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Stream: learning: questions

Topic: Coequalizers in the double category of profunctors

fosco (Jun 22 2025 at 09:56):

I'm having a hard time understanding how coequalizers are constructed in the double category of profunctors (where cells are functors, in one direction, and profunctors, in the other). All (few) references that I find analyzing the matter resort to a slick, indirect proof that I don't know how to adapt to my $\bf Prof$-like double category.

Alternatively, I would like to see a construction of filtered colimits (e.g. colimits indexed over the ordinal $\omega$) and reflexive coequalizers, which should both be simpler and entail cocompleteness, together with coproducts.

Do you have an explicit reference, or would you like to spend some time showing me how to construct them?

Nathanael Arkor (Jun 22 2025 at 10:05):

If you mean the coequaliser of two cells in the double category Prof, Grandis and Paré give a completely explicit description on page 204 of Limits in double categories: view the two cells as functors between the collages of their domain/codomain distributors, and take their coequaliser (as functors). This produces a new category, which is also a collage. The distributor they correspond to is precisely the coequaliser of the two cells.

Nathanael Arkor (Jun 22 2025 at 10:06):

In other words: coequalisers in Prof are computed just as in Cat.

fosco (Jun 22 2025 at 10:08):

But this is the part of the paper we went through together, and it wasn't very helpful, no?

fosco (Jun 22 2025 at 10:12):

This is the situation:

$\begin{array}{ccc}A & \rightsquigarrow &B \\ \downdownarrows && \downdownarrows\\ C & \rightsquigarrow &D\end{array}$

coequalize left, coequalize right. You get

$\begin{array}{ccc}A & \rightsquigarrow &B \\ \downdownarrows && \downdownarrows\\ C & \rightsquigarrow &D\\ \downarrow && \downarrow\\ P & \underset{?}\rightsquigarrow &Q \end{array}$

How do you define the profunctor marked as "?", and the cell filling the lower square?

Nathanael Arkor (Jun 22 2025 at 10:15):

Take the [[collage]] of $A \rightsquigarrow B$ and $C \rightsquigarrow D$ (giving categories sliced over 2), view the transformations between them as functors between their collages, and then coequalise the functors.

fosco (Jun 22 2025 at 11:17):

I see, I probably misunderstood what you told me in person. Ultimately, the two cells $\alpha,\beta$ in the original squares induce a diagram

$A\uplus_s B \rightrightarrows C\uplus_t D$

Than one coequalizes. And the coequalizer $Q$ of this diagram defines a profunctor $P\uplus_r Q \to \Delta^1$.

fosco (Jun 22 2025 at 11:18):

At least this is a slick proof that I can understand. I have to adapt it to my setting. I hope it does!