Re: [HoTT] Definitions of equivalence satisfying judgmental/strict groupoid laws?

["Martín Hötzel Escardó" <escardo.martin@gmail.com>]

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On 17/08/2019 01:14, Jason Gross wrote:
> Resurrecting this thread from many years ago, because I was thinking
> about it again recently, it seems to me that although { f &
> isTrackedByRelEquiv(f) } satisfies the rule sym (sym e) = e
> judgmentally, it doesn't satisfy the rule that sym id = id
> judgmentally.  (In particular, I am not sure what relational equivalence
> to use for the identity equivalence which does not change judgmentally
> when I flip the order of its arguments.)  Is there a version of
> equivalence which simultaneously satisfies that the inverse of the
> identity is judgmentally the identity, and that inverting an equivalence
> twice judgmentally gives you what you started with?
>
> -Jason


I am not sure this answer will be of the kind you are expecting.

First I will consider the identity type and then the equivalence type
using the same idea.

I will use "=" for definitional equality.

(1) If you have some identity type (_≡_, refl , J) with

    _≡_ : {X : 𝓤} → X → X → 𝓤

    refl : {X : 𝓤 } (x : X) → x ≣ x

    J : (X : 𝓤) (A : (x y : X) → x ≡ y → 𝓤)
     → ((x : X) → A x x (refl x))
     → (x y : X) (p : x ≡ y) → A x y p

then you can define another identity type (_≡'_ , refl' , J') by


    x ≡' y = Σ (z : X), (z ≡ x) × (z ≡ y)

(which is equivalent to the diagonal fiber of (x,y) and also to x ≡ y
(both in pure MLTT))


    refl' x = x , refl x , refl x


    J' X A f x y (z , p , q) = γ z x p y q
     where
      φ :  (x y : X) (q : x ≡ y) → A x y (x , refl x , q)
      φ = J X (λ x y q → A x y (x , refl x , q)) f

      γ : (z x : X) (p : z ≡ x) (y : X) (q : z ≡ y) → A x y (z , p , q)
      γ = J X (λ z x p → (y : X) (q : z ≡ y) → A x y (z , p , q)) φ

so that

    J' X A f x x (refl' x) = f x

definitionally using the original J's computation rule.


For this new identity type, we can define

    ≡'-sym : {X : 𝓤 ̇ } {x y : X} → x ≡' y → y ≡' x
    ≡'-sym (a , p , q) = (a , q , p)

so that

   ≡'-sym (refl' x) = refl' x

and

  ≡'-sym (≡'-sym p') = p',

both definitionally.

(2) Similarly we can define an equivalence type _≃'_ from an equivalence 
type _≃_ by

    X ≃' Y = Σ (Z : 𝓤), (Z ≃ X) × (Z ≃ X)

with the analogous identity equivalence and symmetrization operation
so that the definitional equalities you want hold.

This is related to the relational definition of equivalence as
follows. The type

   X × Y → 𝓤

of type-valued relations is in canonical bijection with the "slice type"

  Σ (Z : 𝓤), Z → X × Y

by a well-known construction, and in turn to

  Σ (Z : 𝓤), (Z → X) × (Z → Y).

When you restrict to relations R such that for all x:X and y:Y the types

  Σ (x : X), R x y

and

  Σ (y : Y), R x y

are contractible, the type Σ (Z : 𝓤), (Z → X) × (Z → Y) gets restricted to

  Σ (Z : 𝓤), (Z ≃ X) × (Z ≃ X)

by the canonical equivalence

 (X × Y → 𝓤) ≃ Σ (Z : 𝓤), (Z → X) × (Z → Y).

(3) If you want composition of identifications to be definitionally
associative with refl definitionally neutral on both sides, you can
consider the alternative identity type defined by

   x ≡'' y  = (z : X) → (z ≡ x) → (z ≡ y)

which is again equivalent to the original identity type x ≡ y (which
amounts to Yoneda),

   refl'' x = (z : X) → λ (p : x ≡ z) , p

because composition of identifications is given by function
composition, which is definitionally associative with the identity
function as its definitionally neutral element (assuming the η rule).

  (Exercise: write down J''.)

(4) I don't know how to get the definitional equalities of (1) and (3)
together by a suitable modification of the identity type. The
constructions (1) and (3) are kind of dual.

Martin

On Thu, Nov 13, 2014 at 12:59 PM Vladimir Voevodsky <vlad...@ias.edu 
> <javascript:>> wrote:
>
>> In general no. But their model is essentially syntactic and more or less 
>> complete. Or, to be more precise, I would expect it to 
>> be more or less complete. 
>>
>> V.
>>
>>
>> On Nov 13, 2014, at 9:55 PM, Peter LeFanu Lumsdaine <p.l.l...@gmail.com 
>> <javascript:>> wrote:
>>
>> On Thu, Nov 13, 2014 at 12:04 PM, Vladimir Voevodsky <vlad...@ias.edu 
>> <javascript:>> wrote:
>>
>>> The question is about how you interpret this operation for the univalent 
>>> universe. If there is an interpretation of such an operation then there is 
>>> a way to define equivalences between types in an involutionary way.
>>>
>>
>> I don’t follow why this should be the case.  This shows that there is 
>> some notion of equivalence *in the model* (i.e. constructed in the 
>> meta-theory) which is strictly involutive.  But there is no reason that 
>> this notion need be definable in the syntax of the object theory, is there?
>>
>> –p. 
>>
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