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Stream: learning: questions

Topic: How to work with a labeled concrete category?

Naso (Apr 15 2021 at 12:33):

I'm trying to work with some labeled concrete categories, by which I mean for a concrete category $C$ and a set $L$ of labels, the labeled category is the comma category $U/L$ where $U$ is the forgetful $C \to \mathsf{Set}$ and $L$ is the constant functor on $L$.

My issue is that since the objects of these categories are set functions, when I do constructions like pullback, I just get a set function that I then need to manually show will lift to a structure-preserving map in $C$.

Is there something I can use to make sure the constructions on labelled objects automatically preserve/create the structure in $C$?

As an example: if $C$ is the category of posets and $f : X \to Y$ a set function and $l : UP \to Y$ a $Y$-labelled poset, then I would like to say that the pullback of $f, l$ is an $X$-labeled poset but I'm not sure how.

John Baez (Apr 15 2021 at 16:29):

I'm not sure, but another way of tackling your example might be to say that the forgetful functor $U: \mathsf{Poset} \to \mathsf{Set}$ is a Grothendieck fibration. One still however needs to prove this is true.

Naso (Apr 16 2021 at 04:08):

@John Baez thank you!!!

John Baez (Apr 16 2021 at 04:09):

Sure.

Mike Shulman (Apr 16 2021 at 15:20):

More precisely, if $Z$ is the pullback, then you have a function $Z\to UP$, and a cartesian lifting of this function with respect to $U$ would induce a poset structure on $Z$.

Naso (Apr 16 2021 at 15:26):

Actually I'm not sure how $U$ being a fibration would help?
Since I want posets labeled by sets, I thought I should need a fibration $\pi : U/\mathsf{Set} \to \mathsf{Set}$ so that the fiber over a set $X$ is the $X$-labelled posets, and then a map $f : X \to Y$ should induce a map $\pi^{-1} Y \to \pi^{-1} X$ ? @Mike Shulman I'm not sure I understand your comment either.

Naso (Apr 16 2021 at 15:27):

I did find the following exercise in Jacobs' "Categorical logic and type theory" which seems to say that the $\pi$ I mentioned would be an opfibration?: image.png

Mike Shulman (Apr 16 2021 at 15:32):

Here's another way to put it. Your $U/\rm Set$ is the pullback of ${\rm dom} : {\rm Set}^{\to} \to \rm Set$ and $U$. Thus, if $U$ is a fibration, then so is the projection $U/{\rm Set} \to {\rm Set}^{\to}$. But the functor ${\rm cod} : {\rm Set}^{\to} \to \rm Set$ is also a fibration, and the composite of fibrations is a fibration; thus the functor $U/{\rm Set} \to \rm Set$ is also a fibration.

Naso (Apr 16 2021 at 15:35):

Ok! I will need to think about this for a bit. So would Jacobs' exercise imply $U/Set \to Set$ is also an opfibration (a bifibration?)

Mike Shulman (Apr 16 2021 at 15:36):

Yes, it is.

Naso (Apr 16 2021 at 15:37):

Thank you very much Mike!!

Naso (Apr 22 2021 at 11:23):

I still have a doubt about why $U$ is a fibration... given a $Z \to U P$ we need to find a cartesian arrow above it.
In terms of the isomorphism between posets and T0 Alexandrov spaces, a cartesian arrow above should correspond to putting an initial topology on $Z$.
But I think the initial topology on a T0 space is not necessarily T0.
In terms of orders, the initial order could make $Z$ a preorder rather than a partial order?

Mike Shulman (Apr 22 2021 at 14:42):

You want to distinguish between posets and preorders?!? (-:O

John Baez (Apr 22 2021 at 14:49):

That's either a smiling angel with a halo over his head or a screaming guy with a furrowed brow.

Joachim Kock (Apr 22 2021 at 21:42):

Mike Shulman said:

You want to distinguish between posets and preorders?!? (-:O

I know that reaction :-) It has also been my own instinct for a long time.

But in combinatorics, the distinction is considered similar to the distinction between bijections and surjections. Now that sounds crazy to a category theorist!

The explanation is that preorders are like posets except that you allow repetition of elements, just like a surjection is like a bijection with repetition of elements.

The second statement probably requires further explanation: in combinatorics it is common to represent finite sets as standard ordinals, such as $\mathbf{n} = \{1,2,3,\ldots,n\}$. Then surjections are represented as packed words: a packed word is a word (in the alphabet $\mathbb{N}$) such that if a letter $n$ appears in the word, then all previous letters $k<n$ are required to appear too. So for example the packed word $1142533$ represents the surjection $f:\mathbf{7} \to \mathbf{5}$ where $f(i)$ is the letter in position $i$ in the word. So it's just a word where you are required to use all letters up to some point, but are allowed to repeat letters. In contrast, a bijection is just a packed word without repetition.

For a category theorist, it takes quite an effort to get used to this rather 'reductionist' way of thinking and writing, and there are tons of crazy (and useful) constructions that are not invariant under 'real' bijections. But it is a highly optimised language for what it is designed for (such as the theory of symmetric functions, just to mention one huge area).

Naso (Apr 22 2021 at 23:55):

Mike Shulman said:

You want to distinguish between posets and preorders?!? (-:O

Ha! :) Well to put it in context, I'm trying to represent behaviours that evolve in space time, the labels representing states and the partial order representing causal/chronological dependence. If two points were equal in the order that would indicate simultaneity, which I thought is bad due to the "relativity of simultaneity".

Anyway, If I used preorders would it be easier to work with the slice categories $\mathsf{Ord}/\mathcal{R} L$ where where $\mathcal{R} : \mathsf{Set} \to \mathsf{Ord}$ is right adjoint to the forgetful $\mathcal{U} : \mathsf{Ord} \to \mathsf{Set}$ instead of fibrations? (That option was not available for posets since there was no right adjoint).

Mike Shulman (Apr 23 2021 at 03:00):

For personal reasons I write all my emoticons left-handed, so (-:O is a smiling face with a halo. I generally use it to mean "innocent face". For instance, in this case, I know very well that there are sometimes good reasons to distinguish between posets and preorders, so my comment was meant as a sort of tongue-in-cheek way to point out that since category theorists often don't make that distinction, the answers up to that point may not have been making it either.

Mike Shulman (Apr 23 2021 at 03:02):

However, if you really do mean posets and not preorders, then your original example:
Nasos Evangelou-Oost said:

if $C$ is the category of posets and $f : X \to Y$ a set function and $l : UP \to Y$ a $Y$-labelled poset, then I would like to say that the pullback of $f, l$ is an $X$-labeled poset but I'm not sure how.

is false. For instance, let $P$ be the terminal poset and $l$ the identity function; then the pullback is just $X$, and the only sensible preorder that can be induced on it is the indiscrete one, which is not antisymmetric.

Mike Shulman (Apr 23 2021 at 03:03):

And yes, for preorders you can use the right adjoint to the forgetful. Then the fact that it's a fibration just follows from the existence of pullbacks in $\mathsf{Ord}$.

Naso (Apr 23 2021 at 05:33):

Mike Shulman said:

However, if you really do mean posets and not preorders, then your original example:
Nasos Evangelou-Oost said:

if $C$ is the category of posets and $f : X \to Y$ a set function and $l : UP \to Y$ a $Y$-labelled poset, then I would like to say that the pullback of $f, l$ is an $X$-labeled poset but I'm not sure how.

is false. For instance, let $P$ be the terminal poset and $l$ the identity function; then the pullback is just $X$, and the only sensible preorder that can be induced on it is the indiscrete one, which is not antisymmetric.

Oh! I guess I had in mind the case when the maps $f$ are always injective (these seem to be the only important maps in my context), I wonder if that changes things?

I also learnt that the property assuring the existence of "initial structures" like the initial order is being a topological category. So $\mathsf{Ord}$ is topological but $\mathsf{Pos}$ is only monotopological (ref. The Joy of Cats, defn 21.38)

Mike Shulman (Apr 23 2021 at 13:05):

Yes, restricting to monos should make it work.

Naso (Apr 23 2021 at 13:09):

Thank you Mike :)