Let $D: \mathcal{C} \to \mathbf{Set}$ be a diagram of sets, then we can obtain the colimit of $D$ as the set of connected components of the category of elements of $F$, which we denote by $\mathrm{el}(F)$. The category $\mathrm{el}(F)$ in turn is the oplax colimit of the composition of $F$ and the inclusion $\mathbf{Set} \hookrightarrow \mathbf{Cat}$.
Now let us consider a functor $F: \mathcal{C}^{\mathrm{op}} \times \mathcal{C} \to \mathbf{Set}$, then the coend of $F$ can again be constructed as the set of connected components of a category $\mathrm{el}^\wedge(F)$ which we now describe:

  • The objects are pairs $(X,x)$ with $X \in \mathcal{C}$ and $x \in F(X,X)$.
  • A morphism $(X,x) \to (Y,y)$ is given by a pair $(f,a)$, where $f: X \to Y$ is a morphism in $\mathcal{C}$, and $a \in F(X,Y)$ such that $f_*(x) = f^*(y) = a$.
  • Finally composition of two morphisms $(X,x) \xrightarrow{(f,a)} (Y,y) \xrightarrow{(g,b)} (Z,z)$ is given by $(g \circ f, g_*(a)) = (g \circ f, f^*(b))$.

Does the category $\mathrm{el}^\wedge F$ satisfy a similar universal property as $\mathrm{el}F$?

The above description of the colimit of $D$ feels very natural to me, and I was hoping that this perspective may shed some light on the nature of coends*.

Finally I would like to note that it may be that there are other categories $\mathcal{E}$ than $\mathrm{el}^\wedge F$ such that $\pi_0(\mathcal{E})$ corresponds to the coend of $F$, which are more natural than my construction. If this is the case, I would be grateful for these to be pointed out.

*It is like first taking the quotient of a set by a discrete group in the $(\infty,1)$-category of spaces (or in this case equivalently in the $(2,1)$-category of groupoids) and then applying the left adjoint to $\mathbf{Set} \hookrightarrow \mathbf{Spaces}$ (resp. $\mathbf{Set} \hookrightarrow \mathbf{Groupoids}$).

Warning: As mentioned in the comments, the category I describe does probably not have the correct set of connected components. I hope I will come up with a fix some time soon.

  • $\begingroup$ Are you using $D$ and $F$ interchangeably? $\endgroup$ Sep 30, 2017 at 21:44
  • $\begingroup$ I'm using $D$ for diagrams indexed by $\mathcal{C}$, and $F$ for diagrams indexed by $\mathcal{C}^{\mathrm{op}} \times \mathcal{C}$. $\endgroup$ Sep 30, 2017 at 21:48
  • 1
    $\begingroup$ Are you sure morphisms of $\mathrm{el}^\land F$ must be as stated? I believe a morphism from $(X,x)$ to $(Y,y)$ must be a pair $(f,a)$ where $f:Y\to X$ and $a\in F(X,Y)$ with $f_*(a)=x$ and $f^*(a)=y$, no? $\endgroup$ Oct 1, 2017 at 6:27
  • 1
    $\begingroup$ I mean, the coend is the "multiple pushout" of the collection$$F(Y,Y)\xleftarrow{f^*}F(X,Y)\xrightarrow{f_*}F(X,X)$$indexed by $f:Y\to X$, so $f^*a$ and $f_*a$ must be in the same connected component for each $a\in F(X,Y)$; in view of this maybe it is slightly more natural to view what I described above as a morphism $(Y,y)\to(X,x)$ rather; but this gives the same connected components. $\endgroup$ Oct 1, 2017 at 6:34
  • 1
    $\begingroup$ But on the other hand, the fix you suggest, does not give a category: If I have three objects $(X,x),(Y,y),(Z,z)$ with morphisms $(f:Y\to X, a)$ and $(g:Z \to Y, b)$, then I don't see any canonical way of determining an element $c \in F(Z,X)$ such that $(f \circ g)^*(c) = x$ and $(f \circ g)_*(c) = z$. $\endgroup$ Oct 1, 2017 at 21:34


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.