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Stream: learning: questions

Topic: products in a monoid

Matteo Capucci (he/him) (Feb 26 2021 at 21:39):

Suppose $M$ is a monoid, seen as a one-object category. Let's call $\bullet$ such an object.
What's $\bullet \times \bullet$? The obvious choice ($\bullet$ with identities as projections) seems to only work for the trivial monoid.
Then in general suppose $(x,y)$ is a product for $M$. Considering the cone $(1,1)$ one sees $x$ and $y$ need to be left invertible. If $M$ is finite, this means $x$ and $y$ are invertible and equal, hence given a cone $(a,b)$ and a universal arrow $z$, the condition $a=zx$ and $b=zx$ implies $a=b$, therefore the monoid is trivial.
It remains the case $M$ is infinite, but I can't do much about that. Any thoughts?

David Michael Roberts (Feb 26 2021 at 21:47):

Are you thinking of $M$ as a one-object monoidal category? Any abelian monoid will work, surely?

Matteo Capucci (he/him) (Feb 26 2021 at 21:53):

I don't understand your question, sorry. A one-object monoidal category would be very boring, no?

Matteo Capucci (he/him) (Feb 26 2021 at 21:54):

Commutative is probably enough to enable the proof of triviality I gave for finite monoids. The it remains only non-commutative infinite monoids.

Cole Comfort (Feb 26 2021 at 21:56):

Matteo Capucci (he/him) said:

I don't understand your question, sorry. A one-object monoidal category would be very boring, no?

In a one object monoidal category, musn't every object be the tensor unit, so that all maps commute with each other.

James Wood (Feb 26 2021 at 22:28):

Cole Comfort said:

Matteo Capucci (he/him) said:

I don't understand your question, sorry. A one-object monoidal category would be very boring, no?

In a one object monoidal category, musn't every object be the tensor unit, so that all maps commute with each other.

I think you're thinking of free monoidal categories on one object. If you take the category of $K$-vector spaces, and then take the monoidal subcategory containing only the tensor unit ($K$), then you still have $\lvert K \rvert$-many morphisms.

Cole Comfort (Feb 26 2021 at 22:31):

James Wood said:

Cole Comfort said:

Matteo Capucci (he/him) said:

I don't understand your question, sorry. A one-object monoidal category would be very boring, no?

In a one object monoidal category, musn't every object be the tensor unit, so that all maps commute with each other.

I think you're thinking of free monoidal categories on one object. If you take the category of $K$-vector spaces, and then take the monoidal subcategory containing only the tensor unit ($K$), then you still have $\lvert K \rvert$-many morphisms.

Which commute with each other because a field has the structure of an commutative monoid. Isn't a one object monoidal category just a commutative monoid?

James Wood (Feb 26 2021 at 22:32):

Ah, sorry, yeah, I misunderstood your remark.

James Wood (Feb 26 2021 at 22:34):

It feels like that could be using symmetry, but I don't really know non-symmetric monoidal categories all that well.

Cole Comfort (Feb 26 2021 at 22:37):

It seems to me like one object monoidal categories and one object symmetric monoidal categories ought to be the same, because you can just braid things with the interchange laws

Cole Comfort (Feb 26 2021 at 22:37):

Up to coherence data these must just be commutative monoids.

Mike Shulman (Feb 27 2021 at 02:45):

While a one-object monoidal category is just a commutative monoid, and a one-object cartesian monoidal category is trivial, a one-object semigroupal category can be more interesting, and even a "cartesian semigroupal" category. For instance, the full subcategory of Set determined by any single infinite set has binary products, since any infinite set is isomorphic to its own cartesian square.

John Baez (Feb 27 2021 at 02:59):

Cole Comfort said:

It seems to me like one object monoidal categories and one object symmetric monoidal categories ought to be the same...

Just for extra confirmation: yes, they are both just commutative monoids.

Matteo Capucci (he/him) (Feb 27 2021 at 07:58):

Mike Shulman said:

[...] a one-object cartesian monoidal category is trivial [...]

Thanks @Mike Shulman. Do you have any reference for this? Or some easy way to complete my proof attempt above?
Also, I'm toying with this questions (limits in a monoid) as a preparation for trying to understand bilimits in the delooping of a monoidal category. I guess products could be more interesting then.

John Baez (Feb 27 2021 at 08:07):

The unit object in a cartesian monoidal category is terminal, more or less by definition. So, a one-object cartesian monoidal category also has just one morphism.

Matteo Capucci (he/him) (Feb 27 2021 at 08:07):

Oooh I see. Right, way easier than I thought XD

Matteo Capucci (he/him) (Feb 27 2021 at 08:09):

I guess one can do a similar trick for pullbacks? Since pullbacks are products in the slices, and the slice over a one-object category is still a one-object category, we deduce every slice is trivial, so is the monoid we start with?

Mike Shulman (Feb 27 2021 at 15:31):

No, a slice of a one-object category will generally have more than one object. Plus, a slice category automatically always has a terminal object.

Fawzi Hreiki (Feb 27 2021 at 19:05):

The under category $\star / M$ of a (cancellative) monoid $M$ is its divisibility category (poset)

Fawzi Hreiki (Feb 27 2021 at 19:12):

Which, like Mike said, is usually not trivial. For example, in the 'divisibility' (subtractibility) poset of $(\mathbb{N}, +)$, we have $n \rightarrow m$ iff $n \leq m$ whereas in the divisibility poset of $(\mathbb{N}, \times)$, $n \rightarrow m$ iff $n$ divides $m$.

John Baez (Feb 27 2021 at 20:15):

Fawzi Hreiki said:

The under category $\star / M$ of a (cancellative) monoid $M$ is its divisibility category (poset)

Can you explain why that's true? I'm having some sort of mental block on this, and I don't have the time to sit and calculate. It should be the sort of thing one just "sees".

Martti Karvonen (Feb 27 2021 at 20:27):

I don't know a nicer answer than just doing the calculation, but the objects of $\star/M$ are maps out of $\star$ i.e. elements of the monoid, and maps $m\to n$ are given by elements $k\in M$ with $km=n$. Cancellativity implies that there's at most one such $k$ for given $m$ and $n$, so this is exactly the divisibility poset.

John Baez (Feb 27 2021 at 20:46):

I don't know a nicer answer than just doing the calculation

Luckily you just did the calculation at exactly the level of detail I needed to fill in the rest without writing anything. The thing I was missing was cancellativity - that's why we're getting a poset.

John Baez (Feb 27 2021 at 20:46):

That was my mental block.

Fawzi Hreiki (Feb 27 2021 at 20:46):

More generally, you can think of any extension as a (not necessarily unique) 'division' of one map by another.

John Baez (Feb 27 2021 at 20:47):

This gives a sort of fun definition of "cancellative monoid" for category theorists who want other mathematicians to hate them: a cancellative monoid M is one such that the under category $\star/M$ is a poset.

Fawzi Hreiki (Feb 27 2021 at 20:48):

Ah no that's false I think

Fawzi Hreiki (Feb 27 2021 at 20:48):

There may be cancellative monoids where the over category is not a poset no?

Fawzi Hreiki (Feb 27 2021 at 20:49):

Actually, no you're right

Todd Trimble (Feb 27 2021 at 21:28):

Looks like this depends on whether left or right cancellative is meant. Looks like mx = my implies x = y is the version meant here.

Matteo Capucci (he/him) (Feb 28 2021 at 08:56):

Mike Shulman said:

No, a slice of a one-object category will generally have more than one object. Plus, a slice category automatically always has a terminal object.

Looks like I can't stop embarassing myself...