# What's the underlying $2$-category of a commutative monoid?

Write $$\mathbb N$$ for the natural numbers (with $$0$$) seen as a monoid under $$+$$.

Then for any monoid $$M$$ the underlying set of $$M$$ can be calculated as $$\mathrm{Hom}(\mathbb N,M)$$.

But if we view monoids as one object categories, via their deloopings, then $$\mathrm{Hom}(\mathrm B\mathbb N,\mathrm BM)$$ is a category whose objects are the elements of $$M$$ and whose morphisms $$a\to b$$ are elements $$c \in M$$ with $$ca = bc$$. Hence the isomorphism classes of this category are the conjugacy classes.

So the conjugacy classes are in this sense a categorified version of the underlying set of $$M$$.

Now suppose $$M$$ is commutative. We can now take the double delooping of $$M$$ to get a $$2$$-category with one object and only the identity morphism on that object and with $$M$$ as its $$2$$-morphisms. Question: What $$2$$-category is now given by $$\mathrm{Hom}(\mathrm B^2\mathbb N,\mathrm B^2M)$$?

This should be easy to figure out from the definitions, but every time I try I find that there's simply too much to hold in my head at one time.

• What is your definition of a conjugacy class in monoids? – J.-E. Pin Sep 10 at 12:29
• @J.-E.Pin In this case it's forced to be that $a\sim b$ if and only if there exist $c$ and $c^{-1}$ with $cc^{-1}=\mathrm{id}_M=c^{-1}c$ and $b = cac^{-1}$. That's precisely what falls out if you look at the isomorphism classes of the $1$-category. – Oscar Cunningham Sep 10 at 12:41
• Thanks, I understand now. – J.-E. Pin Sep 10 at 16:28
• I really would not call these constructions the "underlying category" or "underlying 2-category." $\mathbb{N}$ is the free monoid on a point, but its delooping $B \mathbb{N}$ is certainly not the free category on a point (it behaves more like a circle, so $[B \mathbb{N}, -]$ is like a free loop space), and its double delooping $B^2 \mathbb{N}$ is certainly not the free 2-category on a point. – Qiaochu Yuan Sep 10 at 22:05
• @QiaochuYuan I was being lazy in skipping the $\mathrm B$s, I've added them now. I'm keeping "underlying" though, since the motivation for the question is that $\mathrm{Hom}(\mathrm B^2\mathbb N,\mathrm B^2M)$ is analogous to $\mathrm{Hom}(\mathbb N,M)$. – Oscar Cunningham Sep 11 at 10:45