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Stream: theory: category theory

Topic: displaying transformations

Christian Williams (Jun 09 2022 at 05:48):

Today I realized the following idea. Inverse image along a functor $f:X\to A$ is a lax functor from $A$ to Prof; see displayed category. Completing the picture, inverse image along a transformation $\gamma:Q\Rightarrow R(f,g)$ is a lax module transformation from $R$ to $Prof(f^*,g^*)$.

Christian Williams (Jun 09 2022 at 18:44):

In the picture below, $\gamma^*_{a,b}(r): X(a)\to Y(b)$ is a profunctor with the values
$\gamma^*_{a,b}r(x,y)=\{q:Q(x,y) \; | \; \gamma(q)=r\}$.

Christian Williams (Jun 09 2022 at 18:47):

($Q^*$ here is the same as $\gamma^*$)
disp-trans.png
disp-trans1.jpg

Christian Williams (Jun 09 2022 at 18:48):

Blue is $A$ (and blue string/bead is $f^*$), green is $B$ (string/bead is $g^*$), the black string is $R$.

Orange is $Cat_0$, the category of categories and functors, and the orange string is $Cat_1$, the category of profunctors and transformations.

The black bead is $\gamma^*$.

Christian Williams (Jun 09 2022 at 18:50):

Preimage along a transformation is lax, for the same reason that preimage along a functor is lax: if a composable pair lies over a composable pair, then their composite lies over the composite (but there is no reason for there to be an inverse).

Christian Williams (Jun 09 2022 at 19:20):

I'm happy to explain in as much detail as anyone wants. There are plenty of pictures that can make each part clear.

Morgan Rogers (he/him) (Jun 10 2022 at 07:32):

@Christian Williams this exposition is missing a motivating ingredient, I think. What do you get out of thinking of inverse image in this way? More bluntly: "so what?"

Graham Manuell (Jun 10 2022 at 10:57):

@Christian Williams I'm not an expert in double categories, but is this perhaps related to the Grothendieck construction for double fibrations (see https://arxiv.org/abs/2205.15240)?

Christian Williams (Jun 10 2022 at 16:26):

I haven't done any real exposition yet, because nobody had said anything. But I can soon.

Christian Williams (Jun 10 2022 at 16:30):

Short version: why do we care about inverse image? It's how we model dependent types, which are everywhere. Expanding this theory from sets to categories is an immense jump in expressive power, and it starts by asking "how do we take the inverse image along a functor?" You get something fairly loose, and you tighten it up a certain way to get fibrations. Now I'm doing the same thing for transformations between profunctors, to determine a natural concept of a "relation" between fibrations.

Christian Williams (Jun 10 2022 at 16:34):

Graham Manuell said:

Christian Williams I'm not an expert in double categories, but is this perhaps related to the Grothendieck construction for double fibrations (see https://arxiv.org/abs/2205.15240)?

Well, Fib over Cat should be a canonical double fibration, but so far it has only been described as a fibration of 2-categories. so I think I'm filling that gap in the literature.

Tobias Schmude (Jul 05 2022 at 15:15):

I'm interested in pretty much anything displayed, and I think I roughly get your construction. I don't understand the intuition behind it though: Why does the inverse image construction apply to transformations of specifically this form, the domain being a profunctor $X \to Y$, and the codomain being the composition of a profunctor $A \to A$ with $(f, g)$?
This would be answered by making the inverse image construction functorial, such that your construction is in some way the action on morphisms or something similar. I don't see anything like that though.

About your motivation: we know the following:

• morphisms of fibrations over some category $\mathcal{B}$ correspond to pseudonatural transformations between the corresponding pseudofunctors $\mathcal{B} \to \mathrm{Cat}$ (or pseudonatural transformations with representable components between the corresponding pseudofunctors $\mathcal{B} \to \mathrm{Prof}$), and
• morphisms in $\mathrm{Cat}/\mathcal{B}$ correspond to oplax transformations with representable components between the corresponding normal lax functors $\mathcal{B} \to \mathrm{Prof}$.

So wouldn't a natural notion of "relation" between fibrations (or arbitrary functors into $\mathcal{B}$) be given by oplax transformations with not necessarily representable components? I guess I just haven't understood how your construction ties in with your motivation :oh_no:

Tobias Schmude (Jul 05 2022 at 15:28):

I haven't checked it, but these (I mean oplax transformations with not necessarily representable components) might correspond to profunctors $X \to Y$ over $A$ instead of functors. At least that looks plausible to me, and also looks like a sensible notion of "relation" between fibrations.

Christian Williams (Jul 05 2022 at 23:20):

thanks for the interest! I'm happy to discuss

Tobias Schmude said:

I don't understand the intuition behind it though: Why does the inverse image construction apply to transformations of specifically this form, the domain being a profunctor $X \to Y$, and the codomain being the composition of a profunctor $A \to A$ with $(f, g)$?

A transformation $\gamma: Q \to R(f,g)$ is an arbitrary square in the double category of categories, functors, profunctors, and transformations. The profunctor $R$ does not go $A\to A$ but $A\to B$, and $f:X\to A$ and $g:Y\to B$ are substituted in, like substituting functions on either side of a relation.

Just as displaying a functor gives an equivalence between the arrow 2-category of Cat and the lax-functor 2-slice over Prof, the above gives the same kind of equivalence, now with "lax modules" between those lax functors. So it's an equivalence of the "arrow double category" of Cat and a "lax double slice" over Cat.

Christian Williams (Jul 05 2022 at 23:27):

Tobias Schmude said:

• morphisms in $\mathrm{Cat}/\mathcal{B}$ correspond to oplax transformations with representable components between the corresponding normal lax functors $\mathcal{B} \to \mathrm{Prof}$.

So wouldn't a natural notion of "relation" between fibrations (or arbitrary functors into $\mathcal{B}$) be given by oplax transformations with not necessarily representable components? I guess I just haven't understood how your construction ties in with your motivation :oh_no:

interesting idea. in my mind, double categories are more natural / basic that 2-categories. I don't think about (op/lax) transformations with non-representable (horizontal) components, in particular because they do not compose horizontally. that obstacle is one reason the double perspective is really necessary.

Christian Williams (Jul 05 2022 at 23:31):

but anyway, I am certain that the result is correct, and I hope I can help clarify. here's a picture with better colors: preimage.png preimage-actions.png

Christian Williams (Jul 05 2022 at 23:37):

The left is a square in Cat, and the right is a square in DblCat, a lax transformation between lax double profunctors. it sounds fancy, but the diagrams on the right just say:

Given a $q:Q(x,y_0)$ with $\gamma(q)=r:R(a,b_0)$ and a morphism $y:Y(y_0,y_1)$ with $g(y)=b:B(b_0,b_1)$, then for the "composite" $p\cdot y:Q(x,y_1)$ we have $\gamma(q\cdot y) = r\cdot b:R(a,b_1)$.

Similarly for the left action, with $f:X\to A$.

Christian Williams (Jul 05 2022 at 23:51):

(oops, the green bead should be yellow and the blue bead should be red, since they represent the functor actions on morphisms. but the diagrams explain.)

Tobias Schmude (Jul 06 2022 at 06:13):

Christian Williams said:

thanks for the interest! I'm happy to discuss

Tobias Schmude said:

I don't understand the intuition behind it though: Why does the inverse image construction apply to transformations of specifically this form, the domain being a profunctor $X \to Y$, and the codomain being the composition of a profunctor $A \to A$ with $(f, g)$?

A transformation $\gamma: Q \to R(f,g)$ is an arbitrary square in the double category of categories, functors, profunctors, and transformations. The profunctor $R$ does not go $A\to A$ but $A\to B$, and $f:X\to A$ and $g:Y\to B$ are substituted in, like substituting functions on either side of a relation.

Just as displaying a functor gives an equivalence between the arrow 2-category of Cat and the lax-functor 2-slice over Prof, the above gives the same kind of equivalence, now with "lax modules" between those lax functors. So it's an equivalence of the "arrow double category" of Cat and a "lax double slice" over Cat.

Ah, I didn't get that you're considering $f, g$ with different codomains $A, B$. Guess I should have known by the variable names $a$ and $b$ :upside_down:
Or by the second diagram. Sorry, should have read it more carefully!

Tobias Schmude (Jul 06 2022 at 09:31):

An observation: your construction generalizes the action of the "usual" generalized Grothendieck equivalence $\mathrm{Cat}/A \cong \mathrm{Lax}_{\mathrm{norm}}(A, \mathrm{Prof})$ on morphisms.

For $R = \mathrm{id}_A$, $Q$ represented by a functor $\bar{Q}$ such that $g \circ \bar{Q} = f$, and $\gamma$ given by the canonical transformation, the $\gamma^*_{a, a'}$ you construct correspond to the component transformations of the oplax natural transformation $\bar{Q}^*: f^* \Rightarrow g^*$ that the non-double categorical version yields.