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Stream: learning: questions

Topic: On the definition of a monoid

Davi Sales Barreira (Mar 01 2023 at 16:35):

Hello, friends. For quite some time, I've been confused with the seemingly many definitions of a monoid in category theory.
The first definition I was introduced to was simply the regular set theoretic definition of $(M,\cdot,e)$. Then, to see this as a category, one would say that set $M$ was the only object, the morphisms would be elements of $M$, and the composition would be $\cdot$.
(This definition would seem to imply that a monoid would be any small category with a single object).

After reading more, we get to a second definition, where a monoid is an object $M \in \text{Set}$, together with a pair of morphisms
$\mu: M \times M \to M$ and $\eta: 1 \to M$, which satisfy a commutative diagram. This diagram encodes the idea of associativity and neutral element. From this idea, we get again a triple $(M, \mu, \eta)$, but now we have avoided talking about elements. Yet, our definition of a monoid is not a category anymore.

How do we then reconcile these two definitions?

Joe Moeller (Mar 01 2023 at 16:44):

I think this this gap is best filled by thinking of the "internal" definition of category:
a category consists of

• a set Ob of "objects"
• a set Mor of "morphisms"
• functions $s, t \colon Mor \to Ob$ assigning to each morphism a "source object" and "target object" respectively
• a function $\circ \colon Mor \times_{Ob} Mor \to Mor$ called "composition", where $Mor \times_{Ob} Mor$ is the pullback of s and t

satisfying some relations.

Joe Moeller (Mar 01 2023 at 16:45):

Taking the pullback in the definition of composition means you're only defining the composition operation on pairs of maps which happen to coincide: source of one is target of the other.

Joe Moeller (Mar 01 2023 at 16:46):

Then if you ask what happens when |Ob|=1, you'll see they must always coincide, so composition is just a map $\circ \colon Mor \times Mor \to Mor$, and this is your $\mu$.

Joe Moeller (Mar 01 2023 at 16:47):

Oh, and I forgot to mention units, but I hope the point is clear.

Davi Sales Barreira (Mar 01 2023 at 16:48):

Thanks for the response, @Joe Moeller

Davi Sales Barreira (Mar 01 2023 at 16:49):

BTW, it seems that every locally small category with a single object is a monoid.

Davi Sales Barreira (Mar 01 2023 at 16:49):

I suppose that the other implication is not true, right?

Davi Sales Barreira (Mar 01 2023 at 16:49):

I mean, there are monoids that are not locally small categories with a single object...

Davi Sales Barreira (Mar 01 2023 at 16:50):

I guess this would clarify in part the confusion in my part. Because when I was first presented the notion of a monoid, this idea of a category with a single object and morphisms was stuck as "this IS what a category that is a monoid looks like".

Joe Moeller (Mar 01 2023 at 16:53):

They're equivalent. For any monoid, you can make up a one-object category. You described it above. Maybe I don't understand your confusion.

Davi Sales Barreira (Mar 01 2023 at 16:53):

Hmm. So for example, what is the definition of a monad as a single object category?

Davi Sales Barreira (Mar 01 2023 at 16:54):

My confusion stems in part from the fact that a monad seems very distinct from this single object with morphisms.

Davi Sales Barreira (Mar 01 2023 at 16:54):

At least the definitions I saw, they said that a monad was a monoid in the category of endofunctors.

Joe Moeller (Mar 01 2023 at 16:54):

Ah, yeah, the point is being obscured by terminology. I might say "set monoid" and "monoid object" to separate the two things.

Joe Moeller (Mar 01 2023 at 16:55):

A set monoid is a monoid object in Set. A monad is a monoid object in End(C).

Davi Sales Barreira (Mar 01 2023 at 16:56):

But, can a monad be described as a locally small category with a single object?

Davi Sales Barreira (Mar 01 2023 at 16:56):

I mean, a single object $F$ which is the endofunctor, a natural transformation that is the identity, and some composition...

Joe Moeller (Mar 01 2023 at 16:56):

No, that's a thing that works for set monoids specifically. This gap can be bridged too though.

Reid Barton (Mar 01 2023 at 16:57):

This is sort of a "hot take" but I think the idea that a monoid is a category with a single object is more confusing than helpful. Rather, there is a correspondence between monoids and categories with a single object.

Reid Barton (Mar 01 2023 at 16:57):

So, I would reconcile your two definitions by throwing away the first one.

Davi Sales Barreira (Mar 01 2023 at 16:58):

So... They are not equivalent, only in for monoids in the Set category?

Davi Sales Barreira (Mar 01 2023 at 16:59):

If the definition of a monoid (set, topological, etc) is NOT equivalent to a category with a single object. Then I agree with @Reid Barton

Reid Barton (Mar 01 2023 at 16:59):

I didn't say that they aren't "equivalent". Only that they're not the same thing.

Davi Sales Barreira (Mar 01 2023 at 16:59):

So they are equivalent??

Davi Sales Barreira (Mar 01 2023 at 16:59):

I mean, not only set monoids,as @Joe Moeller pointed out.

Joe Moeller (Mar 01 2023 at 17:00):

Monoid objects in Set are equivalent to one-object categories.

Joe Moeller (Mar 01 2023 at 17:00):

Monoid objects in other monoidal categories $(V, \otimes)$ are equivalent to one-object $V$-enriched categories.

Nicolas Blanco (Mar 01 2023 at 17:01):

When you think of a set monoid as a one-object category the monoid is not the object but the set of endomorphisms on this object, with multiplication given by composition. So if you want to see a monoid internal to a category C as a one-object "category" you need form an object in C. That's were enriched categories come in the picture.

Joe Moeller (Mar 01 2023 at 17:03):

Yes, Nicolas's point is I think the most common stumbling block for understanding this stuff. The one object of a monoid is nothing at all, or perhaps completely arbitrary. Looking at it from another angle, the one object is the thing which your monoid is the endomorphism monoid of.

Davi Sales Barreira (Mar 01 2023 at 17:05):

Haven't studied enriched categories yet (I'll try to take a look)... A monad is a "monoid" in the category of endofunctors. Can one "categorify" this monad such that it becomes a category with a single object $F$ (functor), with natural transformations and morphism and a composition operator between them?

Reid Barton (Mar 01 2023 at 17:09):

Well it can't become an ordinary category, since then it would be an ordinary monoid (= monoid object of Set).

Nathanael Arkor (Mar 01 2023 at 17:10):

A monad on a category $C$ is a $[C, C]$-enriched category with one object.

Nathanael Arkor (Mar 01 2023 at 17:11):

Though this is not a perspective that one encounters very often.

Davi Sales Barreira (Mar 01 2023 at 17:11):

Ok, thanks!

Davi Sales Barreira (Mar 01 2023 at 17:11):

This is helping me understand what I was not understanding.

Davi Sales Barreira (Mar 01 2023 at 17:11):

I always tried to see how I could write a monad as an ordinary category with a single object.

Davi Sales Barreira (Mar 01 2023 at 17:12):

It's good to know that such view is limited to Sets and alike. Hence, it makes sense that one cannot do this with monads.

Davi Sales Barreira (Mar 01 2023 at 17:13):

Any suggestions on an easy intro to enriched categories?

Reid Barton (Mar 01 2023 at 17:13):

In careful writing it would be good to distinguish between "monoid" (= monoid object of Set) and "monoid object" (= potentially in a different monoidal category), in my view.

Davi Sales Barreira (Mar 01 2023 at 17:15):

So a "monoid" object, is defined "universally" like one defines "terminal" or "initial" objects or "products".

Davi Sales Barreira (Mar 01 2023 at 17:15):

Is this correct?

Reid Barton (Mar 01 2023 at 17:26):

Davi Sales Barreira said:

Any suggestions on an easy intro to enriched categories?

Do you already know what a monoidal category is?

Davi Sales Barreira (Mar 01 2023 at 17:29):

Reid Barton said:

Davi Sales Barreira said:

Any suggestions on an easy intro to enriched categories?

Do you already know what a monoidal category is?

Not really.

Davi Sales Barreira (Mar 01 2023 at 17:30):

I mean, I have seen the defintion, with the tensor functor and the pentagram. I think I understood the idea behind it. Like how one wants to say that two sets $(A,B) \times C$ and $A\times(B,C)$ are equivalent.

Davi Sales Barreira (Mar 01 2023 at 17:30):

Hence, the use of natural transformations to apply this "soft" equality between such sets.

Davi Sales Barreira (Mar 01 2023 at 17:31):

But I'm not very versed in the subject.

Nicolas Blanco (Mar 01 2023 at 17:52):

Davi Sales Barreira said:

So a "monoid" object, is defined "universally" like one defines "terminal" or "initial" objects or "products".

Not really universally. When we say that an object is define universally it implies that it is unique (up to unique isomorphism) in a given category. While there are many monoid objects in a given category.

Ralph Sarkis (Mar 01 2023 at 17:52):

Maybe take a look at Seven Sketches, it covers Preorder-enriched categories.

Davi Sales Barreira (Mar 01 2023 at 17:57):

Nicolas Blanco said:

Davi Sales Barreira said:

So a "monoid" object, is defined "universally" like one defines "terminal" or "initial" objects or "products".

Not really universally. When we say that an object is define universally it implies that it is unique (up to unique isomorphism) in a given category. While there are many monoid objects in a given category.

Thanks @Nicolas Blanco . I've re-read the definitions, and I see what you are saying.

Davi Sales Barreira (Mar 01 2023 at 17:58):

Thanks, @Ralph Sarkis . I've read Seven Sketches a while ago. I remember that I did not quite understand the whole discussion on preorder enriched-cats. Now that I have more experience, I'll take a look again.

Davi Sales Barreira (Mar 01 2023 at 18:02):

Thanks everyone for all the inputs. Things are starting to click together...
One more question. I've seen this other definition of a monoid:

A monoid is a lax monoidal function from the trivial category $\mathbb 1$ to a monoidal category.

Is this equivalent to saying that a monoid is an object in a monoidal category, or is it yet another overloading of the term "monoid"?

Nicolas Blanco (Mar 01 2023 at 18:06):

Yes it is equivalent. Since the initial category as only one object the functor picks an object into your monoidal category. The lax monoidality of the functor gives you the maps that make it an internal monoid.

Davi Sales Barreira (Mar 01 2023 at 18:09):

I mean, a functor $M: \mathbb 1 \to \mathbf{Set}$ take the object $1$ to a set $A$, and the identity to the identity...

Davi Sales Barreira (Mar 01 2023 at 18:10):

What else is it to act upon if $\mathbb 1$ has only this object and this morphism?

Nicolas Blanco (Mar 01 2023 at 18:25):

It is not just a functor but a lax monoidal functor meaning that you have natural transformations giving functions $M(1)\times M(1) \to M(1 \times 1)$ and $1 \to M(1)$. But since the initial category as only one object $1\times 1 = 1$. So the first function gives you multiplication and the second the unit.

David Egolf (Mar 01 2023 at 18:41):

Isn't the terminal category the one with a single object? On the nlab, "initial category" appears to redirect to the "empty category" article.

Joe Moeller (Mar 01 2023 at 18:43):

yes, the initial category has no objects. the terminal category has one object, and only an identity arrow.

Nicolas Blanco (Mar 01 2023 at 18:48):

Wait, yes sorry. We should replace every initial by terminal in the discussion :speak_no_evil:

Davi Sales Barreira (Mar 01 2023 at 18:56):

David Egolf said:

Isn't the terminal category the one with a single object? On the nlab, "initial category" appears to redirect to the "empty category" article.

My bad. It was "trivial" not initial.

John Baez (Mar 01 2023 at 21:35):

Davi Sales Barreira said:

So... They are not equivalent, only in for monoids in the Set category?

Part of your confusion was due to this: about 98% of mathematicians mean "monoid object in the monoidal category (Set,$\times$)" when they say "monoid". For example that would be the definition on Wikipedia. About 2% of mathematicians, or fewer, mean "monoid object in any monoidal category" when they say "monoid". So if you're reading different things, or talking to different people, you need to make sure which definition they're using!

John Baez (Mar 01 2023 at 21:38):

I'm sure you have it cleared up by now, but anyone who says "a monoid is a one-object category" is using the first definition of "monoid", the one that 95% of mathematicians use. Anyone who says "a monad is a monoid in..." is using the second definition.

Matteo Capucci (he/him) (Mar 02 2023 at 06:56):

Davi Sales Barreira said:

But, can a monad be described as a locally small category with a single object?

Every monoid M in Set gives you a one-object Set-category, as you say. This is sometimes called the 'delooping' of M.
This construction generalizes to the following: every monoid object M in a monoidal category V 'deloops' to a V-[[enriched category]] with only one dummy object and a V-object of morphisms, namely M.
Thus a monad deloops to a locally small category enriched in End(C).

Matteo Capucci (he/him) (Mar 02 2023 at 06:56):

Ah, I see this has already been pointed out