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Stream: learning: questions

Topic: Coproducts of monoids

James Deikun (Oct 03 2024 at 11:26):

I know the concrete construction of coproducts of monoids in $\mathbf{Set}$ but I am unaware of how to do it category-theoretically rather than syntactically even in the special case of $\mathbf{Set}$ never mind anything that works in a more general context. Does anyone know or have a reference to such a construction?

James Deikun (Oct 03 2024 at 11:39):

(I guess I do know a way: I could use the general construction of coproducts for monad algebras from Adamek and Koubek together with the construction of coequalizers from Porst but I was hoping for something maybe a bit more direct.)

Morgan Rogers (he/him) (Oct 03 2024 at 12:23):

@William Troiani looked at this in his Masters thesis ; I'm tagging him since he's produced cleaner exposition of it since, I think.

Morgan Rogers (he/him) (Oct 03 2024 at 12:25):

Oh wait I apparently completely misread your question. William does coproducts of sets, syntactically. Almost entirely unrelated. Please disregard.

Vincent Moreau (Oct 03 2024 at 12:30):

An article that might be related is A unified treatment of transfinite constructions for free algebras, free monoids, colimits, associated sheaves, and so on by Kelly

Morgan Rogers (he/him) (Oct 03 2024 at 12:37):

Let's see... To construct the coproduct of two monoids, you take the coproduct of their underlying sets, take the free structure on that (so we're at $\coprod_{n \geq 0} (A + B)^n$ so far) and quotient by an equivalence relation generated by multiplication of adjacent terms when applicable, so $(A+B)^n \ni uaa'v \sim u(\mu_A(a,a'))v \in (A+B)^{n-1}$, where $u \in (A+B)^k$ and $a,a' \in A$ and $v \in (A+B)^{n-k-2}$ (and similarly for elements of $B$), plus the relation $* \sim 1_A \sim 1_B$, where $*$ is the unique element of $(A+B)^0$. I can see why you would want to streamline that.

Morgan Rogers (he/him) (Oct 03 2024 at 12:48):

Oh right, expanding the syntactic construction, we can do the following:
Present a monoid $M$ as an algebra $TM \to M$ for the free monoid monad $T$ (for $TM$ the free monoid on the underlying set of $M$); taking the kernel pair of this map gives the congruence $R_M \rightrightarrows TM$ defining $M$. Doing this for monoids $A,B$, we have $R_A \rightrightarrows TA \rightarrow T(A+B) \leftarrow TB \leftleftarrows R_B$ defining two relations on $T(A+B)$, of which we can take the union. Taking the standard transitive and reflexive closure of this relation, we obtain the congruence on $T(A+B)$ that I described inefficiently above. It's still cumbersome, but at least it's clearly categorical.

Morgan Rogers (he/him) (Oct 03 2024 at 12:50):

(It feels like I'm missing part of the relation this way, so it needs checking, but that should be the flavour of the construction)

James Deikun (Oct 03 2024 at 13:25):

Seems like that gives the underlying set of the monoid but where does the algebra structure come from?

James Deikun (Oct 03 2024 at 13:28):

Hm, actually maybe it doesn't even give the underlying set, there's no quotienting done on strings that mix $A$ s and $B$ s when you implement this in Set.

Morgan Rogers (he/him) (Oct 03 2024 at 13:41):

Right, you need it to be a congruence, not just an equivalence relation. I suppose that's somehow the crux of the construction

Jean-Baptiste Vienney (Oct 03 2024 at 13:58):

I think you will need a distributivity of products over coproducts for what Morgan said first. You should have to use:
$(A+B)^n \simeq \underset{0 \le k \le n}{\sum}A^kB^{n-k}$
But in fact maybe you can work without distributivity if you directly use the right-hand side expression i.e. the
$M_n=\underset{0 \le k \le n}{\sum}A^kB^{n-k}$

After this, you could try defining a $P_n$ as a coequalizer of a whole bunch of maps from $M_n$ to $M_n$ which do every possible combinations of multiplying and then inserting units both in the $A^k$ and in the $B^{n-k}$ and then take the colimit $P$ of a diagram of the form
$\dots P_n \rightarrow P_{n+1} \rightarrow \dots$
where the arrow from $P_n$ to $P_{n+1}$ is an inclusion map inherited from the inclusion $M_n \rightarrow M_{n+1}$.

Of course making this precise and checking that everything works fine should be a bit of work.

Jean-Baptiste Vienney (Oct 03 2024 at 13:58):

(I'm not even sure that it can work exactly as I wrote.)

Jean-Baptiste Vienney (Oct 03 2024 at 13:59):

EDIT: No, you will need the distributivity in any case when you define the structure of monoid on P I think.

Jean-Baptiste Vienney (Oct 03 2024 at 14:03):

Second of course: defining the multiplication and unit of $P$ will be a bit of work also... and checking the associativity and unitality as well.

Jean-Baptiste Vienney (Oct 03 2024 at 14:11):

Also, for the multiplication, you should have to define maps $M_n \times M_p \rightarrow M_{n+p}$ first, then $P_n \times P_p \rightarrow P_{n+p}$, then $P_n \times P_p \rightarrow P$ (by postcomposing by the inclusion $P_n \rightarrow P$ given by the colimit) and finally $P \times P \rightarrow P$ I think.

Jean-Baptiste Vienney (Oct 03 2024 at 14:13):

I'm not completely sure about this as well, but it must something like this.

Jean-Baptiste Vienney (Oct 03 2024 at 14:16):

Also, sorry I supposed that you wanted to define the coproduct of monoids in an arbitrary category with products but I don't know if it is what you wanted to do.

Jean-Baptiste Vienney (Oct 03 2024 at 14:21):

I've just realized that what I've sketched must work in any monoidal category (no need for the monoidal product to be a cartesian product), producing also the coproduct of non-commutative rings (and non necessarily commutative $k$-algebras as well I guess).

Nathan Corbyn (Oct 03 2024 at 14:22):

A fun related fact: if we’re interested in the coproduct of $A$ and $B$ (monoid objects in a distributive monoidal category) and $B \simeq FV$ is free, their coproduct as monoids has carrier object: $A \otimes \coprod_{n \in \mathbb{N}} (V \otimes A)^{\otimes n}$.

Jean-Baptiste Vienney (Oct 03 2024 at 14:50):

I don't really understand this. I think we will have different copies of the elements of $A$ in
$A \otimes \coprod_{n \in \mathbb{N}} (V \otimes A)^{\otimes n}$ such as $1_A \otimes a$, $a \otimes 1_A$ and $1_A \otimes a \otimes 1_A$. That's why I wanted to take a directed colimit at some point and not just a coproduct: in order to make equal these copies.

Jean-Baptiste Vienney (Oct 03 2024 at 14:55):

But I get it that if $A$ or $B$ is free, then things will be easier. As it is easier if we're looking at commutative monoids.

Nathan Corbyn (Oct 03 2024 at 15:08):

You only get one copy of $A$: the inclusion is $A \xrightarrow{\rho^{-1}_A} A \otimes I \xrightarrow{1_A \otimes \iota_0} A \otimes \coprod_{n\in\mathbb{N}} (V \otimes A)^{\otimes n}$

Nathan Corbyn (Oct 03 2024 at 15:10):

The idea is you’re building ‘formal sums’ by alternating between ‘elements’ of $A$ and ‘variables’ drawn from $V$

Nathan Corbyn (Oct 03 2024 at 15:12):

Jean-Baptiste Vienney said:

But I get it that if $A$ or $B$ is free, then things will be easier. As it is easier if we're looking at commutative monoids.

I think this is unrelated: the coproduct of two commutative monoids as monoids is not the same as their coproduct as commutative monoids

Jean-Baptiste Vienney (Oct 03 2024 at 15:12):

Nathan Corbyn said:

Jean-Baptiste Vienney said:

But I get it that if $A$ or $B$ is free, then things will be easier. As it is easier if we're looking at commutative monoids.

I think this is unrelated: the coproduct of two commutative monoids as monoids is not the same as their coproduct as commutative monoids

Ooh ok. I should have said "the coproduct of commutative monoids in the category of commutative monoids.

Nathan Corbyn (Oct 03 2024 at 15:12):

(Also free monoids aren’t commutative in general)

Jean-Baptiste Vienney (Oct 03 2024 at 15:13):

(yeah, I know this)

Nathan Corbyn (Oct 03 2024 at 15:14):

Then I don’t think I follow :sweat_smile:

Jean-Baptiste Vienney (Oct 03 2024 at 15:17):

Well, I was just saying that the coproduct in a category of commutative of monoids is easier to build that the coproduct in a category of monoids.

Nathan Corbyn (Oct 03 2024 at 15:20):

I was just confused because I haven’t mentioned any commutative monoids—in fact, the monoidal structure I assumed isn’t required to be symmetric or even braided so we don’t actually have an ambient notion of commutative monoid

Jean-Baptiste Vienney (Oct 03 2024 at 15:20):

Nathan Corbyn said:

You only get one copy of $A$: the inclusion is $A \xrightarrow{\rho^{-1}_A} A \otimes I \xrightarrow{1_A \otimes \iota_0} A \otimes \coprod_{n\in\mathbb{N}} (V \otimes A)^{\otimes n}$

Ok. I think I understand this better now. My $a \otimes 1_A$ and $1_A \otimes a \otimes 1_A$ just didn't make sense because of the $V$.

Jean-Baptiste Vienney (Oct 03 2024 at 15:21):

Nathan Corbyn said:

I was just confused because I haven’t mentioned any commutative monoids—in fact, the monoidal structure I assumed isn’t required to be symmetric or even braided so we don’t actually have an ambient notion of commutative monoid

Ok, if the category is symmetric monoidal then the coproduct in the category of commutative monoids is just the tensor coproduct in the base category. This is all I wanted to say.

Jean-Baptiste Vienney (Oct 03 2024 at 15:37):

Jean-Baptiste Vienney said:

I think you will need a distributivity of products over coproducts for what Morgan said first. You should have to use:
$(A+B)^n \simeq \underset{0 \le k \le n}{\sum}A^kB^{n-k}$
But in fact maybe you can work without distributivity if you directly use the right-hand side expression i.e. the
$M_n=\underset{0 \le k \le n}{\sum}A^kB^{n-k}$

After this, you could try defining a $P_n$ as a coequalizer of a whole bunch of maps from $M_n$ to $M_n$ which do every possible combinations of multiplying and then inserting units both in the $A^k$ and in the $B^{n-k}$ and then take the colimit $P$ of a diagram of the form
$\dots P_n \rightarrow P_{n+1} \rightarrow \dots$
where the arrow from $P_n$ to $P_{n+1}$ is an inclusion map inherited from the inclusion $M_n \rightarrow M_{n+1}$.

Of course making this precise and checking that everything works fine should be a bit of work.

Sorry, this is absolutely not correct! :sweat_smile: The right hand side in the first equality is not the good one.

Jean-Baptiste Vienney (Oct 03 2024 at 15:41):

We should have to work with expressions like this:
$\sum A^{k_1}B^{l_1}\dots A^{k_n}B^{l_n}$

Kevin Carlson (Oct 03 2024 at 18:32):

Is the construction here just the Adamek&Koubek/Porst argument you don't want?

James Deikun (Oct 03 2024 at 19:20):

Indeed.