open import Cat.Instances.Shape.Interval
open import Cat.Functor.Base
open import Cat.Univalent
open import Cat.Prelude

import Cat.Functor.Reasoning as Func
import Cat.Reasoning as Cat

open Precategory

module Cat.Functor.Amnestic where

private variable
  o ℓ o' ℓ' : Level
  C D : Precategory o ℓ

Amnestic functors🔗

The notion of amnestic functor was introduced by Adámek, Herrlich and Strecker in their book “The Joy of Cats”¹ as a way to make precise the vibes-based notion of forgetful functor. Classically, the definition reads

A functor $F:\mathcal{C}\to \mathcal{D}$ is amnestic if an isomorphism $f$ is an identity whenever $Ff$ is,

but what does it mean for an isomorphism $f:a\cong b$ to be an identity? The obvious translation would be that $f$ comes from an identification $a\equiv b$ of objects, but this is just rephrasing the condition that $\mathcal{C}$ is univalent, whereas we want a condition on the functor $F\text{.}$ So, we define:

An isomorphism $f:a\cong b$ is an identity if it is an identity in the total space of Hom, i.e. if there is an object $c:\mathcal{C}$ s.t. $(c,c,\mathrm{id})=(a,b,f)$ in the type ${\Sigma }_a{\Sigma }_b(a\to b)\text{.}$ Every isomorphism in a univalent category is an identity, since we can take $c=a\text{,}$ and the identification in Arrows follows from univalence.

module _ (F : Functor C D) where
  private
    module C = Cat C
    module D = Cat D
    module F = Func F

Let $f$ be an identity, i.e. it is such that $(a,b,f)\cong (c,c,\mathrm{id})\text{.}$ Any functor $F$ induces an identification $(Fa,Fb,Ff)\cong (Fc,Fc,id)\text{,}$ meaning that it preserves being an identity. A functor is amnestic if it additionally reflects being an identity: the natural map we have implicitly defined above (called action) is an equivalence.

  action : ∀ {a b : C.Ob} (f : C.Hom a b)
         → Σ[ c ∈ C ] (Hom→Arrow C (C.id {x = c}) ≡ Hom→Arrow C f)
         → Σ[ c ∈ D ] (Hom→Arrow D (D.id {x = c}) ≡ Hom→Arrow D (F.₁ f))
  action f (c , p) = F.₀ c , q where
    q : Hom→Arrow D D.id ≡ Hom→Arrow D (F.₁ f)
    q i .fst = F.₀ (p i .fst)
    q i .snd .fst = F.₀ (p i .snd .fst)
    q i .snd .snd = hcomp (∂ i) λ where
      j (j = i0) → F.₁ (p i .snd .snd)
      j (i = i1) → F.₁ f
      j (i = i0) → F.F-id j

  is-amnestic : Type _
  is-amnestic = ∀ {a b : C.Ob} (f : C.Hom a b)
              → C.is-invertible f → is-equiv (action f)

Who cares?🔗

The amnestic functors are interesting to consider in HoTT because they are exactly those that reflect univalence: If $F:\mathcal{C}\to \mathcal{D}$ is an amnestic functor and $\mathcal{D}$ is a univalent category, then $\mathcal{C}$ is univalent, too!

  reflect-category : is-category D → is-amnestic → is-category C
  reflect-category d-cat forget = record { to-path = A≡B ; to-path-over = q } where
    module _ {A} {B} isom where
      isom' = F-map-iso F isom

For suppose that $i:a\cong b\in \mathcal{C}$ is an isomorphism. $F$ preserves this isomorphism, meaning we have an iso $Fi:Fa\cong Fb\text{,}$ now in $\mathcal{D}\text{.}$ But because $\mathcal{D}$ is a univalent category, $Fi$ is an identity, and by $F\text{’s}$ amnesia, so is $i\text{.}$

      p : Σ[ c ∈ C ] Path (Arrows C) (c , c , C.id) (A , B , isom .C.to)
      p = equiv→inverse (forget (isom .C.to) (C.iso→invertible isom)) $ F.₀ A ,
        Arrows-path D refl (d-cat .to-path isom')
          (Univalent.Hom-pathp-reflr-iso d-cat (D.idr _))

Unfolding, we have an object $x:\mathcal{C}$ and an identification $p:(x,x,\mathrm{id})\equiv (a,b,i)\text{.}$ We may construct an identification $p^{\prime }:a\cong b$ from the components of $p\text{,}$ and this induces an identification $\mathrm{id}\equiv i$ over $p^{\prime }$ in a straightforward way. We’re done!

      A≡B = sym (ap fst (p .snd)) ∙ ap (fst ⊙ snd) (p .snd)

      q : PathP (λ i → A C.≅ A≡B i) C.id-iso isom
      q = C.≅-pathp refl A≡B $ Hom-pathp-reflr C $
           C.idr _
        ·· path→to-∙ C _ _
        ·· ap₂ C._∘_ refl (sym (path→to-sym C (ap fst (p .snd))))
         ∙ Hom-pathp-id C (ap (snd ⊙ snd) (p .snd))