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Stream: theory: category theory

Topic: cofunctors as profunctors

James Deikun (May 09 2022 at 13:51):

Consider a profunctor $P : B ⇸ A$ equipped with the following data:

• for each $a \in A$, an object $b_{a}$ of $B$ and an element $p_{a}$ of $P(a,b_{a})$
• such that each element $p$ of any $P(a,b)$ appears as $P(f,id) p_{a'}$ for some unique $a'$, $f$.

From this data we can reconstruct a cofunctor $\varphi : A \nrightarrow B$ as follows:

• $\varphi_{0}(a) = b_{a}$
• $\varphi_{1}(f)$ for $f : \varphi_{0}(a) \to b$ is the unique $g$ such that $P(id,f) p_{a} = P(g,id) p_{a'}$.

Is there a nicer way to characterize the above data, and/or to reconstruct any of it given some property of $P$?

James Deikun (May 09 2022 at 14:01):

Oh, and you go the other way as follows:

• $P(a,b) = \coprod_{\varphi_{0}(a')=b} A(a,a')$
• $b_{a} = \varphi_{0}(a)$
• $p_{a} = \varphi_{1}(id_{a})$
• $p_a = \varphi_1(id_{\varphi_{0}(a)}) = id_{a}$.

Bryce Clarke (May 10 2022 at 06:47):

This is interesting! I'll have a think about it and post something later today / this week. Ping me if I forget!

Bryce Clarke (May 10 2022 at 09:58):

James Deikun said:

• such that each element $p$ of any $P(a,b)$ appears as $P(f,id) p_{a'}$ for some unique $a'$, $f$.

So I'm a bit confused as to what this bit is trying to say. Could you provide some intuition as to what you mean?

Bryce Clarke (May 10 2022 at 10:00):

$f$ is a morphism from what to what? Below it looks like it is a morphism in $B$, whereas here it appears to be a morphism of $A$?

James Deikun (May 10 2022 at 11:59):

You're correct that below it is a morphism in $B$. Here, though, it is $f : a \to a'$; basically $(a',f)$ is a unique dependent pair.

James Deikun (May 10 2022 at 12:08):

The way I came up with it is thinking of the collage of $P$ where this is a factorization of sorts (not orthogonal) of morphisms that go from the $A$-part to the $B$-part. As seen below in the "other way" part, the $p_{a}$ are the lifts of identity morphisms through each $a$, viewed as "heteromorphisms" from $a$ to $\varphi_{0}(a)$. Because they are lifts of identity morphisms, they are also identity morphisms in $a$. In fact I need to update that part because I messed it up; please check it again.

James Deikun (May 10 2022 at 12:16):

Anyway the way lifts work is you compose with an "identity" on one side to turn the morphism in B into a heteromorphism (this part does most of the actual work) and then factor through an "identity" on the other side to turn it into a morphism in A.

James Deikun (May 10 2022 at 12:22):

As for how $P$ itself is put together from $\varphi$, this is what it looks like when you compose $A(1,F) \circ \Lambda(G,1)$ for $F$ a bijective-on-objects functor and $G$ a discrete opfibration such that the span represents $\varphi$.

Tom Hirschowitz (May 10 2022 at 13:36):

Don't have enough time to look closely enough, but, just in case, this makes me think of [[two-sided discrete fibrations]]. You know about them, I guess?

James Deikun (May 12 2022 at 11:17):

I do know about them and their relevance to profunctors in general, but I'm not sure where you're specifically seeing them here.

Tom Hirschowitz (May 12 2022 at 15:46):

I was not, really. It was just a quick, vague intuition. Sorry if that was unclear, and even more sorry if it was bad intuition!

James Deikun (May 16 2022 at 13:02):

For comparison I decided to try to describe representable profunctors in a similar way:

A representable profunctor is a profunctor $P : A ⇸ B$ equipped with the following data:

• for each $a \in A$, an object $b_a$ of $B$ and an element $p_a$ of $P(b_a,a)$
• such that each element $p$ of any $P(b,a)$ appears as $P(f,id)\,p_{a}$ for some unique $f : b \to b_a$.

From this data we can reconstruct a functor $F : A \to B$ as follows:

• $F_0(a) = b_a$
• $F_1(f)$ for $f : a \to a'$ is the unique $g$ such that $P(id,f)\,p_a = P(g,id)\,p_{a'}$.

Other way:

• $P(b,a) = B(b,F_0(a))$
• $b_a = F_0(a)$
• $p_a = F_1(id_a) = id_{F_0(a)}$.