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    • CommentRowNumber1.
    • CommentAuthorMike Shulman
    • CommentTimeOct 28th 2010
    • CommentRowNumber2.
    • CommentAuthorTobyBartels
    • CommentTimeNov 1st 2010

    I put in my definitions at k-ary factorisation system.

    • CommentRowNumber3.
    • CommentAuthorFinnLawler
    • CommentTimeMar 23rd 2011

    I’ve added to ternary factorization system the example of a span of categories that is both a left and a right fibration, and the condition that makes it a two-sided fibration. There should be a simple common generalization of these and the ambifibrations of the second-last example.

    I think the middle arrow in the three-way factorization has something to do with the unique transformation that makes the right-fibration structure a colax morphism of LL-algebras, where LL is the monad for left fibrations, so that requiring the arrow to be invertible would (via the usual distributive-law business) be the same as requiring the left- and right-fibration structures to underlie an MM-algebra, MM being the two-sided-fibration monad. But it’s fiddly – has anyone seen this explicitly worked out anywhere?

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTimeMar 23rd 2011

    Nice, thanks! Actually, I think that can be regarded as a special case of the ambifibrations example: regard E as over A×BA\times B, where A×BA\times B has the factorization system (E,M) with E being morphisms of the form (1,g) and M being morphisms of the form (f,1). (That factorization system has the amusing property that (M,E) is also a factorization system!)

    What you say about distributive laws sounds reasonable, but I haven’t seen it worked out either. If you work it out, I think it would be worth adding to two-sided fibration.

    • CommentRowNumber5.
    • CommentAuthorFinnLawler
    • CommentTimeMar 24th 2011

    D’oh, the product category – I should have thought of that!

    I think two-sided fibration could use some expanding and rearranging anyway, so I’ll commit myself in public to doing that just so I can’t put it off for too long.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeNov 29th 2012
    • (edited Nov 29th 2012)

    I have added to k-ary factorization system Example-pointers to, first of all, ternary factorization system (that link was missing) and then to

    which I’d think is the archetypical example.

    • CommentRowNumber7.
    • CommentAuthorFosco
    • CommentTimeJan 19th 2014

    I made more explicit the “straightforward exercise in orthogonality”. Hope it helps to visualize the underlying argument! (in a second moment I’ll turn the .png images into real-code diagrams)

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeFeb 22nd 2019

    Updated the broken codecogs pictures to tikzcd.

    diff, v7, current

  1. Nice!

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeJul 6th 2023

    I have believed for 13 years that strict ternary factorization systems were equivalent to triple distributive laws, but today I realized that’s false, because R 2L 1R_2L_1 may not be closed under composition. So I removed this claim from the page and added some facts and an example from the Pultr-Tholen paper showing why it is false, and conjectured that this is the only obstacle.

    diff, v8, current

    • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeJul 6th 2023

    Now I’m curious what the abstract distributive-law-like structure corresponding to a ternary factorization system is. It looks like we should have

    • Three monads A,B,CA,B,C
    • Distributive laws α:BAAB\alpha : B A \to A B and γ:CBBC\gamma : C B \to B C
    • A morphism β:CAABC\beta : C A \to A B C
    • Two unit laws for β\beta, e.g. the composite CCη ACAβABCC \xrightarrow{C \eta_A} C A \xrightarrow{\beta} A B C is equal to Cη Aη BCABCC \xrightarrow{\eta_A \eta_B C} A B C.
    • Two composition laws relating β\beta to α\alpha and γ\gamma, e.g. the composite CAAβAABCAABβABABCAαBCAABBCμ Aμ BCABCC A A \xrightarrow{\beta A} A B C A \xrightarrow{A B \beta} A B A B C \xrightarrow{A \alpha B C} A A B B C \xrightarrow{\mu_A \mu_B C} A B C is equal to the composite CAACμ ACAβABCC A A \xrightarrow{C \mu_A} C A \xrightarrow{\beta} A B C using the multiplication of AA.

    Given these data, we can define a multiplication for the composite ABCA B C as

    ABCABCABβBCABABCBCAαBγCAABBBCCμ Aμ B 2μ CABC A B C A B C \xrightarrow{A B \beta B C} A B A B C B C \xrightarrow{A \alpha B \gamma C} A A B B B C C \xrightarrow{\mu_A \mu_B^2 \mu_C} A B C

    and I hope it should be a monad. Has anyone seen anything like this before?