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Stream: theory: category theory

Topic: Splitting an idempotent to quotient a monad

Ralph Sarkis (Jun 18 2021 at 18:06):

I posted some kind of preface here.

Let $(M,\eta,\mu)$ be a monad on a category where idempotents split. I have shown that a natural family of idempotent homomorphisms of free $M$--algebras $MX \rightarrow MX$ yields a quotient monad by splitting each component. You can specify this family in three different ways:

1. An idempotent natural transformation $K: \mathrm{id}_{\mathbf{C}_M} \Rightarrow \mathrm{id}_{\mathbf{C}_M}$, where $\mathbf{C}_M$ is the Kleisli category of $M$.
2. An idempotent natural transformation $\overline{K}: M \Rightarrow M$ such that each $\overline{K}_X$ is a homomorphism between the free algebras. (EDIT: I changed this condition because I am not sure what I wrote before is equivalent)
3. A natural transformation $K:\mathrm{id}_{\mathbf{C}} \Rightarrow M$ such that $\mu \circ MK$ is idempotent.

I really like this result (mostly because the proof feels like a classic category theoretical argument where you find the right way to state what you have in a categorical way and everything goes smoothly), but unfortunately, I found only one application and a few settings where I thought I could apply it and ended up proving I can't.

I wanted to know if you found the result interesting or if you think it follows easily from some known results or if you have an idea where it can be applied.

Here is a small paper with lots more details: QuotientIdempotent.pdf

Mike Shulman (Jun 18 2021 at 19:59):

Is this different from just having an idempotent on $M$ in the category of monads?

Ralph Sarkis (Jun 18 2021 at 20:29):

In all 3 cases you end up with a natural transformation $\overline{K}:M\Rightarrow M$, but it is not necessarily a monad map. In particular, in my worked out example, it is not a monad map.

Mike Shulman (Jun 18 2021 at 21:40):

Interesting! Doesn't ring any bells for me yet, though.

John Baez (Jun 19 2021 at 00:15):

I really like this result (mostly because the proof feels like a classic category theoretical argument where you find the right way to state what you have in a categorical way and everything goes smoothly), but unfortunately, I found only one application and a few settings where I thought I could apply it and ended up proving I can't.

If it's a nice result with a nice proof you should be able to publish a short paper on it. Not every paper needs to be long and not every paper needs to have applications.

Ralph Sarkis (Jun 19 2021 at 04:07):

Hmm... Any conference/journal suggestions? (I am a CS student so I am not sure where this fits in the mathematics community)

I'll try to look up the ones listed on nlab tomorrow.

John Baez (Jun 19 2021 at 04:17):

Well, the only "pure" category theory journal I publish in is Theory and Applications of Categories, since Journal of Pure and Applied Algebra is published by Elsevier.

John Baez (Jun 19 2021 at 04:19):

(Well, Cahiers is great but I think of that as being more about category theory connected to topology and geometry.)

Mike Shulman (Jun 19 2021 at 05:36):

The full name Cahiers de topologie et géométrie différentielle catégoriques certainly suggests that, but it seems to me that they actually publish plenty of pure category theory.

dusko (Jun 19 2021 at 09:22):

Ralph Sarkis said:

I posted some kind of preface here.

Let $(M,\eta,\mu)$ be a monad on a category where idempotents split. I have shown that a natural family of idempotent homomorphisms of free $M$--algebras $MX \rightarrow MX$ yields a quotient monad by splitting each component. You can specify this family in three different ways:

1. An idempotent natural transformation $K: \mathrm{id}_{\mathbf{C}_M} \Rightarrow \mathrm{id}_{\mathbf{C}_M}$, where $\mathbf{C}_M$ is the Kleisli category of $M$.
2. An idempotent natural transformation $\overline{K}: M \Rightarrow M$ such that each $\overline{K}_X$ is a homomorphism between the free algebras. (EDIT: I changed this condition because I am not sure what I wrote before is equivalent)
3. A natural transformation $K:\mathrm{id}_{\mathbf{C}} \Rightarrow M$ such that $\mu \circ MK$ is idempotent.

I really like this result (mostly because the proof feels like a classic category theoretical argument where you find the right way to state what you have in a categorical way and everything goes smoothly), but unfortunately, I found only one application and a few settings where I thought I could apply it and ended up proving I can't.

I wanted to know if you found the result interesting or if you think it follows easily from some known results or if you have an idea where it can be applied.

Here is a small paper with lots more details: QuotientIdempotent.pdf

nice construction, and nice that you ask about it.

since the idempotents split in $\mathbf{C}$, each of the idempotents $MX\rightarrow MX$ splits in $\mathbf{C}$. consider those whose splitting is $X$. let $^M\mathbf{C}$ be the category where such idempotents are the objects, and the morphisms are the morphisms $X\rightarrow Y$ in $\mathbf{C}$ whose $M$-images commute with the idempotents on $MX$ and $MY$. then $^M\mathbf{C}$ is equivalent to the category of coalgebras on $\mathbf{C}_M$ for the comonad obtained by composing the forgetful functor to $\mathbf{C}$ with the free functor back. this is a consequence of "Quotients in monadic programming" in LICS17 or 18 or better https://arxiv.org/abs/1701.07601.

i am somehow not finding an explicit construction of the components $$KX$ of your natural transformation for an arbitrary monad, which is probably a consequence of a very thick long day on my side. so i am not sure how your construction is related with this. but we seem to be moving around the same set of pebbles.

you ask about applications. the application which drove us to the "Quotients" paper (also called "the coalgebras for a monad") was for programming a family of ciphers functioonaly, and it is sketched in the paper. a slight change of angle opens up a HUGE field of applications: basically the spectral decomposition of adjunctions. that is in
https://arxiv.org/abs/2004.07353
it begins with lots of examples and applications, and it was implemented before it was properly written up. it is an example of a general math result derived from an application. (ross street presented the result in his seminar, which is a math seminar.)

in terms of the above idempotents, the adjunction between $^M\mathbf{C}$ and $\mathbf{C}_M$ also gives a monad on $^M\mathbf{C}$. the category of algebras for this monad is equivalent with the (Eilenberg-Moore) category of algebras $\mathbf{C}^M$ of all $M$-algebras. moreover, the category of coalgebras for the induced monad on $\mathbf{C}^M$ is again equivalent to $^M\mathbf{C}$. the monad on $^M\mathbf{C}$ is the nucleus of the monad $M$. there is also a dual category of idempotents carrying the nucleus of the comonad. the quotient monad that you are getting must be related. applegate and tierney insisted on making the quotient (co)monad idempotent, which then sent them yp transfinite towers...

let me know if you figure out how these things are related.

John Baez (Jun 19 2021 at 20:21):

Mike Shulman said:

The full name Cahiers de topologie et géométrie différentielle catégoriques certainly suggests that, but it seems to me that they actually publish plenty of pure category theory.

Yeah, I don't quite get it. @Joe Moeller was thinking of trying to publish our paper on Schur functors there, but somehow that paper seems too much like representation theory to belong in something called Cahiers de Topologie et Géométrie Différentielle Catégoriques. I think we may try Higher Structures.

John Baez (Jun 19 2021 at 20:23):

There's also a good-looking journal of category I only recently heard about from @David Michael Roberts. It's either Iranian or Iraqi... sorry, I forget.

David Michael Roberts (Jun 19 2021 at 21:30):

Iranian. Here it is https://cgasa.sbu.ac.ir/ It's got a bit more of an algebraic focus

Mike Shulman (Jun 19 2021 at 23:54):

Names often don't get changed, even when the thing being named has changed so that the name doesn't really apply any more.

Ralph Sarkis (May 20 2023 at 21:01):

With @Jean-Baptiste Vienney , we figured out my result above is an instance of the following more abstract result he proved:
image.png

That result seems very natural, so maybe it is already well-known or a consequence of a well-known result. Do you know any reference or examples where this result applies?

Jean-Baptiste Vienney (May 21 2023 at 16:12):

It sounds like it is not something that people or the literature is familiar with. So let me explain how it is very natural, why it can be useful and give examples. I guess that if people find it interesting it will mean that it is worth try to publish this stuff

Jean-Baptiste Vienney (May 21 2023 at 16:13):

As @Ralph Sarkis explained in his preprint, computers struggle with working with quotients.

Jean-Baptiste Vienney (May 21 2023 at 16:13):

That's not very surprising when you see how they are defined.

Jean-Baptiste Vienney (May 21 2023 at 16:16):

Given an instance of some kind of algebraic structure, eg. monoids, groups etc... with underlying set $X$, you take an equivalence relation $R$ on this set, which is compatible in a precise way with this algebraic structure. And then the magic happens when you try to define your operations on $X/R$

Jean-Baptiste Vienney (May 21 2023 at 16:18):

Let say that you are looking at monoids. Suppose that $c,d \in X/R$.

Jean-Baptiste Vienney (May 21 2023 at 16:19):

Then what happens is really difficult to explain, there is a lot of magic

Jean-Baptiste Vienney (May 21 2023 at 16:21):

You somehow define a function on the couple of equivalence classes of elements of $X/R$, but in reality when it's written on an example you see that it is not defined on the couples of elements of $X/R$ but on the couple of elements of $X$

Jean-Baptiste Vienney (May 21 2023 at 16:22):

And you say that "the result doesn't depend on the choice of the representatives of these equivalence classes"

Jean-Baptiste Vienney (May 21 2023 at 16:23):

And you say that "Ok, so I have defined my function from $X/R \times X/R$ to $X/R$ "

Jean-Baptiste Vienney (May 21 2023 at 16:24):

It's really funny that you say the the result doesn't depend on the choice of representatives because I'm wondering if you don't use the axiom of choice at some point and then say that your operation doesn't depend on this choices of representatives

Jean-Baptiste Vienney (May 21 2023 at 16:26):

At the end, what I know is that because of this fuzziness, which can be made clear I'm sure so let's say because of this non-constructivity, you can't program a computer to make computations on some generic quotient.

Jean-Baptiste Vienney (May 21 2023 at 16:27):

Hopefully, some quotients are much better behaved and you can give an algorithm to make the computations on these quotients!

Jean-Baptiste Vienney (May 21 2023 at 16:28):

In the case of quotients of monoids in some monoidal category this is what explains the above proposition.

Jean-Baptiste Vienney (May 21 2023 at 16:29):

Let's see how it works on this example of some group $\mathbb{Z}/n\mathbb{Z}$

Jean-Baptiste Vienney (May 21 2023 at 16:29):

We know that $(\mathbb{Z},+)$ is a group and we have algorithms to compute in this group

Jean-Baptiste Vienney (May 21 2023 at 16:31):

We can look at the function $mod_{n}:\mathbb{Z} \rightarrow \mathbb{Z}$ which takes in an entry some number $x \in \mathbb{Z}$ and gives in output the reduction modulo $n$ of $x$ which is a number $0 \le mod_{n} (x) \le n -1$

Jean-Baptiste Vienney (May 21 2023 at 16:32):

It is an idempotent function ie. $mod_{n}(mod_{n}(x)) = x$.

Jean-Baptiste Vienney (May 21 2023 at 16:33):

Now let's consider the set $[0,n-1] = \{0,...,n-1\}$.

Jean-Baptiste Vienney (May 21 2023 at 16:35):

$mod_{n}$ can be splitted in $mod_{n}=r;s$ where $r:\mathbb{Z} \rightarrow [0,n-1]$ and $s:[0,n-1] \rightarrow \mathbb{Z}$

Jean-Baptiste Vienney (May 21 2023 at 16:35):

and it verifies that $s;r = id_{[0,n-1]}$

Jean-Baptiste Vienney (May 21 2023 at 16:36):

$s:[0,n-1] \rightarrow \mathbb{Z}$ is simply the inclusion.

Jean-Baptiste Vienney (May 21 2023 at 16:37):

and $r:\mathbb{Z} \rightarrow [0,n-1]$ is nothing more than $mod_{n}$ but when we have co-restricted the codomain to the image which is $[0,n-1]$.

Jean-Baptiste Vienney (May 21 2023 at 16:38):

You see that what is going to play the role of $\mathbb{Z}/n\mathbb{Z}$ is really that $[0,n-1]$

Jean-Baptiste Vienney (May 21 2023 at 16:38):

Now I want to define my addition $[0,n-1] \times [0,n-1] \rightarrow [0,n-1]$

Jean-Baptiste Vienney (May 21 2023 at 16:39):

I already know how to do the addition $\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$, abstractly and also on a computer

Jean-Baptiste Vienney (May 21 2023 at 16:40):

It means that I have a program $\nabla:\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$ which perform this addition

Jean-Baptiste Vienney (May 21 2023 at 16:40):

And I have also a program which give me the unit $0$ that I denote $\eta:\ast \rightarrow \mathbb{Z}$

Jean-Baptiste Vienney (May 21 2023 at 16:41):

ie. $\eta(\ast) = 0$. This is a constant program

Jean-Baptiste Vienney (May 21 2023 at 16:42):

My programs $\nabla$ and $\eta$ verify the axioms of a monoid in a monoidal category ie. the diagrams in this definition commute.

Jean-Baptiste Vienney (May 21 2023 at 16:43):

Now how to use my other programs $r,s$ to define the monoid structure on $[0,n-1]$?

Jean-Baptiste Vienney (May 21 2023 at 16:43):

This is explained in the last two diagrams of the proposition.

Jean-Baptiste Vienney (May 21 2023 at 16:44):

That's really the obvious way you compose your programs in the only way you can do it.

Jean-Baptiste Vienney (May 21 2023 at 16:46):

I don't really want to write in phrases because it's going to be too long. Do it yourself on this example if you want to understand completely and you will see this is what we must do

Jean-Baptiste Vienney (May 21 2023 at 16:46):

ie. this exercise $1$

Jean-Baptiste Vienney (May 21 2023 at 16:47):

Now, these operations on $[0,n-1]$ really gives a monoid for some reason, it doesn't work magically!

Jean-Baptiste Vienney (May 21 2023 at 16:48):

This is because the idempotent verifies some conditions!

Jean-Baptiste Vienney (May 21 2023 at 16:49):

These conditions are given by the diagrams $3$ to $5$ in the proposition and they are the obvious conditions to say that "the operations can pass to the quotient", nothing fancy. Check it yourself on this example as exercise 2 if you want to understand it completely

Jean-Baptiste Vienney (May 21 2023 at 16:50):

I think this is enough written, I don't want to write a paper here.

Jean-Baptiste Vienney (May 21 2023 at 16:50):

But I can tell you that there are much more examples

Jean-Baptiste Vienney (May 21 2023 at 16:51):

For instance to define the algebra structure on the symmetric algebra from the symmetrization on the tensor algebra when you are in characteristic $0$

Jean-Baptiste Vienney (May 21 2023 at 16:51):

Same thing for the exterior algebra

Jean-Baptiste Vienney (May 21 2023 at 16:51):

You can also obtain the monad version of the theorem

Jean-Baptiste Vienney (May 21 2023 at 16:52):

And use is to define the monad structure on the symmetric algebra from the one on the tensor algebra in the same conditions

Jean-Baptiste Vienney (May 21 2023 at 16:52):

Finally, it explains how to make a quotient in any situation when you can do it constructively

Jean-Baptiste Vienney (May 21 2023 at 16:52):

I believe that the monad version of the proposition can explain it for all kind of algebras

Jean-Baptiste Vienney (May 21 2023 at 16:53):

I also believe that you can obtain an idempotent like in the hypotheses from any situation where you can do a quotient if you use the axiom of choice

Jean-Baptiste Vienney (May 21 2023 at 16:54):

But what is interesting is of course the situation when you can do it without it in order to have a "constructive quotient"

Jean-Baptiste Vienney (May 21 2023 at 16:55):

That's enough. Tell me if you find this story interesting and we will write a paper!

Jean-Baptiste Vienney (May 21 2023 at 17:09):

Thank you @Ralph Sarkis . I feel like I don't know anything to how to speak to people and I don't know what think the other guys of this

Jean-Baptiste Vienney (May 21 2023 at 17:10):

Like when nobody says anything you never know if they are bored, they find it absurd or very interesting etc...

Ralph Sarkis (May 21 2023 at 17:48):

I think your explanation is great :smiley:

Jean-Baptiste Vienney (May 21 2023 at 18:07):

Yeah thanks, but I would want to know what think the other guys!

Jean-Baptiste Vienney (May 21 2023 at 18:08):

Because you know, we don't know if people find this valuable, or trivial etc...

Jean-Baptiste Vienney (May 21 2023 at 18:10):

That's tricky also to explain stuff but not make it too long because if it's too long nobody read it

Jean-Baptiste Vienney (May 21 2023 at 18:11):

So, if nobody says anything we can try to write the paper and we will see when we submit it :sweat_smile:

John Baez (May 21 2023 at 18:22):

My feeling, when looking at those diagrams, is that they express the concept of lifting an idempotent on some object to some monoid with that underlying object.

John Baez (May 21 2023 at 18:23):

I.e., that the idempotent and its splitting are monoid homomorphisms.

But I didn't look too hard at the diagrams so I didn't think my guess was worth reporting.

Jean-Baptiste Vienney (May 21 2023 at 18:23):

the idempotent is not a monoid homorphism

Jean-Baptiste Vienney (May 21 2023 at 18:23):

I give you an example

Jean-Baptiste Vienney (May 21 2023 at 18:27):

Oh, no I'm not sure, it all depends of what monoid structure you are talking about

Jean-Baptiste Vienney (May 21 2023 at 18:27):

Because maybe in these conditions, $A$ is equipped with two monoid structures

John Baez (May 21 2023 at 18:27):

All your comments came after I looked at that diagram. I'm just explaining why I didn't say anything, since you guys seemed curious about the silence that you encountered here.

Jean-Baptiste Vienney (May 21 2023 at 18:33):

John Baez said:

I.e., that the idempotent and its splitting are monoid homomorphisms.

But I didn't look too hard at the diagrams so I didn't think my guess was worth reporting.

Yes I think it's true

Jean-Baptiste Vienney (May 21 2023 at 18:34):

I think $r$ and $e$ are monoid morphisms, not sure about $s$

Jean-Baptiste Vienney (May 21 2023 at 18:46):

No, $e$ is not a monoid morphism, just $r$

Jean-Baptiste Vienney (May 21 2023 at 18:47):

Example:

Jean-Baptiste Vienney (May 21 2023 at 18:47):

if $e:\mathbb{Z} \rightarrow \mathbb{Z}$ is the function which send $x$ to its representative modulo $n$ in $\{0,...,n-1\}$

Jean-Baptiste Vienney (May 21 2023 at 18:48):

then you don't have $e(x+y)=e(x)+e(y)$

Jean-Baptiste Vienney (May 21 2023 at 18:50):

for instance if $n=5$, then $e(4+4)=e(8)=3 \neq e(4)+e(4)=4+4=8$

Jean-Baptiste Vienney (May 21 2023 at 18:53):

$s:\{0,...,n-1\} \rightarrow \mathbb{Z}$ is neither a monoid morphism

Jean-Baptiste Vienney (May 21 2023 at 18:53):

Example:

Jean-Baptiste Vienney (May 21 2023 at 18:54):

if $n=5$, then $s(4+4)=s(3)=3 \neq s(4)+s(4)=4+4=8$

Jean-Baptiste Vienney (May 21 2023 at 19:09):

I'm still trying to prove that $r$ is a monoid morphism, it should the case

David Egolf (May 21 2023 at 19:10):

In the example you gave, if I understand correctly, you set the idempotent $e: A \to A$ as $mod_n: \mathbb{Z} \to \mathbb{Z}$. The image of $mod_n$ is contained in $\{0, \dots, n-1\}$, which I believe is playing the role of $B$ here.
Do the conditions given on $e$, $r$ and $s$ always force the image of $e$ to be contained in $B$, so that $e$ is a sort of "projection" down to $B$ from $A$?

Jean-Baptiste Vienney (May 21 2023 at 19:13):

Good question, I think I need to know what means "the image of $e$ is contained in B" to answer

David Egolf (May 21 2023 at 19:14):

Yeah, I was just realizing that $B$ is not necessarily a subset of $A$ (I'm thinking of the category of sets together with the cartesian product here, as monoidal categories still scare me!)
Heh, I'm not sure to rephrase my question so it makes sense in a more general setting, just yet.

Jean-Baptiste Vienney (May 21 2023 at 19:15):

You could take $B = \mathbb{Z}/n \mathbb{Z}$ instead but it would give two isomorphic groups.

Jean-Baptiste Vienney (May 21 2023 at 19:21):

(that's what math guys would do but I believe computers would prefer $B=\{0,...,n-1\}$)

Jean-Baptiste Vienney (May 21 2023 at 19:28):

Actually, people in computational number theory prefer to use the second option I believe

David Egolf (May 21 2023 at 19:38):

This seems cool! If I understand, this describes how to make a new monoid from an old monoid and an idempotent that satisfies certain conditions.

I'm curious how varied the resulting produced monoids can be. Continuing the earlier example, what might be a second example of a splitting idempotent $e: \mathbb{Z} \to \mathbb{Z}$ with $e = s \circ r$ and $r \circ s = id_B$? Is the resulting induced monoid on $B$ always isomorphic to $\mathbb{Z}/n\mathbb{Z}$ for some $n$?

I guess I'm wondering if the induced structure associated with $B$ is always a sort of "quotient" of that associated with $A$.

At any rate, thanks for taking the time and energy to share your work!

Jean-Baptiste Vienney (May 21 2023 at 19:49):

David Egolf said:

This seems cool! If I understand, this describes how to make a new monoid from an old monoid and an idempotent that satisfies certain conditions.

I'm curious how varied the resulting produced monoids can be. Continuing the earlier example, what might be a second example of a splitting idempotent $e: \mathbb{Z} \to \mathbb{Z}$ with $e = s \circ r$ and $r \circ s = id_B$? Is the resulting induced monoid on $B$ always isomorphic to $\mathbb{Z}/n\mathbb{Z}$ for some $n$?

I believe it should be the case! But I'm not sure how to prove it now.

Jean-Baptiste Vienney (May 21 2023 at 19:50):

$B$ must be a quotient of $A$ by some subgroup but I don't know how to make this subgroup appear.

Jean-Baptiste Vienney (May 21 2023 at 19:51):

What I already know is that $r:A \rightarrow B$ is a monoid morphism and moreover an epimorphism

Jean-Baptiste Vienney (May 21 2023 at 19:53):

Oh no, I think $B$ is not necessarily isomorphic to a $\mathbb{Z}/ n \mathbb{Z}$!

Jean-Baptiste Vienney (May 21 2023 at 19:54):

Because we are talking of monoids not necessarily groups.

Jean-Baptiste Vienney (May 21 2023 at 19:54):

So I could quotient $\mathbb{Z}$ to obtain $\mathbb{B}=\{0,1\}$ the monoid such that $1+1=1$

Jean-Baptiste Vienney (May 21 2023 at 19:56):

I'm not sure what is the idempotent $e:\mathbb{Z} \rightarrow \mathbb{Z}$ that we should take to obtain $\mathbb{B}$ and not $\mathbb{Z}/2\mathbb{Z}$.

Jean-Baptiste Vienney (May 21 2023 at 20:00):

I'm not sure about that, I don't know if every quotient monoid of $\mathbb{Z}$ is a quotient group of $\mathbb{Z}$

David Egolf (May 21 2023 at 20:03):

Hmm, that's interesting.
To get 1+1=1 in $B$, would that mean that $r(s(1)+s(1)) = 1$? I'm not very comfortable with the monoidal category notation, yet.
If so, I was wondering if this would narrow down what $r$ and $s$ can be (and hopefully that would be helpful for figuring out $e$, if it exists for this case).

And I think we already know that $r(s(1)) = 1$.

Jean-Baptiste Vienney (May 21 2023 at 20:08):

I did some progress. We know that $r:\mathbb{Z} \rightarrow B$ is a monoid morphism from the group $(\mathbb{Z},+)$ and it allows to prove that $B$ is necessarily a group

Jean-Baptiste Vienney (May 21 2023 at 20:09):

For every $b\in B$, we have that $r(-s(b)) + b = r(-s(b))+r(s(b))=r(-s(b)+s(b))=r(0)=0$ so that $r(-s(b))$ is the inverse of $b$.

Jean-Baptiste Vienney (May 21 2023 at 20:09):

(second step because $r \circ s = Id$)

Jean-Baptiste Vienney (May 21 2023 at 20:10):

I also use the fact that $r$ is always a monoid morphism which is a consequence of the hypotheses

David Egolf (May 21 2023 at 20:11):

Very cool! That makes sense!

David Egolf (May 21 2023 at 20:12):

I would guess that means in general if $A$ is a group then $B$ is a group too.

Jean-Baptiste Vienney (May 21 2023 at 20:12):

Exactly!

Jean-Baptiste Vienney (May 21 2023 at 20:15):

I find it much simpler to work with split idempotents like this than with the usual notions of quotient of a monoid by a congruence that I don't even know!

Jean-Baptiste Vienney (May 21 2023 at 20:17):

So know, that we have two groups, we can look at the kernel of the map $r:\mathbb{Z} \rightarrow B$

Jean-Baptiste Vienney (May 21 2023 at 20:18):

which is a subgroup $ker(r) \subseteq \mathbb{Z}$

Jean-Baptiste Vienney (May 21 2023 at 20:18):

And subgroups of $\mathbb{Z}$ are always of the form $n\mathbb{Z}$ for some $n \ge 0$

Jean-Baptiste Vienney (May 21 2023 at 20:19):

So let chose this (unique) $n$ and write $\ker(r) = n\mathbb{Z} \subseteq \mathbb{Z}$

Jean-Baptiste Vienney (May 21 2023 at 20:23):

Jean-Baptiste Vienney said:

So know, that we have two groups, we can look at the kernel of the map $r:\mathbb{Z} \rightarrow B$

which should be a group morphism I believe ie. $r(-x)=-x$...

Jean-Baptiste Vienney (May 21 2023 at 20:30):

By the first isomorphism theorem, we get a morphism of groups $u:\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}$ and a morphism of groups $v:\mathbb{Z}/n\mathbb{Z} \rightarrow B$ such that $r=v\circ u$

Jean-Baptiste Vienney (May 21 2023 at 20:31):

Now, we want to prove that $v$ is an isomorphism.

Jean-Baptiste Vienney (May 21 2023 at 20:32):

We already know that it is a surjection by the theorem of isomorphism

Jean-Baptiste Vienney (May 21 2023 at 20:35):

Oh, the first theorem of isomorphism also say that this $v$ is an isomorphism if $r$ is a surjection which is the case, so we are done.

Jean-Baptiste Vienney (May 21 2023 at 20:37):

To sum up: $B \cong \mathbb{Z}/ n\mathbb{Z}$ where $n\mathbb{Z} = ker(r)$ knowing $B$ is a group and $r:\mathbb{Z} \rightarrow B$ a surjective morphism of group, thanks to the hypotheses on $r,s$ and the fact that $\mathbb{Z}$ is a group.

David Egolf (May 21 2023 at 20:42):

Very cool!

Jean-Baptiste Vienney (May 21 2023 at 21:03):

I was thinking to a more funky application that @John Baez could like. There is a notion of Schur functor which are functors $Vec_{k} \rightarrow Vec_{k}$ for $k$ a field of characteristic $0$. They are defined like this: for every $n \ge 0$ and partition $\lambda \vdash n$, ie. a finite list of numbers $n_{1},...,n_{p}$ such that $n_{1}+...+n_{p}=n$, you have some natural transformation $p_{\lambda}:A^{\otimes n} \rightarrow A^{\otimes n}$ which is an idempotent and the Schur functor $S^{\lambda}:Vec_{k} \rightarrow Vec_{k}$ is defined exactly as the intermediate functor in a splitting $p_{\lambda}=r_{\lambda};s_{\lambda}$ of $p_{\lambda}$ by two natural transformations. Now this is not exactly the same setting, because we now have families of $p_{\lambda},r_{\lambda},s_{\lambda}$, one for every integer partition $\lambda \vdash n$ of some integer $n \ge 0$. But we still have that the the $(A^{\otimes n})_{n \ge 0}$ form a "graded algebra" ie. we have a family of multiplications $\nabla_{n,p}:A^{\otimes n} \otimes A^{\otimes p} \rightarrow A^{\otimes (n+p)}$ and a unit $\eta:I \rightarrow A^{\otimes 0}$ which verify the same diagrams than for a monoid, modulo the grading. There is a definition here. Well, I think I can adapt this proposition to graded monoids and lift the structure of graded algebra on the $(A^{\otimes n})_{n \ge 0}$ to a structure of graded algebra on $(S^{\lambda}(A))_{\lambda}$ which is not a graded algebra that have been explored so far in the litterature as I know.

Jean-Baptiste Vienney (May 21 2023 at 21:05):

knowing that you can define an addition of partitions $\lambda + \mu \vdash n + p$ in the obvious way if $\lambda \vdash n$ and $\mu \vdash p$

Jean-Baptiste Vienney (May 21 2023 at 21:07):

Actually, I think that there is a more complex structure of something like a graded bialgebra which is something I plan to explore during my phd but that would be a first step to know that we have a graded algebra.

Jean-Baptiste Vienney (May 21 2023 at 21:12):

There is a well-known open problem about something named Plethysm which is basically to understand what happens when you compose these functors $S^{\lambda}\circ S^{\mu}$ and I believe that this theorem now in a version "graded monad" could be interesting at least to me to see that you can't apply it to this situation.

Jean-Baptiste Vienney (May 21 2023 at 21:14):

Or maybe, you can, I don't know if the Schur functors give a "graded monad".

Tom Hirschowitz (May 25 2023 at 07:35):

Sorry I'm late to the party and didn't take the time to read in depth, but is your result related to Lemma 6.22 here?
lemma6-22.png

Ralph Sarkis (May 25 2023 at 13:06):

I want to say yes it is related to the original result on quotienting a monad (preprint here). I am in the special case where $X$ and $Z$ are the free $T$-algebra on some $X$, and $m$ is a section of $e$. I do get the same $T$-algebra structure that you define on the splitting which I call $T^KX$, but it is not really explicit there because I only derive that there is a monad structure on $T^K$ (when I have that splitting for every $X$). The free $T^K$ algebra does factor through that $T$-algebra you define, so your lemma could be another starting point to get my result.

Nathanael Arkor (Aug 17 2023 at 11:16):

Ralph Sarkis said:

With Jean-Baptiste Vienney , we figured out my result above is an instance of the following more abstract result he proved:
image.png

That result seems very natural, so maybe it is already well-known or a consequence of a well-known result. Do you know any reference or examples where this result applies?

I recently came across a situation in which one has a retraction of a monoid satisfying some axioms that imply that the retract is also a monoid, and I remembered this thread. However, the situation I have is slightly different. Rather than the two axioms involving $e$ and $r$, I have a single axiom involving $s$ and $e$.
image.png

As far as I can tell, neither set of assumptions implies the other. In my situation, $s$ becomes a monoid morphism (whereas $r$ is not); whereas later in the discussion Jean-Baptiste mentions that in your situation $r$ becomes a monoid morphism (whereas $s$ is not). So perhaps these two results are complementary, in giving natural sufficient conditions for each of $s$ and $r$ to be monoid morphisms.

Ralph Sarkis (Aug 17 2023 at 12:28):

That's great! We have been thinking about this recently with Jean-Baptiste, and we streamlined our thoughts a bit, I'll try to summarize my understanding at this moment.

We work in categories of "algebras" in some category $\mathbf{C}$ (our worked out examples are monoids in $\mathbf{C}$ or $M$-algebras for some monad $M$, and we are not sure if either is more general than the other).

Suppose $A$ is an object of $\mathbf{C}$ underlying an algebra $\mathbb{A}$ and $B$ is a retract of $A$, i.e. there is a split idempotent $A \stackrel{e}{\rightarrow} A = A \stackrel{r}{\rightarrow} B \stackrel{s}{\rightarrow} A$. Intuitively, there is a natural way one could see $B$ as an algebra by embedding it in $A$ with $s$ and going back to $B$ with $r$ whenever necessary. More formally (in our worked out cases), we can define the monoid on $B$ and the $M$-algebra on $B$ respecitively as follows (the second one is exactly like in Tom's result):
image.png
image.png

Now our results state that:

• if $r$ is a homomorphism, then $B$ with this structure defined above is an algebra.
• if $s$ is a homomorphism, then $B$ with this structure defined above is an algebra.

It is weird because we are checking the conditions of a homomorphism without knowing that the (co)domain $B$ is an algebra, and once we know they are true, we conclude that in fact $B$ is an algebra.

Nathanael Arkor (Aug 17 2023 at 12:35):

Right, the assumption I gave is exactly what is necessary for $s$ to be a "tentative homomorphism" in this sense. So it seems like you've already covered my observation! I'll happily cite your paper when it's ready :)

Nathanael Arkor (Aug 17 2023 at 12:36):

(I actually need something that is slightly more general, namely the result for monoids in a multicategory rather than a monoidal category. But it's no more difficult to prove.)

Nathanael Arkor (Aug 17 2023 at 12:43):

Ralph Sarkis said:

Now our results state that:

• if $r$ is a homomorphism, then $B$ with this structure defined above is an algebra.
• if $s$ is a homomorphism, then $B$ with this structure defined above is an algebra.

In fact (as I'm sure you observed), something slightly stronger is also true: the algebra structure on $B$ is unique such that $s$ (and presumably alternatively $r$) is a homomorphism.

Jean-Baptiste Vienney (Aug 17 2023 at 13:58):

Nathanael Arkor said:

Ralph Sarkis said:

With Jean-Baptiste Vienney , we figured out my result above is an instance of the following more abstract result he proved:
image.png

That result seems very natural, so maybe it is already well-known or a consequence of a well-known result. Do you know any reference or examples where this result applies?

I recently came across a situation in which one has a retraction of a monoid satisfying some axioms that imply that the retract is also a monoid, and I remembered this thread. However, the situation I have is slightly different. Rather than the two axioms involving $e$ and $r$, I have a single axiom involving $s$ and $e$.
image.png

As far as I can tell, neither set of assumptions implies the other. In my situation, $s$ becomes a monoid morphism (whereas $r$ is not); whereas later in the discussion Jean-Baptiste mentions that in your situation $r$ becomes a monoid morphism (whereas $s$ is not). So perhaps these two results are complementary, in giving natural sufficient conditions for each of $s$ and $r$ to be monoid morphisms.

We have reduced the axioms to a single diagram since last time:
Screenshot-2023-08-17-at-9.56.31-AM.png
And yes it's not the same than yours. That's really interesting...

Jean-Baptiste Vienney (Aug 17 2023 at 14:03):

Our result can be adapted to three settings: monoids in a monoidal category, monads and algebras over a monad. That's great if it also works for monoids in a multicategory!

Jean-Baptiste Vienney (Aug 17 2023 at 14:05):

I don't think we have covered your observation. Personally, I have never thought to what you say @Nathanael Arkor . Maybe @Ralph Sarkis has, I don't know :upside_down:

Jean-Baptiste Vienney (Aug 17 2023 at 14:18):

I don't know what kind of mysterious duality would permit to go from one result to the other one...

Jean-Baptiste Vienney (Aug 17 2023 at 14:20):

Ralph Sarkis said:

Now our results state that:

• if $r$ is a homomorphism, then $B$ with this structure defined above is an algebra.
• if $s$ is a homomorphism, then $B$ with this structure defined above is an algebra.

It is weird because we are checking the conditions of a homomorphism without knowing that the (co)domain $B$ is an algebra, and once we know they are true, we conclude that in fact $B$ is an algebra.

I completely agree with that, this is the story, but to be precise the first thing is from us, and the second one from Nathanael.

Jean-Baptiste Vienney (Aug 17 2023 at 14:26):

I think we're going to write the paper very quickly with what we have understood. Maybe in one week or two, it will be one the ArXiv and I'm very glad if you're going to cite it and complete the story!

Nathanael Arkor (Aug 17 2023 at 19:44):

I completely agree with that, this is the story, but to be precise the first thing is from us, and the second one from Nathanael.

Ah, I see :)

Nathanael Arkor (Aug 17 2023 at 19:44):

Jean-Baptiste Vienney said:

I think we're going to write the paper very quickly with what we have understood. Maybe in one week or two, it will be one the ArXiv and I'm very glad if you're going to cite it and complete the story!

I'll be very interested to read it!

Nathanael Arkor (Aug 17 2023 at 19:46):

Jean-Baptiste Vienney said:

We have reduced the axioms to a single diagram since last time:
Screenshot-2023-08-17-at-9.56.31-AM.png

Ah, this looks much more "symmetric" with respect to the axiom I mentioned, which is satisfying.

James Deikun (Aug 17 2023 at 19:51):

Doesn't it still work if you use $e$ instead of $r$ in that axiom? Because $s$ is going to be a mono.

James Deikun (Aug 17 2023 at 19:52):

That seems a bit more satisfying because the axiom doesn't have to mention the splitting.

Jean-Baptiste Vienney (Aug 17 2023 at 19:54):

James Deikun said:

Doesn't it still work if you use $e$ instead of $r$ in that axiom? Because $s$ is going to be a mono.

Yes, it works if you use $e$ instead of $r$ (the two diagrams are equivalent), but you still need the splitting in order to define the structure of monoid on $B$

Jean-Baptiste Vienney (Aug 17 2023 at 19:54):

which uses $r$ and $s$ but not $e$

Jean-Baptiste Vienney (Aug 17 2023 at 20:33):

Nathanael Arkor said:

Ralph Sarkis said:

With Jean-Baptiste Vienney , we figured out my result above is an instance of the following more abstract result he proved:
image.png

That result seems very natural, so maybe it is already well-known or a consequence of a well-known result. Do you know any reference or examples where this result applies?

I recently came across a situation in which one has a retraction of a monoid satisfying some axioms that imply that the retract is also a monoid, and I remembered this thread. However, the situation I have is slightly different. Rather than the two axioms involving $e$ and $r$, I have a single axiom involving $s$ and $e$.
image.png

As far as I can tell, neither set of assumptions implies the other. In my situation, $s$ becomes a monoid morphism (whereas $r$ is not); whereas later in the discussion Jean-Baptiste mentions that in your situation $r$ becomes a monoid morphism (whereas $s$ is not). So perhaps these two results are complementary, in giving natural sufficient conditions for each of $s$ and $r$ to be monoid morphisms.

I'm not sure I understand that: I agree that if $B$ is a monoid, with the "tentative structure of monoid", then $s$ is going to be a morphism of monoids. However, I don't see how to prove that $B$ is a monoid?

Jean-Baptiste Vienney (Aug 17 2023 at 20:36):

Maybe what you can prove is instead that if you start with a monoid $B$, and given some axiom, if you define the "tentative structure of monoid" on $A$, you obtain that $A$ is a monoid and $s$ a monoid morphism?

Nathanael Arkor (Aug 17 2023 at 20:47):

Jean-Baptiste Vienney said:

I'm not sure I understand that: I agree that if $B$ is a monoid, with the "tentative structure of monoid", then $s$ is going to be a morphism of monoids. However, I don't see how to prove that $B$ is a monoid?

I haven't checked too carefully yet, so I could have made a mistake. But the unit laws should follow for the same reason as in the case that $r$ is a homomorphism, because the unit laws in both cases are the same. I believe the associativity law should then follow by commutativity of the following diagram (if you interpret the comma as tensor, and elide the structural transformations).
image.png

James Deikun (Aug 17 2023 at 20:49):

It's (a little) clearer in the monad algebra version:

$$\begin{align*} b &= r \circ a \circ Ts \\ \\ b \circ Tb &= r \circ a \circ Ts \circ Tb = r \circ a \circ T (s \circ b) \\ &= r \circ a \circ T (a \circ Ts) = r \circ (a \circ Ta) \circ TTs \\ &= r \circ a \circ (\mu \circ TTs) = (r \circ a \circ Ts) \circ \mu \\ &= b \circ \mu \end{align*}$$

Jean-Baptiste Vienney (Aug 17 2023 at 21:01):

I agree with the fact that the associativity is verified. But I didn't suceed to prove the unitality.

Jean-Baptiste Vienney (Aug 17 2023 at 21:02):

Nathanael Arkor said:

Jean-Baptiste Vienney said:

I'm not sure I understand that: I agree that if $B$ is a monoid, with the "tentative structure of monoid", then $s$ is going to be a morphism of monoids. However, I don't see how to prove that $B$ is a monoid?

But the unit laws should follow for the same reason as in the case that $r$ is a homomorphism, because the unit laws in both cases are the same.
image.png

This is maybe the problem, I tried to prove it in the same way, but it wasn't the same.

Jean-Baptiste Vienney (Aug 17 2023 at 21:07):

Note that with our axiom, this axiom is crucial to prove the unitality, even if it seems weird as this axiom seems to have nothing to do with the units.

Nathanael Arkor (Aug 17 2023 at 21:08):

Jean-Baptiste Vienney said:

Note that with our axiom, this axiom is crucial to prove the unitality, even if it seems weird as this axiom seems to have nothing to do with the units.

Oh, interesting. This is what I had sketched out for left unitality:
image.png

Nathanael Arkor (Aug 17 2023 at 21:09):

Observing that the bottom-left path is equal to the identity.

Jean-Baptiste Vienney (Aug 17 2023 at 21:21):

Hmm, how do you get the commutativity of this triangle?
Screenshot-2023-08-17-at-5.19.14-PM.png

Jean-Baptiste Vienney (Aug 17 2023 at 21:22):

Maybe you need to assume it as an axiom. It was in the list of three axioms that I gave the other time, which is equivalent to the single axiom we gave today.

Nathanael Arkor (Aug 17 2023 at 21:23):

Jean-Baptiste Vienney said:

Maybe you need to assume it as an axiom. It was in the list of three axioms that I gave the other time, which is equivalent to the single axiom we gave today.

Oh, yes, I'm assuming that as an axiom.

Nathanael Arkor (Aug 17 2023 at 21:23):

I overlooked that you had eliminated that axiom.

Nathanael Arkor (Aug 17 2023 at 21:23):

image.png
These are the axioms I'm assuming.

Jean-Baptiste Vienney (Aug 17 2023 at 21:25):

Oh, ok, I see, this is why I was confused. With these two axioms, everything looks good!

Nathanael Arkor (Aug 17 2023 at 21:26):

The asymmetry between the section and retraction cases is curious. But they're still not too dissimilar.

Nathanael Arkor (Aug 17 2023 at 21:26):

Jean-Baptiste Vienney said:

Oh, ok, I see, this is why I was confused. With these two axioms, everything looks good!

Sorry, that was my fault for not checking your screenshot more carefully :sweat_smile:

Jean-Baptiste Vienney (Aug 17 2023 at 21:33):

Oh, but we understood better the differences between the two cases thanks to that. So that was not a too bad mistake ahah

Jean-Baptiste Vienney (Aug 17 2023 at 21:34):

Nathanael Arkor said:

The asymmetry between the section and retraction cases is curious. But they're still not too dissimilar.

I agree that it is curious. The asymmetry comes maybe from the fact that the definition of the unit uses only $r$ and not $s$ contrary to the one of the multiplication.

Jean-Baptiste Vienney (Aug 17 2023 at 21:37):

In our case, you don't need $\eta_A;e=\eta_A$ to prove the proposition, although I maybe used it the first time. And it doesn't seem to be a consequence of our axioms. Nevertheless, I didn't find any example where it is not verified for the moment.

Nathanael Arkor (Aug 17 2023 at 21:38):

Jean-Baptiste Vienney said:

In our case, you don't need $\eta_A;e=\eta_A$ to prove the proposition, although I maybe used it the first time. And it doesn't seem to be a consequence of our axioms. Nevertheless, I didn't find any example where it is not verified for the moment.

Curiously, in the $s$ homomorphism case, you also get that $r$ preserves the unit (though not the multiplication). But I'm not sure it's a useful fact.

Jean-Baptiste Vienney (Aug 17 2023 at 21:40):

Oh, this is because the definition of the unit of $B$ is exactly the equation of the preservation of the unit by $r$!

Tomáš Jakl (Aug 23 2023 at 17:12):

This discussion resembles so much what is in

• Manes, Mulry: "Monad compositions I: general constructions and recursive distributive laws."

Tomáš Jakl (Aug 23 2023 at 17:15):

in fact, one can make a little more concrete situation than that of Definition 2.2.4 therein. For a given monad $M$ we can assume that every object $M(A)$ has a an idempotent split $M(A) \to N(A) \to M(A)$. If this split satisfies conditions similar to those above we obtain that $N(-)$ carries a structure of a monad and that the section part of the splitting is natural and, in fact, forms a monad morphism.

Ralph Sarkis (Aug 23 2023 at 17:50):

I discovered (or maybe invented :laughing:) this paper some time ago but after I stopped intensely working on distributive laws, so maybe I should take the opportunity to read it this time.

Ralph Sarkis (Aug 23 2023 at 17:53):

I guess us and that paper talk about quotient and submonads, so there will be bridges, but there is no direct mention of an idempotent or a splitting there.

Tomáš Jakl said:

in fact, one can make a little more concrete situation than that of Definition 2.2.4 therein. For a given monad $M$ we can assume that every object $M(A)$ has a an idempotent split $M(A) \to N(A) \to M(A)$. If this split satisfies conditions similar to those above we obtain that $N(-)$ carries a structure of a monad and that the section part of the splitting is natural and, in fact, forms a monad morphism.

Where did you get that result from? It was the one that started this whole discussion a couple of years ago.

Jean-Baptiste Vienney (Aug 23 2023 at 20:57):

Jean-Baptiste Vienney said:

It's really funny that you say the the result doesn't depend on the choice of representatives because I'm wondering if you don't use the axiom of choice at some point and then say that your operation doesn't depend on this choices of representatives

This is only know that I have understood what I wrote here. Defining the notion of quotient of any algebraic structure in mathematics requires to use the axiom of the choice. Let $E$ be a set and $\sim$ be an equivalence relation on $E$. Supose that $(E,\eta,\nabla)$ is a monoid. Let $r:E \rightarrow E/\sim$ be the canonical projection. I'm almost sure that the phrase "the result of the multiplication doesn't depend on the choice of representatives" is true if and only if for every choice of a section $s:E/\sim \rightarrow E$ of $r$, this diagram commutes :
Screenshot-2023-08-23-at-5.04.10-PM.png
By using the axiom of choice, we can then define a structure of monoid on $E/\sim$ by chosing a section $s$ of $r$ and defining:
Screenshot-2023-08-23-at-4.53.08-PM.png
We can then show that $\tilde{\eta}$ and $\tilde{\nabla}$ don't depend on the choice of the section $s$.

This is what textbooks do, when they define quotients. Therefore, if you assume the axiom of choice, our paper will actually talk about all kind of quotients in mathematics, not only the especially well-behaved ones. But you don't need the axiom of choice to define the algebraic structures on $\mathbb{Z},\mathbb{Z}/n\mathbb{Z},\mathbb{Q}$ and even $\mathbb{R}$ from the structure of semi-ring of $\mathbb{N}$ because in each of these cases, there is a natural choice of section $s$. We will explain this in the paper.

Jean-Baptiste Vienney (Aug 23 2023 at 21:42):

Apparently, you don't need the axiom of choice to define quotients, but I don't understand how for the moment...

Patrick Nicodemus (Aug 24 2023 at 00:18):

Do you want like, a set theory explanation or a type theory explanation or what? What are your foundations? What is a quotient?

Patrick Nicodemus (Aug 24 2023 at 00:20):

In ZFC set theory, a function is, by definition, a functional relation. So if $X$ and $Y$ are sets, a function from $X$ to $Y$ is defined to be a subset $R$ of $X\times Y$ that satisfies the law $\forall x\in X,\exists ! y \in Y, R(x,y)$.

Patrick Nicodemus (Aug 24 2023 at 00:23):

If $X$ is a set and $\sim$ is an equivalence relation on $X$, then $X/\sim$ is defined to be the set of all equivalence classes of $X$. If $f : X\to Y$ is a function, then we can define a relation $R\subset (X/\sim)\times Y$ by $R(A,y)\iff \exists x\in A, f(x)=y$. This relation can always be defined. If the relation happens to be a functional relation, i.e., it is total and single valued, then we say that "$f$ induces a function $\tilde{f}$ from $X/\sim$ to $Y$" but all that is happening is that we are proving this relationship is total and single-valued.

Patrick Nicodemus (Aug 24 2023 at 00:26):

Using the existential quantifier doesn't depend on the axiom of choice. If you know that $\exists x. A(x)$, you can say "Let $x$ be an element satisfying property $A$". Do you have to "choose" such an $x$? Yes. But this is fundamentally not what the axiom of choice is concerned with.

Jean-Baptiste Vienney (Aug 24 2023 at 00:30):

I want a set theory explanation. The question is how to define the new multiplication on the set of equivalence classes of a monoid by a congruence.

Jean-Baptiste Vienney (Aug 24 2023 at 00:34):

Ok, so you define a relation $R \subset ((X/\sim) \times (X/\sim)) \times (X/\sim)$ by $R((c,d),e) \Leftrightarrow \exists x,y,z \in X, ~\overline{x}=c, \overline{y}=d, \overline{z}=e,~x.y=z \}$ I guess. And then it is a function, and it is your new multiplication.

Patrick Nicodemus (Aug 24 2023 at 00:36):

Yes, that should be correct, I hope this makes sense. It is not necessarily a function, but checking that it is a function is equivalent to what people usually mean when they talk about "well-definedness".

Jean-Baptiste Vienney (Aug 24 2023 at 00:37):

Yes, it makes sense, thanks!

Jean-Baptiste Vienney (Aug 24 2023 at 00:50):

I had never thought that you could start by defining a relation and then prove that it is a function.

Mike Shulman (Aug 24 2023 at 01:12):

More generally, the same proof shows that the quotient of any equivalence relation has the universal property of a [[coequalizer]] in the category of ZFC-sets. And once one has that, the fact that multiplication can be defined on a quotient monoid follows from abstract categorical arguments.

Jean-Baptiste Vienney (Aug 24 2023 at 01:15):

It is the coequalizer of which morphisms?

Mike Shulman (Aug 24 2023 at 01:22):

If $E$ is an equivalence relation on $X$, then $E\subseteq X\times X$, so regarding $E$ as a set in its own right there are two projection maps $E \rightrightarrows X$. The quotient is the coequalizer of those two functions.

Tomáš Jakl (Aug 24 2023 at 08:48):

Ralph Sarkis said:

I guess us and that paper talk about quotient and submonads, so there will be bridges, but there is no direct mention of an idempotent or a splitting there.
...
Where did you get that result from? It was the one that started this whole discussion a couple of years ago.

I invented a version of this for myself in my thesis (Section 4.1.2) some time ago. More recently in the game comonad programme we needed to create new comonads from old and I came up with different idempotent split conditions to those that are in my thesis that also make this work. If you're interested I can try to extract it from my notes.

Ralph Sarkis (Aug 24 2023 at 09:40):

This looks so close :grimacing:. We did not consider going to a subcategory as you did, and for some reason, your section $s$ is not even a natural transformation, but it is in our case. To avoid going back to the start of this topic, here are the three conditions (the first is stronger than the second and third which are equivalent) I had when dealing with monads only:
image.png

Tomáš Jakl (Aug 24 2023 at 11:49):

Interestingly, in my game comonad use case $s$ is a monad morphism.

Tomáš Jakl (Aug 24 2023 at 11:51):

My situation (stated for comonads, so the retract $r$ becomes the comonad morphism):
image.png

Tomáš Jakl (Aug 24 2023 at 11:52):

(it's all stated in terms of Kleisli representations where $\varepsilon$ is the counit of the comonad and $\overline{(-)}$ is the coextension operation)

Graham Manuell (Aug 25 2023 at 18:31):

Nathanael Arkor said:

I recently came across a situation in which one has a retraction of a monoid satisfying some axioms that imply that the retract is also a monoid, and I remembered this thread. However, the situation I have is slightly different. Rather than the two axioms involving $e$ and $r$, I have a single axiom involving $s$ and $e$.
image.png

In case anyone is interested, I have also seen this version come up: I use what is essentially an instance of this in my paper on quotient locales. See the (rather terse) first paragraph on page 7. I am in the category of dcpos and there are two monoid structures (given joins and meets). My version of the condition is stated with inequalities, but the reverse inequality is automatic.

Ralph Sarkis (Aug 25 2023 at 19:19):

Awesome :sunglasses: Are dcpos monadic over Set ?

Graham Manuell (Aug 25 2023 at 19:39):

They are not. I think they are monadic over the category of posets.

Jean-Baptiste Vienney (Aug 25 2023 at 20:43):

Does anyone of you @Nathanael Arkor , @Tomáš Jakl , @Graham Manuell have some intuition on why $s$ is a morphism of algebras and not $r$? In classical algebra, $r$ is the projection to the quotient and preserves the algebraic structure. Why is it $s$ which preserves the structure in these situations and not $r$ and how should I interpret it? It looks like your contexts are more on the topological/geometrical side of mathematics than on the algebraic side. It has maybe something to do with that.

Nathanael Arkor (Aug 25 2023 at 21:02):

I can share what my example is. I don't have an intuition for why it happens to be an example of this phenomenon.

Let $A$ be a category. Then there is a multicategory of endo-distributors (a.k.a. endo-profunctors) $A \not\to A$ and natural transformations. Let $t \colon A \to E$ be a functor. This functor induces a monoid $E(t, t) \colon A \not\to A$ in the aforementioned multicategory (a.k.a. promonad). A [[relative monad]] with carrier $t$ is precisely a section of this monoid whose underlying distributor is "$t$-corepresentable" (i.e. of the form $E(j, t)$ for some functor $j \colon A \to E$).

Nathanael Arkor (Aug 25 2023 at 21:03):

It was quite surprising to me that the axioms for a relative monad turned out to be exactly the axioms for a monoid section.

Jean-Baptiste Vienney (Aug 26 2023 at 01:36):

That's a beautiful example and it was very satisfying to unpack it to see that it gives something that really looks like a relative monad. It's maybe even a better formulation because you replace three axioms by an only one, and you get the morphism of promonads.

Jean-Baptiste Vienney (Aug 26 2023 at 02:14):

I found a simple algebraic example of this situation: In the category of $\mathbb{R}$-vector spaces, consider the two vector spaces $\mathbb{R}$ and $\mathbb{C}$. Consider $(s :=~ \subseteq) :\mathbb{R} \rightarrow \mathbb{C}$ the inclusion and $(r := \mathfrak{R}):\mathbb{C} \rightarrow \mathbb{R}$ the real part map. Then, $s$ and $r$ are $\mathbb{R}$-linear maps. Moreover $s;r=1_{\mathbb{R}}$ (but $r;s \neq 1_{\mathbb{C}}$). Both $\mathbb{R}$ and $\mathbb{C}$ are $\mathbb{R}$-algebras and $s$ is a morphism of $\mathbb{R}$-algebras, but not $r$ because $\mathfrak{R}(ab) \neq \mathfrak{R}(a)\mathfrak{R}(b)$.

Jean-Baptiste Vienney (Aug 26 2023 at 02:15):

In this example, the intution is that $s$ is the inclusion of a sub-object which preserves the algebraic structure. But not necessarily $r$, which is not the projection on a quotient.

Jean-Baptiste Vienney (Aug 26 2023 at 17:11):

In fact, it works with any field extension.

If you have two fields $K,L$ such that $L$ is an extension of $K$, ie. you have a morphism of rings $(s := \rho):K \rightarrow L$, then by restriction of scalars, $L$ is a $K$ vector-space, and $\rho$ is a $K$-linear map. Moreover, both $K$ and $L$ are $K$-algebras and $\rho$ is a morphism of $K$-algebras.

In addition to that, consider a decomposition $L=Im(\rho) \oplus E$ of $K$-vector spaces.
You can define $r:L \rightarrow K$ to be the projection $L \rightarrow Im(\rho)$ followed by the isomorphism $Im(\rho) \cong K$ (recall that any homomorphism of rings with domain a field is injective). They are both $K$-linear maps and $r$ is thus a $K$-linear map. Moreover $s;r=1_{K}$. But $r$ is not necessarily a morphism of $K$-algebras as shown in the previous example.

Nathanael Arkor (Apr 11 2024 at 22:52):

Nathanael Arkor said:

It was quite surprising to me that the axioms for a relative monad turned out to be exactly the axioms for a monoid section.

(By the way, this now appears as §3 of The pullback theorem for relative monads.)

Jean-Baptiste Vienney (Apr 14 2024 at 00:30):

@Ralph Sarkis, I think it would be useful that we finally write our paper now that Nathanael cited us in a preprint. :smile: