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Jan 19, 2017 at 2:32 vote accept John Pardon
Apr 19, 2016 at 8:39 answer added Piotr Achinger timeline score: 11
Apr 19, 2016 at 7:33 answer added მამუკა ჯიბლაძე timeline score: 10
Apr 18, 2016 at 19:30 comment added Dylan Wilson The diagram is produced algorithmically. If you want to do it for n composable morphisms, draw a them in a line. Now extend the line to the right with a zero, and down on the left with a zero. Now fill in this rectangle with pushout diagrams (so you're looking at an [n+1] \times [1] diagram.) Now on the far lower right you can extend by a map to 0, and on the lower left offset by one you can extend by zero... rinse wash and repeat. You get a staircase lookin' thing.
Apr 18, 2016 at 19:26 comment added მამუკა ჯიბლაძე @DylanWilson Are you sure it is on p.24? I see there (TR4) and it is certainly not evident for me how to generalize this.
Apr 18, 2016 at 18:57 comment added Dylan Wilson (It is not known whether any triangulated thing is n-angulated, but any stable $\infty$-category is $n$-angulated for any $n$, and the proof is the same as in loc. cit.)
Apr 18, 2016 at 18:55 comment added Dylan Wilson the picture at the bottom of p.24 here: math.harvard.edu/~lurie/papers/HA.pdf has an evident generalization to any n-tuple of morphisms. If you write it down you'll join the group of m people who have independently discovered the notion of an "n-angulated category". The shape you seek comes from collapsing that unrolled version in some probably not-so-trivial way.
Apr 18, 2016 at 17:30 history edited John Pardon CC BY-SA 3.0
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Apr 17, 2016 at 21:33 history asked John Pardon CC BY-SA 3.0