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Stream: theory: category theory

Topic: An extranatural question

Paolo Perrone (Apr 20 2024 at 00:30):

I have a question about extranatural transformations.

Let $\mathrm{C}$ be a category (let's say small), let $A:\mathrm{C}^\mathrm{op}\to\mathrm{Set}$ be a presheaf, let $B:\mathrm{C}\to\mathrm{Set}$ be a functor, and let $C$ be a set.
Consider now an extranatural transformation $F:A(-)\times B(-)\to C$, i.e. of components as follows:
1.png

Extranatural transformations, by currying, are in bijection to natural transformations $F^\sharp:A(-)\to\mathrm{Set}(B(-),C)$, of components as follows:
2.png

They are also, by the universal property of coends, in bijection with functions $F'$ between the following sets:
3.png

Now I wonder: are $F'$ and $F^\sharp$ related in some way? For example, if all the components of $F^\sharp$ are injective (resp. surjective), can we conclude that $F'$ is injective (resp. surjective)?

Rémy Tuyéras (Apr 20 2024 at 02:51):

It seems that you have

$F^{\sharp} \in \int_x Set(A(x),Set(B(x),C))$

(by characterization of natural transformations as an end) and also the series of iso:

$\int_x Set(A(x),Set(B(x),C)) \cong \int_x \prod_{A(x)}Set(B(x),C)$

$\dots \cong \int_x Set(\coprod_{A(x)}B(x),C)$

$\dots \cong Set(\int^xA(x)\times B(x),C)$

Then, express injectiveness as a lifting problem in Set, or rather translate that characterization in homset language: i.e. a function $X\to Y$ is injective if the cospan:

$Set(\mathbf{2},X) \to Set(\mathbf{2},Y)$

$Set(\mathbf{1},Y) \to Set(\mathbf{2},Y)$

admits $Set(\mathbf{1},X)$ as a weak pullback (maybe there is a better way to say it).

Note: You can generalize the previous characterization with the composition map
$Set(\mathbf{2},X) \times Set(X,Y) \to Set(\mathbf{2},Y)$
to characterize the set of injections $X\to Y$. This characterization should probably look like a lifting problem for a certain diagram where you start by "choosing $\mathbf{1} \to Set(X,Y)$ such that ...".

After this, the sequence of isomorphisms I shown above should lead you to an answer (by successfully transferring -- or failing to transfer -- the lifting problem from one side to another)

Note: You may need to use some smallness property for $\mathbf{2}$ and $\mathbf{1}$ with respect to the colimit $\int^x\coprod_{A(x)} B(x)$. Interestingly, this hints toward a condition involving both $A(x)$ and $B(x)$, hence a little more than just the domain of $F^{\sharp}$.

You can repeat the same argument for surjectiveness using the map $\mathbf{0} \to \mathbf{1}$

Paolo Perrone (Apr 21 2024 at 11:27):

Hi Rémy, nice to hear from you.
I wasn't aware of this characterization of injectivity using cospans, it's quite cool.
I don't have very much intuition for it yet, though. How can I use it in the chain of isomorphisms?

By the way, the chain of isomorphisms that I had in mind, which is very similar to what you wrote down, is the following.
$\mathrm{Set}\big(\int^X A(X)\times B(X), C\big)$
$\cong \int_X \mathrm{Set}\big(A(X)\times B(X), C\big)$
$\cong \int_X \mathrm{Set}\big(A(X), \mathrm{Set}(B(X),C)\big)$
$\cong \mathrm{Nat} \big(A(-), \mathrm{Set}(B(-), C)\big)$.

Rémy Tuyéras (Apr 21 2024 at 13:07):

Hey Paolo :)

The idea is that the lifting problem gives you two 2-dimensional diagrams, and the series of isomorphisms allow you to link these two diagrams in a 3-dimensional picture, but only on certain parts of the two diagrams. The goal of the exercise would be to complete the whole picture into a 3 dimensional commutative diagram.

The lifting problem should be used for the following two homsets:
1) $Set(\int^x A(x) \times B(x) ,C)$
2) $\int_x Set(A(x), Set(B(x) ,C)$

Note: right now, the second homset is more like a limit of homsets, but it is still a homset, namely $Nat(-,-)$

Now, referring to the notations of my previous post, the object $Y$ will be $C$, and the object $X$ will be:

1) either $\int^x A(x) \times B(x)$
2) or $A(x)$

I would recommend starting with 1) as it will be easier to figure out how to use smallness conditions.

Now, this is our lifting problem:

174a9f04-222d-4c6f-853a-634ef0592eea.jpg

For convenience, I will denote $X=\int^x A(x) \times B(x)$, but I will write $Y$ as $C$ for clarity.

Now, we shall try to use the series of iso wherever we can on the diagram of the lifting problem:

1)

$Set(2,X) \times Set(X,C) \to Set(2,C)$

is isomorphic to

$Set(2,X) \times \int_x Set(A(x), Set(B(x) ,C) \to Set(2,C)$

2)

$Set(1,X) \times Set(X,C) \to Set(1,C)$

is isomorphic to

$Set(1,X) \times \int_x Set(A(x), Set(B(x) ,C) \to Set(1,C)$

3)

$Set(1,X) \times Set(X,C) \to Set(2,X) \times Set(X,C)$

is isomorphic to

$Set(1,X) \times \int_x Set(A(x), Set(B(x) ,C) \to Set(2,X) \times \int_x Set(A(x), Set(B(x) ,C)$

4) as you can see, our remaining problem to transfer a lifting solution (see picture above) to the other side of the 3 dimensional diagram is to work on the homsets $Set(1,X)$ and $Set(2,X)$ so that you can link them to

$Nat(\Delta(1),A) \cong \int_x Set(1,A(x))$

$Nat(\Delta(2),A) \cong \int_x Set(2,A(x))$

For this, my intuition is that:

i) You have to use some smallness property to take the colimit of $X$ out of the homset $Set(2,X)$. The catch is that you will have to turn that colimit into a limit $\int_x (-)$ (but note that it is always possible to map a limit to a colimit by first using the projections and then the inclusions)

ii) You will have to go back to the details of how you compose natural transformations when their homsets are defined as an end. You want to do this because you want your isomorphic transfer to commute with the compositions (see the compositions represented by circles in my photo).

Hopefully, this strategy helps you find conditions that help you transfer injectiveness/surjectiveness between the maps (e.g. the shape of your category $\mathcal{X}$, etc.)

Paolo Perrone (Apr 21 2024 at 14:32):

Rémy Tuyéras said:

1)

$Set(2,X) \times Set(X,C) \to Set(2,C)$

is isomorphic to

$Set(2,X) \times \int_x Set(A(x), Set(B(x) ,C) \to Set(2,C)$

This seems promising. I can see why the domains are isomorphic, but what's the second map?

Rémy Tuyéras (Apr 21 2024 at 14:36):

The second map is just the obvious pre-composition of the isomorphism with the composition map.

You should be able to link the second map to the actual composition for natural transformations when you investigate my comment in ii)

Paolo Perrone (Apr 21 2024 at 14:38):

Oh, I see. Let me try.