Question

This is a follow-up question to this coend computation. Here's an attempt at a slightly simpler computation:

$\int^{a \in A} \mbox{hom_A(a,a)}$

This should be similar to the trace operator. In attempting to follow the derivation

$\begin{array}{l}\mbox{Set}(\int^{b \in B}\mbox{hom}(a, b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b) \times F(b), S)\\ \cong \int_b \mbox{Set}(\mbox{hom}(a,b), \mbox{Set}(F(b), S)) \\ \cong \mbox{Nat}(\mbox{hom}(a,-), \mbox{Set}(F(-), S) \\ \cong \mbox{Set}(F(a), S),\end{array}$

I get

$\begin{array}{l}\mbox{Set}(\int^{a \in A} \mbox{hom}_A(a,a), S) \\ \cong \int_{a \in A} \mbox{Set}(\mbox{hom}_A(a,a), S) \\ \cong \mbox{Nat}(\mbox{hom}_A(-,-), S)\end{array}$

So here I guess we have the set of natural transformations from the hom functor to the constant functor $S$. For any first parameter $a$, we have the set of natural transformations from hom$(a,-)$ to $S(a,-)$, which by Yoneda's lemma is isomorphic to $S(a,a) = S$. So I think it goes

$\begin{array}{l}\cong \displaystyle \prod_a \mbox{Nat}(\mbox{hom}_A(a,-), S(a,-)) \\ \cong \prod_a S\\ \cong S^{\mbox{Ob}(A)} \\ \cong \mbox{Set}(\mbox{Ob}(A), S).\end{array}$

So $\int^{a \in A} \mbox{hom_A(a,a)} \cong \mbox{Ob}(A).$ Is that right?

Answer 1

I agree with Reid's answer, but I want to add a bit more.

Putting Reid's calculation into a more general setting, if $A$ is any category then $$ \int^{a \in A} \mathrm{hom}_A (a, a) = (\mathrm{endomorphisms\ in\ } A)/\sim $$ where $\sim$ is the (rather nontrivial) equivalence relation generated by $gh \sim hg$ whenever $g$ and $h$ are arrows for which these composites are defined.

You can see confirmation there that your instinct about traces was right. If we wanted to define a 'trace map' on the endomorphisms in $A$, it should presumably satisfy $\mathrm{tr}(gh) = \mathrm{tr}(hg)$, i.e. it should factor through $\int^a \mathrm{hom}(a, a)$.

In fact, Simon Willerton has done work on 2-traces in which exactly this coend appears. See for instance these slides, especially the last one.

You can see in that slide something about the dual formula, the end $$ \int_{a \in A} \mathrm{hom}_A(a, a). $$ By the "fundamental fact" I mentioned before, this is the set $$ {}[A, A](\mathrm{id}, \mathrm{id}) = \mathrm{Nat}(\mathrm{id}, \mathrm{id}) $$ of natural transformations from the identity functor on $A$ to itself. Evidently this is a monoid, and it's known as the centre of $A$. For example, when $G$ is a group construed as a one-object category, it's the centre in the sense of group theory. So your set might, I suppose, be called the co-centre of $A$.

Answer 2

The step $\int_{a \in A} \mathrm{Set}(\mathrm{hom}_A(a, a), S) = \mathrm{Nat}(\mathrm{hom}_A(-,-),S)$ does not really make sense, because $a \mapsto \mathrm{hom}_A(a,a)$ is not a functor. And $\int^{a \in A} \mathrm{hom}_A(a,a)$ is not equal to $\mathrm{Ob}(A)$ in general. For instance, let $G$ be a group and let $A = BG$ be the groupoid with a single object with automorphism group $G$. Then $\int^{a \in A} \mathrm{hom}_A(a,a)$ can be identified with the abelianization set of conjugacy classes of $G$. In general, $\int^{a \in A} \mathrm{hom}_A(a,a)$ is the "Hochschild homology" of the category $A$.

---

Edit: I was originally thinking of the case of $G$ abelian, and generalized wrongly to the nonabelian case. As atonement, let me write out the computation directly from the definition of coend:

$$\int^{a \in A} \mathrm{hom}_A(a, a) = \operatorname{colim} \left[ \coprod_{f:a \to b} \mathrm{hom}_A(b,a)\rightrightarrows \coprod_{a\in A} \mathrm{hom}_A(a,a)\right] = \operatorname{colim} [G \times G \rightrightarrows G]$$

where the two maps send $(g, h)$ to $gh$ and $hg$ respectively. So the coend is the quotient of $G$ in the category of sets by the relation $gh = hg$, or $g = hgh^{-1}$, i.e., it is the set of conjugacy classes of elements of $G$.