The abelianization of $\operatorname{End}(\{1, \ldots, n\})$

Question

Consider the following categories:

• $\mathsf{Mon}$ of monoids

• $\mathsf{Grp}$ of groups

• $\mathsf{CMon}$ of commutative monoids

• $\mathsf{CGrp}$ of commutative groups (more commonly known as $\mathsf{Ab}$)

Then we have the following commutative square of categories and inclusions:

$$ \require{AMScd} \begin{CD} \mathsf{CGrp} @> \iota_{\mathsf{Grp} \leftarrow \mathsf{CGrp}} >> \mathsf{Grp} \\ @V \iota_{\mathsf{CMon} \leftarrow \mathsf{CGrp}} VV @VV \iota_{\mathsf{Mon} \leftarrow \mathsf{Grp}} V \\ \mathsf{CMon} @>> \iota_{\mathsf{Mon} \leftarrow \mathsf{CMon}} > \mathsf{Mon} \\ \end{CD} $$

It is well-known that all the above inclusion functors admit left adjoints. Hence, we obtain the following commutative square containing the left adjoint functors to the inclusions:

$$ \begin{CD} \mathsf{CGrp} @< L_{\mathsf{CGrp} \leftarrow \mathsf{Grp}} << \mathsf{Grp} \\ @A L_{\mathsf{CGrp} \leftarrow \mathsf{CMon}} AA @AA L_{\mathsf{Grp} \leftarrow \mathsf{Mon}} A \\ \mathsf{CMon} @<< L_{\mathsf{CMon} \leftarrow \mathsf{Mon}} < \mathsf{Mon} \\ \end{CD} $$

My question is: Consider the monoid $(\operatorname{End}(\{1, \ldots, n\}), \circ)$ consisting of all set mappings $\{1, \ldots, n\} \to \{1, \ldots, n\}$. What happens to this monoid under the various left adjoints? Explicitely, what are the following objects?

• $L_{\mathsf{Grp} \leftarrow \mathsf{Mon}}(\operatorname{End}(\{1, \ldots, n\}))$, the groupification of $\operatorname{End}(\{1, \ldots, n\})$

• $L_{\mathsf{CMon} \leftarrow \mathsf{Mon}}(\operatorname{End}(\{1, \ldots, n\}))$, the abelianization of $\operatorname{End}(\{1, \ldots, n\})$

• $L_{\mathsf{CGrp} \leftarrow \mathsf{Mon}}(\operatorname{End}(\{1, \ldots, n\}))$, the commutative group freely constructed from $\operatorname{End}(\{1, \ldots, n\})$

The reason I am interested in this is that I would like to do the following construction: Take a category $\mathcal{C}$ and construct the category $\mathcal{C}^{\operatorname{Com}}$ by forcing all endomorphism monoids to be commutative. In particular, I would suspect the endomorphism monoids of $\mathsf{FinSet}^{\operatorname{Com}}$ to be precisely the commutative monoids $L_{\mathsf{CMon} \leftarrow \mathsf{Mon}}(\operatorname{End}(\{1, \ldots, n\}))$.

Answer 1

To calculate the abelianization we can first of all piggyback off the known calculation of the abelianization of $S_n$, which is that it's given by the sign homomorphism $\text{sgn} : S_n \to \{ \pm 1 \}$. The argument starts by observing that

1. $S_n$ is generated by transpositions, which have order $2$, and
2. All transpositions are conjugate, so they are all identified in the abelianization.

This already implies the abelianization of $S_n$ is either the sign homomorphism or trivial; from here it's a classic exercise to show that the abelianization is nontrivial (for $n \ge 2$), or equivalently that the sign homomorphism really exists, and there's a nice MO thread discussing this.

Second, we can generalize the fact that $S_n$ is generated by transpositions to the following: $M_n = \text{End}([n])$ is generated by transpositions together with the idempotent functions $e_{ij \to i} : [n] \to [n]$ which send $e(i) = e(j) = i$ and fix all other values, for $i \neq j$. All of these functions are conjugate so they are all identified in the abelianization into a single idempotent $e$. This already implies that the abelianization consists of at most $4$ elements $\{ 1, -1, e, -e \}$ where every non-permutation is sent to either $e$ or $-e$, but we can multiply $e_{ij \to i}$ by, say, the transposition $s_{ij}$ (which maps to $-1$) and get another idempotent $e_{ij \to j}$, which is conjugate. So $e = -e$ in the abelianization, and $e$ is absorbing.

So the abelianization is $\{ 0, 1, -1 \}$ and can be identified with the determinant of the corresponding "function matrix" (which shows the abelianization doesn't collapse further); all non-permutations are sent to $0$.

I don't know why you want to force all endomorphism monoids to be commutative; it's a very violent operation. A less violent version of this operation is given by a construction you might call the "Hochschild homology of a category"; I actually don't remember where I saw this but it can be defined as the coend of the hom functor $C^{op} \times C \to \text{Set}$, and it works out to being the quotient of the disjoint union $\bigsqcup_c \text{Hom}(c, c)$ of all the endomorphism monoids by the equivalence relation $fg \sim gf$ where $f : c \to d, g : d \to c$ are morphisms. This is sort of a "universal recipient of traces for endomorphisms in $C$." It is not itself a commutative monoid, though. But it's less trivial than the abelianization for $C = \text{FinSet}$; for example the function that counts the number of fixed points of an endomorphism (the trace of the function matrix) satisfies $\text{fix}(gf) = \text{fix}(fg)$ so it factors through the coend, and more generally we could consider the coefficients of the characteristic polynomial of the function matrix.