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I have added to orthogonal factorization system
in the Definition-section three equivalent explicit formulations of the definition;
in the Properties-section the statement of the cancellability property.
Wanted to add more (and to add the proofs). But have to quit now. Maybe later.
I have added the proof of the Canellation property.
(Not that it’s not almost trivial. But just for the record.)
Our definition 2 of orthogonal factorization system asks that in addition to factorizations being unique up to unique isomorphism, the classes $E$ and $M$ contain all isomorphisms and are closed under composition. However, I’ve seen it claimed elsewhere (such as arXiv:1112.3076) that the latter condition follows automatically. Is that true? I’m doubtful.
Here is a simple counterexample, if I’m not mistaken. Let’s consider the “commutative square” category, generated by the following objects, morphisms and equality:
$\begin{array}[t]{rcl} \A & \stackrel{e}{\to} & B \\ m'\downarrow & = & \downarrow e' \\ C & \stackrel{m}{\to} & D \end{array}$
We put in $E$ all identities, $e, e'$ and the composite $e'e = m m'$, whereas $M$ contains all identities, $m$ and $m'$. Every morphism in the category can be factorized as an $E$ followed by an $M$ in a unique way (up to unique isomorphism, since the only isomorphisms are the identities), but $M$ isn’t closed under composition.
We can correct the result in the following way: two full subcategories $E, M$ of $\mathcal{C}^\mathbb{2}$ form an orthogonal factorisation system iff the composition functor $\Gamma: M\circ E \to \mathcal{C}^\mathbb{2}$ is an equivalence (where $M\circ E$ is the composite of the spans $\mathcal{C}\stackrel{\mathrm{dom}}{\leftarrow} E \stackrel{\mathrm{codom}}{\rightarrow}\mathcal{C}$ and $\mathcal{C}\stackrel{\mathrm{dom}}{\leftarrow} M \stackrel{\mathrm{codom}}{\rightarrow}\mathcal{C}$). The incorrect conjecture would be equivalent to $\Gamma$ being surjective and fully faithful on isomorphisms would be sufficient. (N.B. I only use the belonging up to isomorphism relation between objects and full subcategories of a category, so I may have forgotten to add “replete” before “full”.)
Your counterexample looks right. And your proposed fix amounts essentially to saying $E\perp M$, right?
Yes, essentially, but expressed in the following way: For all commutative diagrams of the form
$\begin{array}[t]{rcl} A & \stackrel{e}{\to} &I &\stackrel{m}{\to} &B\\ u\downarrow & & & & \downarrow v\\ A' & \stackrel{e'}{\to} &I' &\stackrel{m'}{\to} &B'\\ \end{array}$
where $e,e'\in E$ and $m,m'\in M$, there is a unique morphism $d\colon I\to I'$ making the two squares commute. The special case where $u,v$ are isomorphisms amounts essentially (by repleteness) to the uniqueness of the factorisation up to a unique isomorphism.
This is equivalent to $E\bottom M$ if all identities factor as an $E$ followed by an $M$ (so that we can prove that all identities are both in $E$ and $M$).
(I think I first learned about this from On localization and stabilization for factorizations systems by Carboni, Janelidze, Kelly, Paré, Appl.Cat.Str. 5 (1997), paragraph 2.8.)
Page orthogonal factorization system lists the following as an example:
If $F\to C$ is a fibered category in the sense of Grothendieck, then $F$ admits a factorization system $(E,M)$ where $E$ = arrows whose projection to $C$ is invertible, $M$ = cartesian arrows in $F$
Is this correct ? I agree that putting cartesian morphisms in $M$ is a correct choice, but then $E$ must consist of all vertical morphisms, that is the morphisms whose projection to $C$ is an identity and not merely an invertible morphism. I mean every morphism is a composition of a vertical morphism followed by a cartesian morphism.
P.S. while my last sentence is correct, with the definition of the orthogonal factorization system in the entry we must allow isomorphisms as well, and it seems that $E$ proposed in the entry consists of morphisms which are vertical followed by iso. So, the entry seems correct.
Yes, I agree with your PS that the entry is correct; we have to allow the projections to be invertible rather than just identities in order to include all the isomorphisms in both classes. It’s certainly possible to have an OFS in which some of the factorizations are “extra-special” in a way not noticed by the factorization system.
Hi Beppe, welcome, and thanks for the question! Yes, I think you’re right. Instead of eso and eso+full, I think one needs to strictify a bit, to bijective on objects and bijective on objects + full respectively. Let’s fix this, and maybe give a few more details. I’ll not get to it just now; maybe someone else can give it a go?
Well at the moment it has both (eso, fully faithful) and (bo, fully faithful). They can’t both be correct, since the left class should be determined by the right. But the page doesn’t say that it’s only giving examples in 1-categories, so (because Cat is nicer as a 2-category than a 1-category) I think it would be better to just give (eso, fully faithful) and (eso+full, faithful), rather than whatever their strictified versions are.
I think I’m missing something though. Why is bijective-on-obejects the correct analogue of essentially-surjective-on-objects (rather than just surjective-on-objects)? How can every functor (in particular a functor not essentially-injective-on-objects) factor as (bo, fully faithful) when both bo and full imply essentially-injective-on-objects?
The factorisation of $F: A \rightarrow B$ can be taken to $A \rightarrow C \rightarrow B$, where $C$ is defined by $Ob(C) = Ob(A)$ and $Hom_C(a,b) = Hom_B(F(a), F(b))$, where the two functors are the obvious ones.
One needs bijectivity on objects rather than just surjectivity to be able to construct a lift (given $F_1 \circ F_0 = F_3 \circ F_2$ where $F_1$ is fully faithful and $F_2$ is bijective on objects, one constructs the lift on objects as the inverse of $F_2$ on objects composed with $F_0$ on objects; on arrows one uses the inverses coming from the fully faithfulness of $F_1$). Uniqueness is obvious.
Being able to lift up to isomorphism is much weaker. One will, I believe, need the axiom of choice to get the essentially surjective, fully faithful 2-factorisation. Just think about equivalences of categories.
Oh I see. Being bijective-on-objects doesn’t imply being essentially-bijective-on-objects. I got the wrong idea from the fact that being surjective-on-objects does imply being essentially-surjective-on-objects.
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