From: Michael Shulman <firstname.lastname@example.org>
Subject: [HoTT] Are cubical sets hypercomplete?
Date: Tue, 11 Jun 2019 10:02:27 -0700 [thread overview]
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.
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reply other threads:[~2019-06-11 17:02 UTC|newest]
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