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In trying to begin to understand the idea of a $k$-tuply monoidal $n$-category, I'm already a bit stuck on the idea (Baez, nLab) that a commutative monoid can be defined as a monoid object in the category Mon of monoids. So what I have in a monoid object in Mon is a normal old monoid $M$ equipped with a multiplication morphism $\mu: M\times M \rightarrow M$ such that $\mu\circ(id_M\times \mu):M\times M\times M\rightarrow M= \mu\circ(\mu\times id_M): M\times M\times M \rightarrow M$ and $\mu: 1\times M\rightarrow M=\mu: M\times 1\rightarrow M=\pi_1(M\times 1)=\pi_2(1\times M)$.

In the last part the trivial monoid $1$ is identified with its image in $M$, while the formula before just says $\mu$ defines an associative binary operation on $M$. I've also identified $M\otimes M$, the monoidal category tensor, with $M\times M$. That works here, right?

All of this is separate from the multiplication on $M$ that I get just from being in Mon. Now I need somehow to connect this monoid-object-in-Mon multiplication with the original multiplication $\cdot$ on $M$ to get $a\cdot b=b\cdot a$ for each $a,b \in M$. I don't know how to do this, in part because I'm not sure how to think about that last statement categorically. Do I just define $\cdot$ as another monoidal structure on $M$ that follows the same rules as $\mu$, and try to relate them?

I appreciate any help.

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You seem to be missing (at least you don't spell it out) that $\mu$ is assumed to be a morphism in the category of monoids, in particular all the morphisms you write here are actually homomorphisms of monoids (and that's all the connection you need). See the Eckmann-Hilton argument. – t.b. Aug 2 '12 at 7:09
Thanks, t.b., the link/the observation that both multiplications are homomorphisms from $M^2$, were exactly what I needed. – Kevin Carlson Aug 2 '12 at 7:26
the product in the category Mon, is just the direct product. so requiring that $\mu be a monoid homomorphism means that: – David Wheeler Aug 2 '12 at 7:26
up vote 6 down vote accepted

This is the classical situation covered by the Eckmann–Hilton argument.

The point is that the multiplication on $M$ defined by $\mu$ is a homomorphism for the original multiplication $\cdot$ on $M$ and that's enough for the argument given on the Wikipedia page to apply. It seems a bit pointless to repeat it here. A number of references are given on the Wikipedia page (including a nice video by the Catsters).

The original article by Eckmann and Hilton is freely available on the GDZ and I recommend having a look at it:

B. Eckmann, P.J. Hilton, Group-Like Structures in General Categories I. Multiplications and Comultiplications, Mathematische Annalen 145 (1962), 227–255.

PS: Yes, you can use the ordinary product to get a monoidal structure on the category of monoids and that's the one that is usually intended.

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Since you are interested in higher categories, you should see the Eckmann-HIlton argument as a special case of the use of the interchange law, or exchange law. A general formulation of this is for double categories: this is a set, or class, $C$, with two category structures $\circ_1, \circ_2$ each of which is a morphism for the other. This amounts to the rule $$(x \circ_1 y)\circ_2(z\circ_1 w)= (x \circ_2 z) \circ_1 (y \circ_2 w)$$whenver both sides are defined, which can also be represented in matrix terms as there is only one way of evaluating $$\left[ \matrix{ x&z \\y&w }\right] $$ ($\circ_2$ is horizontal, $\circ_1 $ is vertical, like matrices). One then proves that while such a structure is not itself abelian, it does contain a family of abelian monoids. (The fact that such morphisms preserve identities is a crucial part of the argument.) It was this fact that raised the possibility of non abelian higher homotopy groupoids, and indeed of higher groupoids having a potential in wide areas of mathematics and physics, giving a kind of "higher dimensional group theory".

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Thanks, Ronnie-taking the double category as the morphisms in a 2-category, I see this is the general form of the claim that vertical and horizontal composition play nicely together. That helps! – Kevin Carlson Aug 2 '12 at 10:38
I can't resist pointing out the article on "double semigroups" . See also on double inverse semigroups. – Ronnie Brown Mar 25 '15 at 18:59

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