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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeFeb 14th 2012

    I have added to orthogonal factorization system

    1. in the Definition-section three equivalent explicit formulations of the definition;

    2. 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.

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeFeb 15th 2012

    I have added the proof of the Canellation property.

    (Not that it’s not almost trivial. But just for the record.)

    • CommentRowNumber3.
    • CommentAuthorMike Shulman
    • CommentTimeJul 12th 2014

    Our definition 2 of orthogonal factorization system asks that in addition to factorizations being unique up to unique isomorphism, the classes EE and MM 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.

    • CommentRowNumber4.
    • CommentAuthorvej.kse
    • CommentTimeJul 12th 2014
    • (edited Jul 12th 2014)

    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:

    A e B m = e C m D\begin{array}[t]{rcl} \A & \stackrel{e}{\to} & B \\ m'\downarrow & = & \downarrow e' \\ C & \stackrel{m}{\to} & D \end{array}

    We put in EE all identities, e,ee, e' and the composite ee=mme'e = m m', whereas MM contains all identities, mm and mm'. Every morphism in the category can be factorized as an EE followed by an MM in a unique way (up to unique isomorphism, since the only isomorphisms are the identities), but MM isn’t closed under composition.

    We can correct the result in the following way: two full subcategories E,ME, M of 𝒞 𝟚\mathcal{C}^\mathbb{2} form an orthogonal factorisation system iff the composition functor Γ:ME𝒞 𝟚\Gamma: M\circ E \to \mathcal{C}^\mathbb{2} is an equivalence (where MEM\circ E is the composite of the spans 𝒞domEcodom𝒞\mathcal{C}\stackrel{\mathrm{dom}}{\leftarrow} E \stackrel{\mathrm{codom}}{\rightarrow}\mathcal{C} and 𝒞domMcodom𝒞\mathcal{C}\stackrel{\mathrm{dom}}{\leftarrow} M \stackrel{\mathrm{codom}}{\rightarrow}\mathcal{C}). The incorrect conjecture would be equivalent to Γ\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”.)

    • CommentRowNumber5.
    • CommentAuthorMike Shulman
    • CommentTimeJul 12th 2014

    Your counterexample looks right. And your proposed fix amounts essentially to saying EME\perp M, right?

    • CommentRowNumber6.
    • CommentAuthorvej.kse
    • CommentTimeJul 12th 2014

    Yes, essentially, but expressed in the following way: For all commutative diagrams of the form

    A e I m B u v A e I m B \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,eEe,e'\in E and m,mMm,m'\in M, there is a unique morphism d:IId\colon I\to I' making the two squares commute. The special case where u,vu,v are isomorphisms amounts essentially (by repleteness) to the uniqueness of the factorisation up to a unique isomorphism.

    This is equivalent to EME\bottom M if all identities factor as an EE followed by an MM (so that we can prove that all identities are both in EE and MM).

    (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.)

    • CommentRowNumber7.
    • CommentAuthorzskoda
    • CommentTimeFeb 13th 2018
    • (edited Feb 13th 2018)

    Page orthogonal factorization system lists the following as an example:

    If FCF\to C is a fibered category in the sense of Grothendieck, then FF admits a factorization system (E,M)(E,M) where EE = arrows whose projection to CC is invertible, MM = cartesian arrows in FF

    Is this correct ? I agree that putting cartesian morphisms in MM is a correct choice, but then EE must consist of all vertical morphisms, that is the morphisms whose projection to CC 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 EE proposed in the entry consists of morphisms which are vertical followed by iso. So, the entry seems correct.

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 13th 2018

    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.

    • CommentRowNumber9.
    • CommentAuthorbeppe
    • CommentTimeMay 9th 2018

    I was reading the page

    and at the "Examples" section I read that among the classical examples of OFS in Cat, there are:

    (i) Cat, (eso, fully faithful) factorization system

    (ii) Cat, (eso+full, faithful) factorization system

    mmm... aren't those examples of factorization system in a 2-cat?

  1. 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?

  2. 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?

    • CommentRowNumber12.
    • CommentAuthorRichard Williamson
    • CommentTimeMay 9th 2018
    • (edited May 9th 2018)

    The factorisation of F:ABF: A \rightarrow B can be taken to ACBA \rightarrow C \rightarrow B, where CC is defined by Ob(C)=Ob(A)Ob(C) = Ob(A) and Hom C(a,b)=Hom B(F(a),F(b))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 1F 0=F 3F 2F_1 \circ F_0 = F_3 \circ F_2 where F 1F_1 is fully faithful and F 2F_2 is bijective on objects, one constructs the lift on objects as the inverse of F 2F_2 on objects composed with F 0F_0 on objects; on arrows one uses the inverses coming from the fully faithfulness of F 1F_1). Uniqueness is obvious.

  3. 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.

  4. 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.

    • CommentRowNumber15.
    • CommentAuthorMike Shulman
    • CommentTimeMay 9th 2018

    move the 2-categorical examples to a separate section, explicitly marked.

    diff, v34, current

    • CommentRowNumber16.
    • CommentAuthorbeppe
    • CommentTimeMay 10th 2018
    Thanks Mike, and thanks e everybody.

    I could have done this myself, but I must confess I didnt manage to - and I am writing on a smartphone!