Re: [HoTT] Re: Vladimir Voevodsky

[Steve Awodey <steve...@gmail.com>]

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thanks for posting this Dan. 
It’s very interesting to learn the background of the discovery of univalence.

Steve

> On Oct 11, 2017, at 1:47 PM, Daniel R. Grayson <danielrich...@gmail.com <mailto:danielrich...@gmail.com>> wrote:
> 
> At https://en.wikipedia.org/wiki/Homotopy_type_theory <https://en.wikipedia.org/wiki/Homotopy_type_theory> Voevodsky says this:
> 
> "Also in 2009, Voevodsky worked out more of the details of a model of type
> theory in Kan complexes, and observed that the existence of a universal Kan
> fibration could be used to resolve the coherence problems for categorical
> models of type theory. He also proved, using an idea of A. K. Bousfield, that
> this universal fibration was univalent: the associated fibration of pairwise
> homotopy equivalences between the fibers is equivalent to the paths-space
> fibration of the base."
> 
> When I asked him about that, he showed me the email from Bousfield, 
> answering Vladimir's question as forwarded by Peter May, containing
> a very nice description of the idea, and here it is:
> 
> -----------------------------------------------------------------------------
> Date:    Mon, 01 May 2006 10:10:30 CDT
> To:      Peter May <m...@math.uchicago.edu <mailto:m...@math.uchicago.edu>>
> cc:      jar...@uwo.ca <mailto:jar...@uwo.ca>, pgo...@math.northwestern.edu <mailto:pgo...@math.northwestern.edu>
> From:    "A. Bousfield" <bo...@uic.edu <mailto:bo...@uic.edu>>
> Subject: Re: Simplicial question
> 
> Dear Peter,
> 
> I think that the answer to Voevodsky's basic question is "yes," and I'll
> try to sketch a proof.
> 
> Since the Kan complexes X and Y are homotopy  equivalent, they share the
> same minimal complex M, and we have trivial fibrations X -> M  and Y -> M
> by Quillen's main lemma in "The geometric realization of a Kan fibration
> ."  Thus  X + Y -> M + M is also a trivial fibration where "+" gives the
> disjoint union.  We claim that the composition of  X + Y -> M + M with the
> inclusion  M + M >-> M x Delta^1 may be factored as the composition of an
> inclusion X + Y >-> E with a trivial fibration  E -> M x Delta^1 such that
> the counterimage of  M + M is  X + Y.  We may then obtain the desired
> fibration
> 
> E  ->  M x Delta^1  ->  Delta^1
> 
> whose fiber over 0 is X and whose fiber over 1 is Y.
> 
> We have used a case of:
> 
> Claim. The composition of a trivial fibration A -> B with an inclusion B
> -> C may be factored as the composition of an inclusion A >-> E  with a
> trivial fibration  E -> C such that the counterimage of B is A.
> 
> I believe that this claim follows by a version of the usual iterative
> construction of a Quillen factorization for A -> C using the "test"
> cofibrations
> Dot Delta^k >-> Delta^k
> for all k. At each stage, we use maps from the "test" cofibrations
> involving k-simplices of C outside of B.
> 
> I hope that this helps --  I haven't thought about Voevodsky's  more
> general question.
> 
> Best regards,
> Pete
> 
>     On Sun, 30 Apr 2006, Peter May wrote:
> 
>     Here is an extract from an email from Voevodsky (vlad...@ias.edu <mailto:vlad...@ias.edu>)
> 
>     -----------------
> 
> 	Q. Let X and Y be a pair of Kan simplicial sets which are homotopy
> 	equivalent. Is there a Kan fibration E -> \Delta^1 such that its
> 	fiber over 0 is X (up to an iso) and its fiber over 1 is Y (up to an
> 	iso)?
> 
> 	A more advanced version of the same question: let X' -> X  and X'' ->
> 	X be two Kan fibrations which are fiberwise equivalent to each other
> 	over X. Is there a kan fibration over X\times\Delta^1 whose fiber
> 	over X\times 0 is isomorphic to X', fiber over  X\times 1 is
> 	isomorphic to X'' and the homotopy between the two inclusions of X to
> 	X\times\Delta^1 define the original equivalence (up to homotopy)?
> 
> 	I encountered this issue trying to write up a semantics for dependent
> 	type systems with values in the homotopy category. which is in turn
> 	related to the problem of creating computer programs for proof
> 	verification.
> -----------------------------------------------------------------------------
> 
> 
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