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Stream: learning: questions

Topic: fibred category of quotients

Jules Hedges (Sep 15 2023 at 16:51):

Let $\mathbf{Equiv}$ be the category whose objects are sets equipped with an equivalence relation, and whose morphisms are compatible functions, ie. if $x \sim_X x'$ then $f (x) \sim_y f (x')$ (ie. the category of setoids, but during this paragraph I'm being classical). The forgetful functor $U : \mathbf{Equiv} \to \mathbf{Set}$ is a fibration if I'm not mistaken: given $(Y, \sim_Y)$ and $f : X \to Y$ we can define a "pullback" equivalence relation on $X$ by $x \sim_X x'$ iff $f (x) \sim_Y f (y)$, and this satisfies the universal property of a cartesian lift

I'm looking for a decently general version of this fibration. I feel like it should be {standard, well known, folkloric} but I failed to find any google hits

This is my guess: If $\mathcal C$ is a regular category, let $\mathbf{E} (\mathcal C)$ be full subcategory of $\mathbf{Arr} (\mathcal C)$ whose objects are regular epis, where I'm thinking of a regular epi $X \twoheadrightarrow X'$ as playing the same role as an equivalence relation on $X$. Let $\mathrm{dom} : \mathbf{E} (\mathcal C) \to \mathcal C$ be the domain functor, which I would like to upgrade to a fibration. Given a regular epi $Y \twoheadrightarrow Y'$ and a morphism $X \to Y$, we can "pull back" by taking the image factorisation of the composite $X \to Y \twoheadrightarrow Y'$ to give $X \twoheadrightarrow X' \to Y'$. But, I can't figure out how to prove the universal property of a cartesian lift for this

Jules Hedges (Sep 15 2023 at 16:56):

(I'm not looking for someone to write my proof for me, just hoping for an answer of the form "look in this section of this book")

Kevin Arlin (Sep 15 2023 at 18:04):

It's explicitly not what you asked for, but maybe it's at least a lead to observe that this seems to follow from the fact that regular epis in a regular category are strong:
jh.png

I guess this is dual to the question of when the subobject fibration admits pushforwards/existential quantification, except for the difference between arbitrary epi/monos and regular/strong/extremal ones, but the slight gap in duality seems interesting.

John Baez (Sep 15 2023 at 18:18):

Hmm. I can never remember this stuff, but I read that strong epimorphisms are precisely those left orthogonal to any monomorphism. Furthermore I read that they obey the left two-out-of-three property: if $g$ and $g f$ are strong epis then so is $f$. (In fact something stronger is true.)

David Jaz Myers has a paper in which he defines a cartesian factorization system to be an orthogonal factorization system in which the left class (which are like the monos) satisfies the left two-out-of-three and is closed under pullback along the right class (which are like the strong epis). He proves that any such factorization gives a Street fibration (which is the non-evil version of a fibration). So he may be doing what you want, @Jules Hedges! See Proposition 2.4 and the diagram underneath.

John Baez (Sep 15 2023 at 18:19):

Well, you can't always get what you want, but if you try sometimes, you might find you get what you need.

John Baez (Sep 15 2023 at 18:32):

It looks like the one thing I didn't check is the pullback condition.

Dylan McDermott (Sep 15 2023 at 18:54):

If $(\mathcal{E}, \mathcal{M})$ is an orthogonal factorization system on $\mathcal{C}$, then the codomain functor $\mathcal{M} \to \mathcal{C}$ is always an opfibration, nothing else is needed. So if I've dualized correctly, this implies the domain functor $\mathcal{E} \to \mathcal{C}$ is a fibration. The square on the right of @Kevin Arlin's picture is an instance of the diagonal fill-in property.

Dylan McDermott (Sep 15 2023 at 18:57):

I don't know of a good reference for this, but Hughes and Jacobs (bottom of page 160) have weaker assumptions than David Jaz Myers (they only assume existence of pullbacks).

Jules Hedges (Sep 15 2023 at 20:37):

Dylan McDermott said:

If $(\mathcal{E}, \mathcal{M})$ is an orthogonal factorization system on $\mathcal{C}$, then the codomain functor $\mathcal{M} \to \mathcal{C}$ is always an opfibration, nothing else is needed. So if I've dualized correctly, this implies the domain functor $\mathcal{E} \to \mathcal{C}$ is a fibration. The square on the right of Kevin Arlin's picture is an instance of the diagonal fill-in property.

Bryce Clarke gave me the same answer over on Mastodon, I think this looks like the best approach