From mboxrd@z Thu Jan 1 00:00:00 1970 Date: Fri, 13 Oct 2017 07:12:27 -0700 (PDT) From: Robin Adams To: Homotopy Type Theory Message-Id: In-Reply-To: References: Subject: Re: A small observation on cumulativity and the failure of initiality MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="----=_Part_10493_1481627462.1507903948045" ------=_Part_10493_1481627462.1507903948045 Content-Type: multipart/alternative; boundary="----=_Part_10494_89993186.1507903948045" ------=_Part_10494_89993186.1507903948045 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 7bit On Thursday, 12 October 2017 20:43:01 UTC+2, Dimitris Tsementzis wrote: > > > But there are two distinct TT-model homomorphisms from C_TT to C_TT*, one > which sends p(t0) to pq(t0) and one which sends p(t0) to qp(t0) (where > p(t0) is regarded as an element of Tm_{C_TT} (empty, B(B(T0))), i.e. of the > set of terms of B(B(T0)) in the empty context as they are interpreted in > the term model C_TT). > There seems to be a gap in the proof here. In a term model we quotient out by judgemental equality (correct me if this is wrong), so this step does not give a contradiction: rather, we conclude |- qp(t0) = pq(t0) : B(B(T0)) I would expect this equality to hold in the examples you have in mind. If T = Type_0, B(T) = Type_1, and p(t) = t -> t, then the equality is q(t -> t) = q(t) -> q(t). This holds if TT* includes the rule G |- A : Type_0 G |- B : Type_0 ------------------------------------------------------- G |- q(A -> B) = q(A) -> q(B) : Type_1 which I would expect you need in order for the category of TT-models to be isomorphic to the category of TT*-models. So I suggest you also check the proof of this isomorphism in more detail. -- Robin ------=_Part_10494_89993186.1507903948045 Content-Type: text/html; charset=utf-8 Content-Transfer-Encoding: quoted-printable


On Thursday, 12 October 2017 20:43:01 UTC+2, Dimit= ris Tsementzis wrote:

But there are two distinct TT-model ho= momorphisms from C_TT to C_TT*, one which sends p(t0) to pq(t0) and one whi= ch sends p(t0) to qp(t0) (where p(t0) is regarded as an element of Tm_{C_TT= } (empty, B(B(T0))), i.e. of the set of terms of B(B(T0)) in the empty cont= ext as they are interpreted in the term model C_TT).=C2=A0

There seems to be a gap in the proof here.=C2=A0 In a term mode= l we quotient out by judgemental equality (correct me if this is wrong), so= this step does not give a contradiction: rather, we conclude |- qp(t0) =3D= pq(t0) : B(B(T0))

I would expect this equality to hold in the examp= les you have in mind.=C2=A0 If T =3D Type_0, B(T) =3D Type_1, and p(t) =3D = t -> t, then the equality is q(t -> t) =3D q(t) -> q(t).=C2=A0 Thi= s holds if TT* includes the rule

G |- A : Type_0=C2=A0=C2=A0=C2=A0= =C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0=C2=A0 G |- B : Type_0
-------------= ------------------------------------------
G |- q(A -> B) =3D q(A) -&= gt; q(B) : Type_1

which I would expect you need in order for the cat= egory of TT-models to be isomorphic to the category of TT*-models.=C2=A0 So= I suggest you also check the proof of this isomorphism in more detail.
=
--
Robin

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