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.

  • $\begingroup$ What is your definition of a conjugacy class in monoids? $\endgroup$ – J.-E. Pin Sep 10 at 12:29
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    $\begingroup$ @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. $\endgroup$ – Oscar Cunningham Sep 10 at 12:41
  • $\begingroup$ Thanks, I understand now. $\endgroup$ – J.-E. Pin Sep 10 at 16:28
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    $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Sep 10 at 22:05
  • $\begingroup$ @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 $\mathrm{Hom}(\mathbb N,M)$. $\endgroup$ – Oscar Cunningham Sep 11 at 10:45

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