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?

slightlymore stand-alone? – Theo Johnson-Freyd Apr 7 '10 at 4:03