6
$\begingroup$

Let $I$ be a small category and let $D : I^{\mathrm{op}} \times I \to \mathsf{Ab}$ be a functor. The coend $$\int^{i \in I} D(i,i)$$ can be constructed as the direct sum $\bigoplus_{i \in I} D(i,i)$ modulo the equivalence relation generated by $$D(i,i) \ni D(f,\mathrm{id}_i)(x) \sim D(\mathrm{id}_j,f)(x) \in D(j,j)$$ for $f : i \to j$, $x \in D(j,i)$.

Question. Under which conditions on $I$ and $D$ is every element of $\int^{i \in I} D(i,i)$ the class of some element in some $D(i,i)$? In other words, when $U : \mathsf{Ab} \to \mathsf{Set}$ is the forgetful functor, I ask for a condition that the canonical map $$\int^{i \in I} U(D(i,i)) \to U\left(\int^{i \in I} D(i,i)\right)$$ is surjective. This is clearly equivalent to the condition that the image of this map is closed under sums.

For example, this is the case when for all $x \in D(i,i)$, $y \in D(j,j)$ there is some span $i \leftarrow k \rightarrow j$ such that $x$ has a preimage in $D(i,k)$ and $y$ has a preimage in $D(j,k)$, since then $x$ and $ y$ are equivalent to elements in $D(k,k)$. This condition is already useful in practice, but obviously it is quite strong. Probably weaker assumptions are also sufficient.

Answers for the following special case are appreaciated as well: Let $A : I^{\mathrm{op}} \to \mathsf{Ab}$, $B : I \to \mathsf{Ab}$ be two functors and let $D(i,j) = A(i) \otimes B(j)$, so that the coend is the tensor product of functors $A \otimes_I B$. In this case, the question is when every element of $A \otimes_I B$ is the class of some element in some $A(i) \otimes B(i)$.

Background. I have to do some computations with some complicated coends, and it won't be sufficient to use the universal property. I really need a better idea of the element structure of the coend. The next step would be to find a description of the kernel of $D(i,i) \to \int^{i \in I} D(i,i)$.

$\endgroup$
5
  • 1
    $\begingroup$ Doesn't the special case of the tensor product (even for $I=1$) show the answer is "almost never"? I mean, it might still be interesting to work out when this happens, but I wouldn't expect to it too in useful cases. $\endgroup$ Commented Feb 14, 2020 at 3:38
  • $\begingroup$ @OmarAntolín-Camarena Actually, for $I=1$ the sufficient condition I mentioned is satisfied. My question is not when every tensor is pure (I corrected that in the post), but when every element belongs to a single tensor product then, or in the general case, to a single $D(i,i)$. I have clarified that in the post as well. $\endgroup$ Commented Feb 15, 2020 at 7:08
  • 1
    $\begingroup$ @MartinBrandeburg Oh, right! I was a little careless, I did think it meant the analogue of every tensor being pure. The correct thing has a much better chance of happening. :) $\endgroup$ Commented Feb 15, 2020 at 16:49
  • $\begingroup$ When doing coends of abelian groups, I'm usually thinking of $Ab$-enriched coends, which I think are relevantly different from the $Set$-enriched coends you seem to be using here. $\endgroup$ Commented Apr 28, 2021 at 16:38
  • $\begingroup$ @TimCampion Good point. I need to think about the context again, but there was probably a reason that the coends where normal ($\mathbf{Set}$-enriched) coends. They appeared in the context of certain categories of algebras, which are not $\mathbf{Ab}$-enriched, for instance. $\endgroup$ Commented Apr 28, 2021 at 17:00

0

You must log in to answer this question.