open import Cat.Functor.Adjoint.Properties
open import Cat.Functor.Adjoint.Monadic
open import Cat.Functor.Equivalence
open import Cat.Functor.Properties
open import Cat.Instances.Functor
open import Cat.Displayed.Total
open import Cat.Functor.Adjoint
open import Cat.Diagram.Monad
open import Cat.Prelude

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

module Cat.Functor.Adjoint.Reflective where

private variable
  o o' ℓ ℓ' : Level
  C D : Precategory o ℓ
  F G : Functor C D
open Functor
open _=>_
open ∫Hom

Reflective subcategories🔗

Occasionally, full subcategory inclusions (hence fully faithful functors — like the inclusion of abelian groups into the category of all groups, or the inclusion $\mathbf{Props}\hookrightarrow \mathbf{Sets}\text{)}$ participate in an adjunction

is-reflective : F ⊣ G → Type _
is-reflective {G = G} adj = is-fully-faithful G

The first thing we will prove is that the counit map $\epsilon :FGo\to o$ of a reflexive subcategory inclusion is invertible. Luckily, we have developed enough general theory to make this almost immediate:

• $G$ is full, so the counit must be a split monomorphism.
• $G$ is faithful, so the counit must be a epimorphism.
• Every morphism that is simultaneously split monic and epic is invertible.

module
  _ {C : Precategory o ℓ} {D : Precategory o' ℓ'} {F : Functor C D} {G : Functor D C}
    (adj : F ⊣ G) (g-ff : is-reflective adj)
  where
  private
    module DD = Cat.Reasoning Cat[ D , D ]
    module C = Cat.Reasoning C
    module D = Cat.Reasoning D
    module F = Func F
    module G = Func G
    module GF = Func (G F∘ F)
    module FG = Func (F F∘ G)
    module g-ff {x} {y} = Equiv (_ , g-ff {x} {y})
  open _⊣_ adj

  is-reflective→counit-is-invertible : ∀ {o} → D.is-invertible (ε o)
  is-reflective→counit-is-invertible {o} =
    D.split-monic+epic→invertible
      (right-full→counit-split-monic adj (ff→full {F = G} g-ff))
      (right-faithful→counit-epic adj (ff→faithful {F = G} g-ff))

  is-reflective→counit-is-iso : ∀ {o} → FG.₀ o D.≅ o
  is-reflective→counit-is-iso {o} = morp where
    morp : F.₀ (G.₀ o) D.≅ o
    morp =
      D.invertible→iso (ε _) $
      D.split-monic+epic→invertible
        (right-full→counit-split-monic adj (ff→full {F = G} g-ff))
        (right-faithful→counit-epic adj (ff→faithful {F = G} g-ff))

  is-reflective→counit-iso : (F F∘ G) ≅ⁿ Id
  is-reflective→counit-iso = DD.invertible→iso counit invs where
    invs = invertible→invertibleⁿ counit λ x →
      is-reflective→counit-is-invertible

  η-comonad-commute : ∀ {x} → unit.η (G.₀ (F.₀ x)) ≡ G.₁ (F.₁ (unit.η x))
  η-comonad-commute {x} = C.right-inv-unique
    (F-map-iso G is-reflective→counit-is-iso)
    zag
    (sym (G.F-∘ _ _) ∙ ap G.₁ zig ∙ G.F-id)

  is-reflective→unit-G-is-iso : ∀ {o} → C.is-invertible (unit.η (G.₀ o))
  is-reflective→unit-G-is-iso {o} = C.make-invertible (g-ff.to (ε _))
    (unit.is-natural _ _ _ ∙∙ ap₂ C._∘_ refl η-comonad-commute ∙∙ GF.annihilate zag)
    zag

  is-reflective→F-unit-is-iso : ∀ {o} → D.is-invertible (F.₁ (unit.η o))
  is-reflective→F-unit-is-iso {o} = D.make-invertible
    (ε _)
    (sym (counit.is-natural _ _ _) ∙ ap₂ D._∘_ refl (ap F.₁ (sym η-comonad-commute)) ∙ zig)
    zig

We can now prove that the adjunction $L⊣\iota$ is monadic.

is-reflective→is-monadic
  : ∀ {F : Functor C D} {G : Functor D C}
  → (adj : F ⊣ G) → is-reflective adj → is-monadic adj
is-reflective→is-monadic {C = C} {D = D} {F = F} {G} adj g-ff = eqv where

  module EM = Cat.Reasoning (Eilenberg-Moore (R∘L adj))
  module C = Cat.Reasoning C
  module D = Cat.Reasoning D
  module F = Functor F
  module G = Functor G
  open Algebra-on
  open _⊣_ adj

  Comp : Functor D (Eilenberg-Moore (R∘L adj))
  Comp = Comparison-EM adj
  module Comp = Functor Comp

It suffices to show that the comparison functor $D\to C^{G}F$ is fully faithful and split essentially surjective. For full faithfulness, observe that it’s always faithful; The fullness comes from the assumption that $G$ is ff.

  comp-ff : is-fully-faithful Comp
  comp-ff {x} {y} = is-iso→is-equiv λ where
    .is-iso.from alg → equiv→inverse g-ff (alg .fst)
    .is-iso.rinv x → ext (equiv→counit g-ff _)
    .is-iso.linv x → equiv→unit g-ff _

To show that the comparison functor is split essentially surjective, suppose we have an object $o$ admitting the structure of an $GF\text{-algebra;}$ We will show that $o\cong GFo$ as $GF\text{-algebras}$ — note that $GF(o)$ admits a canonical (free) algebra structure. The algebra map $\nu :GF(o)\to o$ provides an algebra morphism from $GF(o)\to o\text{,}$ and the morphism $o\to GF(o)$ is can be taken to be adjunction unit $\eta \text{.}$

The crucial lemma in establishing that these are inverses is that ${\eta }_{GFx}=GF({\eta }_x)\text{,}$ which follows because both of those morphisms are right inverses to $G{\epsilon }_x\text{,}$ which is an isomorphism because $\epsilon$ is.

  comp-seso : is-split-eso Comp
  comp-seso (ob , alg) = F.₀ ob , isom where
    Fo→o : Algebra-hom (R∘L adj) (Comp.₀ (F.₀ ob)) (ob , alg)
    Fo→o .fst = alg .ν
    Fo→o .snd = alg .ν-mult

    o→Fo : Algebra-hom (R∘L adj) (ob , alg) (Comp.₀ (F.₀ ob))
    o→Fo .fst = unit.η _
    o→Fo .snd =
        unit.is-natural _ _ _
      ∙ ap₂ C._∘_ refl (η-comonad-commute adj g-ff)
      ∙ sym (G.F-∘ _ _)
      ∙ ap G.₁ (sym (F.F-∘ _ _) ∙∙ ap F.₁ (alg .ν-unit) ∙∙ F.F-id)
      ∙ sym (ap₂ C._∘_ refl (sym (η-comonad-commute adj g-ff)) ∙ zag ∙ sym G.F-id)

    isom : Comp.₀ (F.₀ ob) EM.≅ (ob , alg)
    isom = EM.make-iso Fo→o o→Fo
      (ext (alg .ν-unit))
      (ext (
          unit.is-natural _ _ _
        ∙∙ ap₂ C._∘_ refl (η-comonad-commute adj g-ff)
        ∙∙ sym (G.F-∘ _ _)
        ∙∙ ap G.₁ (sym (F.F-∘ _ _) ∙∙ ap F.₁ (alg .ν-unit) ∙∙ F.F-id)
        ∙∙ G.F-id))

  eqv : is-equivalence Comp
  eqv = ff+split-eso→is-equivalence comp-ff comp-seso

Constructing reflective subcategories🔗

Earlier, we saw that any reflective subcategory has an invertible counit. We will now prove the converse: if the counit of an adjunction is invertible, then the left adjoint is a reflector.

module _
  {C : Precategory o ℓ} {D : Precategory o' ℓ'}
  {F : Functor C D} {G : Functor D C}
  (adj : F ⊣ G)
  where
  private
    module C = Cat.Reasoning C
    module D = Cat.Reasoning D
    module [D,D] = Cat.Reasoning Cat[ D , D ]
    module F = Func F
    module G = Func G
    module GF = Func (G F∘ F)
    module FG = Func (F F∘ G)
    open _⊣_ adj

Again, the sea has risen to meet us:

• A right adjoint is faithful if and only if the counit is epic.
• A right adjoint full if and only if the counit is split monic.

If the counit is invertible, then it is clearly both split monic and epic, and thus the corresponding right adjoint must be fully faithful.

  is-counit-iso→is-reflective : is-invertibleⁿ counit → is-reflective adj
  is-counit-iso→is-reflective counit-iso =
    full+faithful→ff G
      G-full
      G-faithful
    where
      G-full : is-full G
      G-full =
        counit-split-monic→right-full adj $
        D.invertible→to-split-monic $
        is-invertibleⁿ→is-invertible counit-iso _

      G-faithful : is-faithful G
      G-faithful =
        counit-epic→right-faithful adj $
        D.invertible→epic $
        is-invertibleⁿ→is-invertible counit-iso _

Furthermore, if we have any natural isomorphism $\alpha :FG\cong \mathrm{Id}\text{,}$ then the left adjoint is a reflector! To show this, we will construct an inverse to the counit; our previous result will then ensure that $F$ is fully faithful.

To begin, recall that isos have the 2-out-of-3 property, so it suffices to show that $\epsilon \circ \alpha$ is invertible. Next, note that we can transfer the comonad structure on $FG$ onto a comonad structure on $\mathrm{Id}$ by repeatedly composing with $\alpha \text{;}$ this yields a natural transformation $\delta :\mathrm{Id}\to \mathrm{Id}$ that is a right inverse to $\epsilon \circ \alpha \text{.}$

Finally, all natural transformations $\mathrm{Id}\to \mathrm{Id}$ commute with one another, so $\delta$ is also a right inverse, and $\epsilon \circ \alpha$ is invertible.

  FG-iso→is-reflective : (F F∘ G) ≅ⁿ Id → is-reflective adj
  FG-iso→is-reflective α =
    is-counit-iso→is-reflective $
    [D,D].invertible-cancell
      ([D,D].iso→invertible (α [D,D].Iso⁻¹))
      ([D,D].make-invertible δ right-ident right-ident⁻¹)
    where
      module α = Isoⁿ α

      δ : Id {C = D} => Id
      δ .η x = α.to .η x D.∘ α.to .η (F.F₀ (G.₀ x)) D.∘ F.₁ (unit.η (G.₀ x)) D.∘ α.from .η x
      δ .is-natural x y f =
          D.extendr (D.extendr (D.extendr (α.from .is-natural _ _ _)))
        ∙ D.pushl (D.pushr (D.pushr (F.weave (unit .is-natural _ _ _))))
        ∙ D.pushl (D.pushr (α.to .is-natural _ _ _))
        ∙ D.pushl (α.to .is-natural _ _ _)

      right-ident : (counit ∘nt α.from) ∘nt δ ≡ idnt
      right-ident = ext λ x →
          D.cancel-inner (α.invr ηₚ _)
        ∙ D.pulll (sym $ α.to .is-natural _ _ _)
        ∙ D.cancel-inner (F.annihilate zag)
        ∙ α.invl ηₚ _

      right-ident⁻¹ : δ ∘nt (counit ∘nt α.from) ≡ idnt
      right-ident⁻¹ = id-nat-commute δ (counit ∘nt α.from) ∙ right-ident