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Reading this beautiful paper on Model Theory (somebody posted the link yesterday, could not find): https://www.irif.fr/~mellies/mpri/mpri-ens/articles/hyland-power-lawvere-theories-and-monads.pdf learned from this paper that if you model group theory in a category of groups, you get abelian groups ("and this explains why the higher homotopy groups are abelian") - I wonder if the same holds for monoids.
Yes! It's called the Eckmann–Hilton argument :-) https://ncatlab.org/nlab/show/Eckmann-Hilton+argument
Yes, a monoid in the category of monoids is exactly a commutative monoid. This is called the Eckmann-Hilton argument.
And as my link shows, you don't even need associativity! A unital magma (a thing with a binary operation having an element that's both a left and right unit) in the category of unital magmas is a commutative unital magma.
Puzzle. Is a semigroup in the category of semigroups necessarily a commutative semigroup? If so give a proof. If not, find a simple counterexample.
John Baez said:
Puzzle. Is a semigroup in the category of semigroups necessarily a commutative semigroup? If so give a proof. If not, find a simple counterexample.
OK, I think I have a counterexample: consider the free semigroup of non-empty sequences from an alphabet with concat. as multiplication (). Now consider a binary operation that simply discards the second argument. It is clear that is a semigroup homomorphism (from the "paired" semigroup to the single one) as . However, .
Hmmm, what's the intuition here...
Very very nice! Here's a puzzle for anyone except Stelios, since it'll be too easy for him:
Puzzle. Let be the free monoid on the set . This is the set of finite sequences of elements of , with concatenation as the monoid operation (let's call that operation ). It has a second semigroup structure that simply discards the second argument: that is, .
Using these two operations, can we see a monoid in the category of semigroups, or perhaps a semigroup in the category of monoids?
Here we are trying to skirt as close as possible to the Eckmann-Hilton argument without letting it apply: neither operation is commutative.
John Baez said:
Puzzle. Is a semigroup in the category of semigroups necessarily a commutative semigroup? If so give a proof.
Well, it is only slightly commutative. On this page: http://mat.uab.cat/~kock/cat/commutativity.html there is a more precise statement and a proof given by an animated gif.
Wow, that's great, @Joachim Kock! I'd never seen that.
Clearly you're even a bigger Eckmann-Hilton addict than I am.
I assume you can easily handle this (if you can, don't give away the answer):
Puzzle. What's a monoidal category in the 2-category of monoidal categories?
More precisely:
Puzzle. What's a pseudomonoid in the monoidal 2-category where objects are monoidal categories, morphisms are (strong) monoidal functors and 2-morphisms are monoidal natural transformations?
@Joachim Kock: that's a really pretty presentation of a proof
A few years ago, De Wolf found a tiling with only 10 tiles, with room enough for the same move. And then Bremner and Madariaga found one with only 9 tiles and proved that it cannot be smaller.
i have a tiny bit of experience w/ model theory & category theory intersecting
im familiar with some very nice properties of horn clause theories
to be precise, universal horn theories always have initial models if they have any models
actually i think they have finite products, but initial models was relevant for my purposes
so the thing is that adding new operations or relation symbols and constraining by universal horn sentences keeps you in the realm of universal horn theories, and that corresponds to, like, certain subcategories of slice categories and stuff, which is what i was doing
man, i should learn something about prolog one of these days...
Yes, and this middle-interchange-law looks like a triviality when drawn as a quantum circuit, see eg.
"Quantum Computing for Computer Scientists", Noson Yanofsky et al. , and is somehow 'isomorphic' to the analogue
law in a 2-category ??