrary objects $A$. Morphisms in this category are morphisms of bundles making the triangle commute. If $A\times Y\to Y$ and $B\times Y\to Y$ are two trivial bundles, maps between them must have the form $(f, \pi_Y)$ where $f: A\times Y\to B$ is arbitrary.

If $X$ is an arbitrary object in $\mathcal{C}$ and $(s,t) : X\to A\times Y$ is a morphism, then given two morphisms of bundles $(f,\pi_Y)$ and $(g,\pi_Y)$ from $A\times Y$ to $B\times Y$, it is clear that to prove $(f,\pi_Y)\circ (s,t)=(g,\pi_Y)\circ (s,t)$ it would be necessary and sufficient to prove $f\circ (s,t)=g\circ (s,t)$. Somehow the map $\pi_Y$ carries no interesting information, it merely has to be carried along for structural reasons, but it doesn't affect the reasoning about the diagram.

I need this observation for a proof and I have been trying to think about how to generalize it to comonads other than $G(A)=A\times Y$. This condition is what I have come up with so far.

Patrick Nicodemus (May 22 2023 at 02:16):

This is all true for obvious reasons, because the projections of a Cartesian product are jointly monic. But it's difficult to express this in the language of comonads, and in the problem I'm working on I think working with "an arbitrary comonad $G$ satisfying some nice properties" will pay off over working specifically with the comonad $-\times Y$ if I can do this successfully.
(One cheap out would be to assume a terminal object and then just use $G(1)$ as a proxy for $Y$ but I'm not super happy with that approach.)

Patrick Nicodemus (May 22 2023 at 02:19):

The sharp operator refers to passing across the bijection $Hom(A, GB)\cong Hom(FA,B)$ of the adjunction. So if $f : A\to GB$ then $f^\sharp: FA\to B$ is its "transpose."

Patrick Nicodemus (May 22 2023 at 02:20):

Hope this helps.

Patrick Nicodemus (May 22 2023 at 02:27):

I have asked Chat-GPT for suggestions out of boredom, and it fabricated some convincing nonsense about this being a special case of the Beck-Chevalley condition for an adjunction, as well as giving LaTeX code for a commutative diagram using the xymatrix package. It was well-written enough that I googled "Beck-Chevalley condition for an adjunction" for 5 minutes wondering if this existed before concluding it does not.

David Egolf (May 22 2023 at 02:30):

Out of curiosity, was this chatGPT 4 or the free version of chatGPT? I'm very curious how helpful chatGPT can be for math, but I haven't tried to really test it yet.

Googling "Beck-Chevalley" I found this: https://ncatlab.org/nlab/show/Beck-Chevalley+condition However, I don't know if it's relevant; I haven't tried to read the article.

Jean-Baptiste Vienney (May 22 2023 at 02:31):

Concerning the math, thanks for giving more context, that looks interesting. I just need some time to familiarize with these things, so I'm not going to say something interesting like in the next two minutes.

Patrick Nicodemus (May 22 2023 at 02:35):

David Egolf said:

Out of curiosity, was this chatGPT 4 or the free version of chatGPT? I'm very curious how helpful chatGPT can be for math, but I haven't tried to really test it yet.

Googling "Beck-Chevalley" I found this: https://ncatlab.org/nlab/show/Beck-Chevalley+condition However, I don't know if it's relevant; I haven't tried to read the article.

Yes, the Beck-Chevalley condition is indeed an important criterion, it just isn't relevant to this setting. It is nonsensical to speak about the Beck-Chevalley holding for a single adjunction; rather it relates multiple distinct adjunctions and describes how they commute with each other.
I used the free version of ChatGPT. It actually was somewhat helpful in that it pointed out three concepts in category theory which were superficially connected to what I was doing and provided a different perspective, but you have to do some translation:
"This is an instance of X" = This vaguely looks like X
"You can apply this theorem" = This theorem has similar words in it to your question
"This is equivalent to Y" = This is loosely analogous to Y
and so on.

Jean-Baptiste Vienney (May 22 2023 at 02:49):

For starters, I don't even know how $- \times Y$ is a comonad. What are the comultiplication $A \times Y \rightarrow (A \times Y) \times Y$ and the counit $A \times Y \rightarrow A$?

Jean-Baptiste Vienney (May 22 2023 at 02:51):

I would say that the comultiplication is the pairing $(id_{A \times Y},p_{2})$ where $p_{2}: A \times Y \rightarrow Y$ that's just a random try

David Egolf (May 22 2023 at 02:56):

I think p. 159 of Perrone (https://arxiv.org/pdf/1912.10642.pdf ) discusses this

Jean-Baptiste Vienney (May 22 2023 at 03:31):

Is the adjunction generated by the co-Kleisli category of your comonad $- \times Y$ one which verifies the property you are looking for?

Jean-Baptiste Vienney (May 22 2023 at 03:35):

And why are you looking for other ones which verify this property? What is interesting about this property?

Jean-Baptiste Vienney (May 22 2023 at 03:38):

Oh sorry, one of your message had mysteriously disappeared (everything from "Let $\mathcal{C}$ be a category" to "about the diagram") and now I can see it again so maybe the answers I'm looking for are in this message

Jean-Baptiste Vienney (May 22 2023 at 03:39):

I'll try harder to understand this later.

Patrick Nicodemus (May 22 2023 at 03:59):

David Egolf said:

I think p. 159 of Perrone (https://arxiv.org/pdf/1912.10642.pdf ) discusses this

This is indeed what I was referring to, yes. Jean-Baptiste, you are correct about the comultiplication. The counit is the projection $\pi_A: A\times Y\to A$.

Patrick Nicodemus (May 22 2023 at 04:00):

Jean-Baptiste Vienney said:

Is the adjunction generated by the co-Kleisli category of your comonad $- \times Y$ one which verifies the property you are looking for?

Yes, that's right.

Patrick Nicodemus (May 22 2023 at 04:05):

Jean-Baptiste Vienney said:

And why are you looking for other ones which verify this property? What is interesting about this property?

I am not sure if I am seeking other examples, although it couldn't hurt. I am writing a paper about strong monads (with monoidal product = the Cartesian product) and it hinges on the observation that a strength for a monad $T$ is equivalently a distributivity law of the comonad $-\times Y$ over $T$, for all $Y$, naturally in $Y$. Therefore in many situations I benefit from thinking of $-\times Y$ as a comonad and I am looking for proofs which hold for other comonads which have a distributivity law over $T$. But I do not know how relevant this is to the question.