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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 $L$-algebras, where $L$ 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 $M$-algebra, $M$ 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 $A\times B$, where $A\times B$ 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.

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 $R_2L_1$ 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.

• 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 $A,B,C$
• Distributive laws $\alpha : B A \to A B$ and $\gamma : C B \to B C$
• A morphism $\beta : C A \to A B C$
• Two unit laws for $\beta$, e.g. the composite $C \xrightarrow{C \eta_A} C A \xrightarrow{\beta} A B C$ is equal to $C \xrightarrow{\eta_A \eta_B C} A B C$.
• Two composition laws relating $\beta$ to $\alpha$ and $\gamma$, e.g. the composite $C A A \xrightarrow{\beta A} A B C A \xrightarrow{A B \beta} A B A B C \xrightarrow{A \alpha B C} A A B B C \xrightarrow{\mu_A \mu_B C} A B C$ is equal to the composite $C A A \xrightarrow{C \mu_A} C A \xrightarrow{\beta} A B C$ using the multiplication of $A$.

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

$A B C A B C \xrightarrow{A B \beta B C} A B A B C B C \xrightarrow{A \alpha B \gamma C} A A B B B C C \xrightarrow{\mu_A \mu_B^2 \mu_C} A B C$

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