# The $2$-category of monoids

People sometimes say that monoids are "categories with one object". In fact people sometimes suggest that this is the natural definition of a monoid (and likewise "groupoid with one object" as the definition of a group).

But categories naturally form a $2$-category $\mathbf{Cat}$. So if we took the above definition seriously then we would view monoids as forming a $2$-category $\mathbf{Mon}$. The objects would be monoids and the morphisms would be monoid homomorphisms, but there would also be $2$-morphisms between homomorphisms. A $2$-morphism between $f,g:M\to N$ is an $n\in N$ such that $nf(m)=g(m)n$ for all $m\in M$.

If one takes the principle of equivalence seriously then this poses a problem because we lose the ability to talk about the "underlying set" of a monoid. There's no $2$-functor $U:\mathbf{Mon}\to\mathbf{Set}$ (treating $\mathbf{Set}$ as a $2$-category with no nontrivial $2$-morphisms) that sends each monoid to its underlying set and each homomorphism to its underlying function. In the $1$-category of monoids this would be given by applying the functor $\mathrm{Hom}(\Bbb N,-)$. But in the $2$-category $\mathbf{Mon}$ two homomorphisms $f,g:\Bbb N\to M$ are isomorphic whenever $f(1)=mg(1)m^{-1}$ for some $m\in M$, so this construction only gives us the set of conjugacy classes of $M$ rather than its set of elements.

Clearly this poses a problem if we want to work with monoids and groups. In particular proofs involving finite groups often require the ability to count the number of elements in some subset of a group. It becomes impossible to state Lagrange's Theorem. We also lose the ability to talk about the free group on a set, since we can't construct the adjoint to the nonexistent functor $U$.

In light of this, I want to know if it's actually possible to take "category with one object" as our definition of monoid, and still be able to prove things in a practical way. I can see two ways to do this:

1) Recover the $1$-category of monoids from $\mathbf{Mon}$ in some natural way

or

2) Show that we can reconstruct group theory in a way that never uses concepts like "order of a group" or "free group on a set"

Does anybody know a way to do either of these?

First of all, even having one object is not invariant under equivalence. So perhaps a monoid is actually a category with a unique isomorphism class of objects. Then the way this issue is handled in topology is to consider a different 2-category: that of pointed categories, that is, categories with a distinguished object, functors preserving that object, and natural transformations which are the identity on that object. This fixes your problem: the category of pointed functors between two pointed monoids is discrete.

• This (bi)equivalence between monoids and pointed connected categories is spelt out in "Lectures on n-Categories and Cohomology" by Baez and Shulman, arxiv.org/abs/math/0608420, in the section "Pointedness versus connectedness" – Dap Apr 25 '18 at 12:46
• Thanks! This completely solves the issue. I added my own answer that explores some of the details. – Oscar Cunningham Apr 25 '18 at 21:23

Kevin Carlson answered the question, but I thought I'd add my own answer based on his, with some more details.

The ($$1$$-)category $$\mathbf{Set}$$ lives inside the $$2$$-category $$\mathbf{Cat}$$, as the full sub-$$2$$-category on the discrete categories. The inclusion $$F:\mathbf{Set}\to\mathbf{Cat}$$ has a right adjoint $$U:\mathbf{Cat}\to\mathbf{Set}$$ that sends a category to its set of isomorphism classes.

So a "category with one object" (or, to better respect the principle of equivalence, a "category with one isomorphism class of objects") is precisely a category $$\mathcal{M}$$ such that there's a bijection $$1\to U\mathcal M$$. Since there's at most one such bijection we could equally well say that it's a category equipped with a bijection $$a:1\to U\mathcal M$$. But as I said in the question, this gives a $$2$$-category with unwanted $$2$$-morphisms.

Instead, the correct definition is to look at categories equipped with a particular object to which every other object is isomorphic. An object is precisely a functor from the terminal category, and the terminal category is equivalent to $$F1$$. So we define a monoid to be a category $$\mathcal M$$ equipped with a functor $$a:F1\to\mathcal M$$ which corresponds to a bijection $$1\to U\mathcal M$$ under the isomorphism $$\mathrm{Hom}(F1,\mathcal M)\cong\mathrm{Hom}(1,U\mathcal M)$$ given by the adjunction.

Based on this definition, it makes sense to say that a morphism between monoids $$(\mathcal M,a)\to(\mathcal N,b)$$ is a functor $$f:\mathcal M\to\mathcal N$$ such that $$f\circ a\simeq b$$, and that a 2-morphism $$f\to g$$ is given by a natural transformation $$\alpha:f\to g$$ such that $$(\alpha\circ f)_\bullet = g(\mathrm{id}_\bullet)$$ (where $$\bullet$$ is the object of $$1$$). Of course there is in fact only one such natural transformation, so this version of $$\mathbf{Mon}$$ is indeed a $$1$$-category.

The sort of definition we gave above is actually quite common in mathematics. Two similar definitions arise from the usual adjunction between $$\mathbf{Set}$$ and $$\mathbf{Vect}$$. A basis $$S$$ of a vector space $$V$$ is precisely a function $$f:S\to UV$$ that the corresponding function $$FS\to V$$ is an isomorphism. Dually a vector space structure $$V$$ on a set $$S$$ is a function $$FS\to V$$ such that the corresponding function $$S\to UV$$ is a bijection.

By analogy, we could say that a monoid is not a "category with one object" but rather a "category structure on the set with one element". This gives some intuition for why monoids only form a $$1$$-category. Categories naturally form $$2$$-categories, but structures based on sets are only sophisticated enough to form $$1$$-categories.

In fact, I believe that if we allow any set $$S$$ in place of $$1$$ in the above definition of a monoid then we get a definition of the $$1$$-category of categories. So we could also define monoids by first passing to this $$1$$-category, and then looking at the "categories with one object" within it.