From: Thomas Streicher <stre...@mathematik.tu-darmstadt.de>
To: Andrew Polonsky <andrew....@gmail.com>
Cc: Andrej Bauer <andrej...@andrej.com>,
Martin Escardo <escardo...@googlemail.com>,
"HomotopyT...@googlegroups.com" <HomotopyT...@googlegroups.com>
Subject: Re: [HoTT] Meta-conjecture about MLTT
Date: Mon, 12 Sep 2016 13:41:19 +0200 [thread overview]
Message-ID: <20160912114119.GA15456@mathematik.tu-darmstadt.de> (raw)
In-Reply-To: <CABcT7WDW4nj7kpeoEoTS0nB5X0MB04ZCf8t3g027=NE9H9NVOg@mail.gmail.com>
the fibres will just be equivalent in general
but for Fam(Set) they are isomorphic
On Mon, Sep 12, 2016 at 01:23:09PM +0200, Andrew Polonsky wrote:
> Perhaps one does not need to take a skeleton.
>
> A dependent type x:U |- P(X) : Type gives rise, in the *standard*
> set-theoretic model, to a family of sets indexed by [[U]].
>
> Since the standard type constructor preserve isomorphism, two fibers
> [[P(A)]] and [[P(B)]] will be isomorphic if A and B are.
>
> Saying "A provably has property P" translates into "[[P(A)]] is
> inhabited by a definable element".
>
> In principle, one could conceive that [[P(B)]] has no definable
> elements, even if the two are isomorphic.
>
> But saying that "B provably does not have property P" translates into
> "[[P(B)]] is empty", which cannot be the case.
>
> Cheers,
> Andrew
>
> On Mon, Sep 12, 2016 at 1:14 PM, Thomas Streicher
> <stre...@mathematik.tu-darmstadt.de> wrote:
> > I am afraid this skeleton model is not split and spliiting it it wan't
> > be a skeleton anymore.
> >
> > Thomas
> >
> >
> > On Mon, Sep 12, 2016 at 01:02:24PM +0200, Andrej Bauer wrote:
> >> Let me try.
> >>
> >> Consider the model of MLTT in the skeleton of sets, i.e., types are
> >> interpreted as cardinal numbers and functions are the set-theoretic
> >> functions between cardinals. The identity type is just what you'd
> >> expect in a set-theoretic model (in fact, there is *no* choice about
> >> i). There are no problems of coherence anywhere.
> >>
> >> This model validates equality reflection. If MLTT+reflection
> >> distinguished isomorphic types, then there would exist in this model
> >> two isomorphic types which are not equal, but since we started with a
> >> skeleton this will not do.
> >>
> >> It is a bit amusing that the model also validates things like Id(U,
> >> Nat -> Nat, Nat -> Bool) and is thus quite educational.
> >>
> >> It is a model, right?
> >>
> >> With kind regards,
> >>
> >> Andrej
> >>
> >
next prev parent reply other threads:[~2016-09-12 11:41 UTC|newest]
Thread overview: 17+ messages / expand[flat|nested] mbox.gz Atom feed top
2016-09-11 20:47 Martin Escardo
2016-09-11 22:08 ` [HoTT] " Peter LeFanu Lumsdaine
2016-09-11 22:26 ` Martin Escardo
2016-09-12 9:11 ` Thomas Streicher
2016-09-12 5:20 ` Andrew Polonsky
2016-09-12 5:59 ` [HoTT] " Jason Gross
2016-09-12 9:55 ` Andrew Polonsky
2016-09-12 10:07 ` Andrew Polonsky
2016-09-12 10:35 ` Nicolai Kraus
2016-09-12 10:16 ` Peter LeFanu Lumsdaine
2016-09-12 10:44 ` [HoTT] " Nicolai Kraus
2016-09-12 11:02 ` Andrej Bauer
2016-09-12 11:14 ` Thomas Streicher
2016-09-12 11:23 ` Andrew Polonsky
2016-09-12 11:41 ` Thomas Streicher [this message]
2016-09-12 12:47 ` Thomas Streicher
2016-09-12 13:01 ` Martin Escardo
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