TITLE: Cofinality for coends?
QUESTION [13 upvotes]: Recall that a functor $I \xrightarrow u J$ is cofinal if it has the property that for any functor $J \xrightarrow F C$, we have that $\varinjlim F \cong \varinjlim Fu$ via the canonical map, either side of the equation existing if the other does. This notion is very useful: it admits various reformulations which can be checked directly; there are numerous practical examples, and once a functor is known to be cofinal, the property which I've just treated as a definition becomes a great tool for computing colimits.
When passing from colimits to coends, it would be nice to have similarly powerful tools available. Coends can be reduced to colimits -- e.g. we have $\int^{j \in J} F(j,j) = \varinjlim(Tw(J) \to J^{op} \times J \xrightarrow F C)$, where $Tw(J)$ is the twisted arrow category. But from examples I've tried in the past, my impression is that it's relatively uncommon for a functor $I \xrightarrow u J$ to induce a cofinal functor $Tw(I) \xrightarrow{Tw(u)} Tw(J)$.

Question 1: What are some examples of functors $I \xrightarrow u J$ which induce cofinal functors $Tw(I) \xrightarrow {Tw(u)} Tw(J)$? Are they really quite rare?

Even if I'm right in thinking that such functors are rare, we shouldn't be deterred. After all, a coend is not the colimit of an arbitrary functor out of $Tw(J)$ --  but rather one which factors through $J^{op} \times J$. So there may still be functors $I \xrightarrow u J$ out there which always induce equivalences of coends, even if $Tw(I) \xrightarrow{Tw(u)} Tw(J)$ is not cofinal.

Question 2: What are some examples of functors $I \xrightarrow u J$ such that for any $J^{op} \times J \xrightarrow F C$, we have $\int^{j \in J} F(j,j) \cong \int^{i \in I} F(ui,ui)$ (via the canonical map), either side existing if the other does?

Finally, a more systematic question about these functors:

Question 3: Is it possible to give a direct combinatorial characterization of functors $u$ of the form described in Question 1 or Question 2 (analogous to the usual characterization of cofinal functors via connectedndess of slice categories)?

I'm also interested in versions of these questions other settings like enriched category theory or $\infty$-category theory.
Obviously, everything should have a dual story about limits and ends, too.

REPLY [8 votes]: Offline, Alex Campbell independently suggested a similar approach to the one Roald mentions in the comments, and worked it out. Here are the results -- we work with ends rather than coends for simplicity:
We observe that if $I^{op} \times I \xrightarrow F C$ is a functor, then the end $\int_{i \in I} F(i,i)$ is precisely the limit $\varprojlim_{Hom_I} F$ of $F$ weighted by the Hom-functor $Hom_I: I^{op} \times I \to Set$. Thus, we can apply the weighted version of initiality (the "limit" version of cofinality -- the more modern thing seems to be to say "final" for what I called "cofinal" above), which says in general that

Initiality for Weighted Limits: Let $I \xrightarrow u J$ be a functor, let $\phi: I \to Set$ and $\psi: J \to Set$ be functors (which we regard as "weights" for weighted limits), and let $\eta: \phi u \Rightarrow \psi$ be a natural transformation. Then the following are equivalent:

For any $C$ and any functor $J \xrightarrow F C$, we have $\varprojlim_\psi F \cong \varprojlim_\phi F u$ via the canonical map induced by $\eta$, either side existing if the other does.
$\eta$ exhibits $\psi$ as the Left Kan extension $\psi = Lan_u \phi$ of $\phi$ along $u$.


In particular, we can apply this in the case where $\phi = Hom_I$, $\psi = Hom_J$, and $\eta$ is given by the action of the functor $u$. The left Kan extension can be computed explicitly via a coend formula, and the result is the following:

Proposition (Initiality for Ends): Let $I \xrightarrow u J$ be a functor. The following are equivalent:

For every functor $F: J^{\mathrm{op}} \times J \to C$, we have $\int_{j \in J} F(j,j) \cong \int_{i \in I} F(ui,ui)$ via the canonical map, either side existing if the other does.
For every $j,j' \in J$, the canonical map $\int^{i \in I}Hom_J(j,ui) \times Hom_J(ui,j') \to Hom_J(j,j')$ is an isomorphism.


There are various ways to reformulate (2). For instance,


The composite of profunctors $Hom_J(1,u) \circ_I Hom_J(u,1)$ is canonically isomorphic to $Hom_J$.
For every $j\xrightarrow \beta j' \in J$, the "category of $u$-factorizations" of $\beta$  -- whose objects consist of triples $i \in I, j \xrightarrow \alpha ui \xrightarrow {\alpha'} j'$ composing to $\beta$ (morphisms are the obvious thing) -- is connected.
[ABSV] For any $C$, the functor $Fun(u,C): Fun(J,C) \to Fun(I,C)$, given by precomposition with $u$, is fully faithful.
[ABSV again] The functor $u$ is absolutely dense, i.e. for any $j \in J$ we have $j = \varinjlim (u / j \to J)$ and the colimit is absolute.


In the ABSV paper linked to above, such functors are called "lax epimorphisms" in light of (5) above (the idea being that a "pseudo-epimorphism" is a functor $u$ such that $F(u,C)$ is always a pseudo-monomorphism, which has something to do with the core of the categories involved, but here we take into account non-invertible 2-cells of $Cat$). In light of (5) above, one might also say "co-fully-faithful" or something like that.
Any localization is an example of such a functor. So is any composite or transfinite composite of localizations. The transfinite composites of localizations form the left half of a factorization system on $Cat$ whose right half is the conservative functors, and it's not hard to see that if $u$ is co-fully-faithful, in the factorization $u = wv$ with $v$ being a transfinite composite of localizations and $w$ being conservative, both $v$ and $w$ are co-fully-faithful. Thus when we look beyond localizations, it seems the appropriate thing to ask is "which conservative functors are co-fully-faithful?". For example, I think the functor from a category to its idempotent completion is co-fully-faithful (while also being fully faithful and in particular conservative). I think that's about all there is to say about co-fully-faithful functors which are also fully-faithful -- any such functor induces an equivalence of idempotent completions (one way to see this is to use the absolute density condition above with the Yoneda embedding). But of course, there may be quite a lot of daylight between co-fully-faithful functors which are conservative and those which are fully faithful.
The co-fully-faithful functors also seem related to the "liberal" functors (functors $u$ such that $Fun(u,C)$ is always conservative) of CJSV: co-fully-faithful implies liberal but not conversely.
Of course, the enriched and $\infty$-categorical counterparts of all of this should be clear at a conceptual level, at any rate.
Note also that everything is self-dual: a functor $u$ is co-fully-faithful iff $u^{op}$ is, so the questions about ends and coends are actually equivalent (and not just dual).