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Mar 27 at 3:53 comment added Tim Campion There’s also some yoga about most of the pushouts which exists in $\Delta$ being absolute — they are preserved by any function whatsoever. For one thing, the pushout of any epimorphism along any epimorphism exists and is absolute (i.e. $\Delta$ is an elegant Reedy category). For another, many of the pushouts of monomorphisms which exist are absolute — I believe this fact is used in Cisinski’s Homotopical Algebra book
Mar 26 at 11:44 vote accept Calin Tataru
Mar 26 at 2:34 answer added David Gao timeline score: 3
Mar 25 at 23:53 comment added Calin Tataru No, it doesn't have most colimits (e.g. no coproducts) but it does have some. Actually once you show that $\Delta \to \mathbf{Pos}$ preserves colimits, you get that a diagram in $\Delta$ has a colimit iff its colimit in $\mathbf{Pos}$ is totally ordered, so it's easy to check which diagrams in $\Delta$ have colimits.
Mar 25 at 23:30 comment added Andrej Bauer $\Delta$ doesn't have too many colimits, does it?
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S Mar 25 at 21:39 history asked Calin Tataru CC BY-SA 4.0