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

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

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

  • $\begingroup$ Who is "you"? Presumably one of the commenters on your previous question. But please revise to make this slightly more stand-alone? $\endgroup$ Apr 7 '10 at 4:03
  • 1
    $\begingroup$ I copied over the answer from the previous question; hopefully it's clearer now. $\endgroup$
    – Mike Stay
    Apr 7 '10 at 4:42
  • $\begingroup$ Watch Bartosz Milewski himself explaining the Coend Computation theory link below. dausel.co/57fyLh $\endgroup$ Nov 16 '19 at 2:38

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 $A$.

  • 1
    $\begingroup$ (endomorphisms in A)/~. Ie. closed loops of arrows without regard to start/endpoint. So this is telling us something about the topology of A considered as a directed graph. Is that right? $\endgroup$
    – Dan Piponi
    Apr 7 '10 at 18:07
  • $\begingroup$ Yes, I think it must be. At the moment I can't see quite what it's telling us about the topology of A, but it must be telling us something. I hadn't thought of it that way before. $\endgroup$ Apr 7 '10 at 22:02
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    $\begingroup$ The topological intuition is as follows: $\int^a \hom(a,a)$ is the set of connected components of the arrow category $(A^\to)_{end}$ where you take only endomorphisms as objects; more generally, $\int^a \hom(Fa, Ga)$ is the set of conn. comp. of the comma category $(F/G)_{end}$. $\endgroup$
    – fosco
    Aug 3 '19 at 19:54

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 $G$.

  • $\begingroup$ Yeah, I clearly got that wrong. At best I've got a dinatural transformation from a constant functor to hom that picks out the identity. $\endgroup$
    – Mike Stay
    Apr 7 '10 at 5:03
  • $\begingroup$ If you treat $\text{hom}_A$ as a functor $A^{op} \times A \to \text{Monoid}$, instead of $A^{op} \times A \to \text{Set}$, then I think the coend does give you the abelianization. So it depends on what you're taking to be the codomain of the profunctor. A similar story holds if you replace $\text{Set}$ by $\text{Vect}_k$ and $\text{Monoid}$ by $k\text{-Alg}$, and that might make the connection to Hochschild homology clearer. $\endgroup$
    – L.Z. Wong
    May 20 '17 at 22:45

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