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Stream: deprecated: id my structure

Topic: "Monoidification" of a category

Moana Jubert (Aug 16 2023 at 19:12):

Let $\mathcal{C}$ be a small category.

Consider two subsets $X, Y \subseteq \coprod \mathcal{C}(c, d)$ of arrows of $\mathcal{C}$. I can define a "pointwise composition" of $X$ and $Y$:

$X \circ Y \coloneqq \{f \circ g \mid f \in X, g \in Y, \operatorname{cod} g = \operatorname{dom} f\}$

Now inductively define a set $M\mathcal{C}$ with the following:

1. For any arrow $f : c \to d$ of $\mathcal{C}$, the set $\{f\} \cup \{\mathrm{id}_x \mid x \in \mathcal{C}\}$ is an element of $M\mathcal{C}$
2. Given $X, Y \in M\mathcal{C}$, then $X \circ Y \in M\mathcal{C}$

It turns out $(M\mathcal{C}, \circ)$ is a monoid with unit element $\{\mathrm{id}_x \mid x \in \mathcal{C}\}$, and the construction surely is simple enough for it to have been formulated by someone else.

Is it the case? Does this have a name?

Additional points:

• $M\mathcal{C}$ weirdly behaves, as for example if $\mathcal{C}$ is already a monoid $\mathbb{M}$ seen as a one-object category then $M\mathcal{C} \ncong \mathbb{M}$ in general
• I am actually looking for a "reasonable" way to turn a category into a monoid which is "well-behaved enough", in the sense that everything that I need to know about a category is reflected by the structure of the associated monoid...

Ivan Di Liberti (Aug 16 2023 at 22:30):

Hello Moana! How are you doing?

I hope to say more tomorrow on this, but I cannot guarantee it. For the moment, check out "How Comprehensive is the Category of Semigroups?". It does not give you precisely what you want, but I like it. :)

Jean-Baptiste Vienney (Aug 17 2023 at 01:07):

More simply, I would define, $M\mathcal{C}$ to be the set of non-empty subsets of $Mor(\mathcal{C})$ with composition as you defined it. I think, it's not exactly the same thing that what you obtain with your definition by induction?

Now, if $\mathcal{C}$ is already a monoid ie. a category with a single object, $M\mathcal{C}$ is still equal to the set of all the non-empty subsets of $Mor(\mathcal{C})$. It is not the same thing than $\mathcal{C}$ but you have an injective function $i:Mor(\mathcal{C}) \rightarrow M\mathcal{C}$ which send an element $m$ of the monoid to $\{m\}$ and you can identify your monoid $\mathcal{C}$ with the image $i(\mathcal{C})$.

Jean-Baptiste Vienney (Aug 17 2023 at 01:16):

I got a better idea: define $M_{0}\mathcal{C}$ to be the subset of $Ob(\mathcal{C})\times Ob(\mathcal{C}) \times Mor(\mathcal{C})$ such that $(a,b,f) \in M_{0}\mathcal{C}$ if and only if $dom(f)=a$ and $cod(f)=b$. Now define $M\mathcal{C} := M_{0}\mathcal{C} \sqcup \{0,1\}$.

Jean-Baptiste Vienney (Aug 17 2023 at 01:20):

Define $(a,b,f) * (b,c,g) := (a,c, f;g)$, $(a,b,f) * (c,d,h) = 0$ if $b \neq c$, $(a,b,f) * 0 = 0 * (c,d,g) := 0$, $(a,b,f)*1=1*(a,b,f)=(a,b,f)$, $1*1:=1$ and $1*0=0*1=0$.

Jean-Baptiste Vienney (Aug 17 2023 at 01:26):

It looks like it's really a monoid. And if you start with a monoid, ie. a category $\mathcal{C}$ with a single object $\star$ then $\mathcal{M}_{0}(\mathcal{C}) = \{(\star,\star,f), f \in Mor(\mathcal{C})\}$, you obtain $\mathcal{M}(\mathcal{C}) = \{(\star,\star,f), f \in Mor(\mathcal{C})\} \sqcup \{0,1\}$ and it's almost isomorphic to $\mathcal{C}$ except that you have two new useless elements $0,1$.

Jean-Baptiste Vienney (Aug 17 2023 at 01:36):

The general idea is that your monoid, by multiplication, composes the morphisms like in $\mathcal{C}$ and there is also an element "0 ie. error" for when you multiply morphisms which are not composable, and a unit $1$ which is here because you want a unit.

Jean-Baptiste Vienney (Aug 17 2023 at 02:05):

Remark that in the case where $\mathcal{C}$ is a monoid, then $M_{0}\mathcal{C}$ is a sub-semi-group of $M\mathcal{C}$ which is isomorphic to $\mathcal{C}$. A solution to obtain that $M$ preserves monoids would therefore be to do the $- \sqcup \{0,1\}$ if and only if $M_{0}\mathcal{C}$ don't become a sub-semi-group of $M\mathcal{C}$ which is a monoid if you do the $- \sqcup \{0,1\}$. But it is maybe simpler not to care about that.

Moana Jubert (Aug 17 2023 at 09:45):

Ivan Di Liberti said:

Hello Moana! How are you doing?

I hope to say more tomorrow on this, but I cannot guarantee it. For the moment, check out "How Comprehensive is the Category of Semigroups?". It does not give you precisely what you want, but I like it. :)

Hello Ivan :) How are you?

It sure looks very interesting! I will definitely put it at the top of my to-read list :)

Moana Jubert (Aug 17 2023 at 10:20):

@Jean-Baptiste Vienney Thank you :) One solution indeed would be to add an absorbing element $0$ that would "skip" composition when it's impossible. I see two consequences to this:

• Morphisms won't "aggregate" (which is something perfectly reasonable actually) i.e. if $f$ and $g$ are not composable, then the set $\{f, g\}$ has no reason to play any role as $\{f\} \circ \{g\} = 0$
• I don't think there would be any reasonable way to build a unit element out of the identity morphisms (e.g. something like $\{\mathrm{id}_x \mid x \in \mathcal{C}\}$) because of the above point, and inevitably you need to formally add a unit element $1$. This is less desirable, I think, as I would expect a monoid that "looks like" the composition of morphisms to have a unit element that "looks like" the identity morphisms...

In any case I'll think about this!

Carlos Zapata-Carratala (Aug 20 2023 at 17:40):

Moana Jubert said:

Let $\mathcal{C}$ be a small category.

Consider two subsets $X, Y \subseteq \coprod \mathcal{C}(c, d)$ of arrows of $\mathcal{C}$. I can define a "pointwise composition" of $X$ and $Y$:

$X \circ Y \coloneqq \{f \circ g \mid f \in X, g \in Y, \operatorname{cod} g = \operatorname{dom} f\}$

Now inductively define a set $M\mathcal{C}$ with the following:

1. For any arrow $f : c \to d$ of $\mathcal{C}$, the set $\{f\} \cup \{\mathrm{id}_x \mid x \in \mathcal{C}\}$ is an element of $M\mathcal{C}$
2. Given $X, Y \in M\mathcal{C}$, then $X \circ Y \in M\mathcal{C}$

It turns out $(M\mathcal{C}, \circ)$ is a monoid with unit element $\{\mathrm{id}_x \mid x \in \mathcal{C}\}$, and the construction surely is simple enough for it to have been formulated by someone else.

Is it the case? Does this have a name?

Additional points:

• $M\mathcal{C}$ weirdly behaves, as for example if $\mathcal{C}$ is already a monoid $\mathbb{M}$ seen as a one-object category then $M\mathcal{C} \ncong \mathbb{M}$ in general
• I am actually looking for a "reasonable" way to turn a category into a monoid which is "well-behaved enough", in the sense that everything that I need to know about a category is reflected by the structure of the associated monoid...

Hi Moana,
I cannot add much more to your original question about the construction you propose but I find the topic intriguing. I was wondering, what motivates you to want to capture "everything you need to know about a category" into a monoid? Why that particular (total) algebraic structure?

Moana Jubert (Aug 21 2023 at 09:23):

Hello Carlos! Here's what I have been thinking about:

Let $M$ be a monoid seen as a one-object category $\mathcal{M}$. A functor $X : \mathcal{M} \to \mathbf{Set}$ amounts to a set $X$ equipped with an action $M \times X \to X$ when we unfold the necessary data. Now, a natural transformation $X \Rightarrow Y$ between $X, Y : \mathcal{M} \to \mathbf{Set}$ merely is a function $X \to Y$ on the underlying sets which preserves the action of $M$.

$M$ canonically acts on itself "on the left" via the only (co)representable functor $M \coloneqq \mathcal{M}(\ast, -) : \mathcal{M} \to \mathbf{Set}$.

I then have two questions:

1. When is a natural transformation $X \Rightarrow M$ between $X : \mathcal{M} \to \mathbf{Set}$ and $M \coloneqq \mathcal{M}(\ast, -) : \mathcal{M} \to \mathbf{Set}$ equivalent to the data of a discrete fibration over $\mathcal{M}$?
2. Given any small category $\mathcal{C}$ and a "reasonable monoidification" $M\mathcal{C}$, when is a natural transformation $X \Rightarrow M\mathcal{C}$ equivalent to the data of a discrete fibration over $\mathcal{C}$?

Carlos Zapata-Carratala (Aug 21 2023 at 23:31):

I see! So you are looking for constructions of such a $MC$ right? Why should we expect it to be a total algebraic structure?

Moana Jubert (Aug 23 2023 at 06:19):

Yes! Well because I'm curious if it would work!

But for example, a module is a particular case of monoid, yet it is more "natural" to define a module in terms of an action over an abelian group rather than as a single monoid with some prescribed properties...

In any case monoids are such a basic structure that maybe, maybe there's some way to encode all of, or many of the properties of a category

Carlos Zapata-Carratala (Aug 23 2023 at 12:17):

I can see the pattern for (total) associative-like algebras to be encoded into monoids, like the case of modules you point out, but I am having a hard time thinking how you are going to encode the partial/typed nature of categories. In a sense, that's precisely what categories do beyond just being monoids. If one insists that monoid be the name of an algebra with an associative binary and identity, then categories should be called "monoidoids" in the same naming convention that gives "groupoids" from groups ;) Maybe I am missing something... I do find this question interesting since I am exploring exotic higher-arity algebras and I have just focused on the "-oid" version from the start, since they are easier to define in a typed framework (totality is a specific property of certain diagrams or type constraints).