From: Michael Shulman <firstname.lastname@example.org> To: "HomotopyTypeTheory@googlegroups.com" <email@example.com> Subject: [HoTT] Are cubical sets hypercomplete? Date: Tue, 11 Jun 2019 10:02:27 -0700 Message-ID: <CAOvivQyPJsVzRtJw7uWX=LJH0-3r7TarVm=CSaqfoFU4k7foqw@mail.gmail.com> (raw) I have always assumed that cubical set models, like the simplicial set model, satisfy Whitehead's principle (one form of which is the statement that if all n-truncations of a type are contractible, then it is contractible). However, since cubical set models aren't known to have an underlying model structure that's equivalent to simplicial sets (and, as discussed previously on this list, at least one model structure for cubical sets is known to be *not* equivalent to simplicial sets), it's not completely obvious to me how to prove this. Has anyone checked carefully that one or more cubical set models satisfy Whitehead's principle -- and in particular, is the argument fully constructive? I could imagine that it might require something like countable choice. -- You received this message because you are subscribed to the Google Groups "Homotopy Type Theory" group. To unsubscribe from this group and stop receiving emails from it, send an email to HomotopyTypeTheoryfirstname.lastname@example.org. To view this discussion on the web visit https://groups.google.com/d/msgid/HomotopyTypeTheory/CAOvivQyPJsVzRtJw7uWX%3DLJH0-3r7TarVm%3DCSaqfoFU4k7foqw%40mail.gmail.com. For more options, visit https://groups.google.com/d/optout.
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