open import Cat.Morphism.Strong.Epi
open import Cat.Functor.Properties
open import Cat.Morphism.Duality
open import Cat.Functor.Adjoint
open import Cat.Prelude

import Cat.Functor.Reasoning
import Cat.Reasoning

module Cat.Functor.Morphism
  {o ℓ o' ℓ'}
  {𝒞 : Precategory o ℓ} {𝒟 : Precategory o' ℓ'}
  (F : Functor 𝒞 𝒟)
  where

private
  module 𝒞 = Cat.Reasoning 𝒞
  module 𝒟 = Cat.Reasoning 𝒟
open Cat.Functor.Reasoning F

private variable
  A B C : 𝒞.Ob
  a b c d : 𝒞.Hom A B
  X Y Z : 𝒟.Ob
  f g h i : 𝒟.Hom X Y

Actions of functors on morphisms🔗

This module describes how various classes of functors act on designated collections of morphisms.

First, some general definitions. Let $H$ be a collection of morphisms in $\mathcal{C}\text{.}$ A functor $F:\mathcal{C}\to \mathcal{D}$ preserves $H\text{-morphisms}$ if $f\in H$ implies that $F(f)\in H\text{.}$

preserves-monos : Type _
preserves-monos =
  ∀ {a b : 𝒞.Ob} {f : 𝒞.Hom a b} → 𝒞.is-monic f → 𝒟.is-monic (F₁ f)

preserves-epis : Type _
preserves-epis =
  ∀ {a b : 𝒞.Ob} {f : 𝒞.Hom a b} → 𝒞.is-epic f → 𝒟.is-epic (F₁ f)

preserves-strong-epis : Type _
preserves-strong-epis =
  ∀ {a b : 𝒞.Ob} {f : 𝒞.Hom a b} → is-strong-epi 𝒞 f → is-strong-epi 𝒟 (F₁ f)

Likewise, a functor $F:\mathcal{C}\to \mathcal{D}$ reflects $H\text{-morphisms}$ if $F(f)\in H$ implies that $f\in H\text{.}$

reflects-monos : Type _
reflects-monos =
  ∀ {a b : 𝒞.Ob} {f : 𝒞.Hom a b} → 𝒟.is-monic (F₁ f) → 𝒞.is-monic f

reflects-epis : Type _
reflects-epis =
  ∀ {a b : 𝒞.Ob} {f : 𝒞.Hom a b} → 𝒟.is-epic (F₁ f) → 𝒞.is-epic f

Functors that reflect invertible morphisms are called conservative, and are notable enough to deserve their own name and page!

Faithful functors🔗

Faithful functors reflect monomorphisms and epimorphisms. We will only comment on the proof regarding monomorphisms, since the argument for epimorphisms is formally dual. Let $F(a)$ be monic in $\mathcal{D}\text{,}$ and let $b,c$ be a pair of morphisms in $\mathcal{C}$ such that $a\circ b=a\circ c\text{.}$ Because $F$ preserves all commutative diagrams, $F(a)\circ F(b)=F(a)\circ F(c)\text{.}$ $F(a)$ is monic, so $F(b)=F(c)\text{.}$ Finally, $F$ is faithful, so we can deduce $b=c\text{.}$

module _ (faithful : is-faithful F) where
  faithful→reflects-mono : 𝒟.is-monic (F₁ a) → 𝒞.is-monic a
  faithful→reflects-mono {a = a} F[a]-monic b c p =
    faithful (F[a]-monic (F₁ b) (F₁ c) (weave p))

  faithful→reflects-epi : 𝒟.is-epic (F₁ a) → 𝒞.is-epic a
  faithful→reflects-epi {a = a} F[a]-epic b c p =
    faithful (F[a]-epic (F₁ b) (F₁ c) (weave p))

Likewise, faithful functors reflect all diagrams: this means that if $F(a)$ and $F(b)$ either form a section/retraction pair or an isomorphism, then it must have been the case that $a$ and $b$ already did.

  faithful→reflects-section-of : (F₁ a) 𝒟.section-of (F₁ b) → a 𝒞.section-of b
  faithful→reflects-section-of p = faithful (F-∘ _ _ ∙ p ∙ sym F-id)

  faithful→reflects-retract-of : (F₁ a) 𝒟.retract-of (F₁ b) → a 𝒞.retract-of b
  faithful→reflects-retract-of p = faithful→reflects-section-of p

  faithful→reflects-inverses : 𝒟.Inverses (F₁ a) (F₁ b) → 𝒞.Inverses a b
  faithful→reflects-inverses ab-inv .𝒞.Inverses.invl =
    faithful→reflects-section-of (𝒟.Inverses.invl ab-inv)
  faithful→reflects-inverses ab-inv .𝒞.Inverses.invr =
    faithful→reflects-section-of (𝒟.Inverses.invr ab-inv)

Fully faithful, essentially surjective functors🔗

If a functor $F$ is fully faithful and essentially surjective, then it preserves all mono- and epimorphisms. Keep in mind that, since we have not assumed that the categories involved are univalent, this condition is slightly weaker than being an equivalence of categories.

Let $a:\mathcal{C}(A,B)$ be a mono, and let $f,g:\mathcal{D}(X,F(A))$ be a pair of morphisms in $\mathcal{D}\text{,}$ satisfying that $F(a)\circ f=F(a)\circ g\text{.}$ Since $F$ is eso, there merely exists a $C:\mathcal{C}$ with $i:F(C)\cong X\text{.}$ Because $F$ is also full, there must merely exist a pair of morphisms $f^{\prime },g^{\prime }:\mathcal{C}(C,A)\text{,}$ satisfying $F(f^{\prime })=f\circ i\text{,}$ and $F(g^{\prime })=g\circ i\text{.}$

module _ (ff : is-fully-faithful F) (eso : is-eso F) where
  ff+eso→preserves-mono : 𝒞.is-monic a → 𝒟.is-monic (F₁ a)
  ff+eso→preserves-mono {a = a} a-monic {x} f g p = ∥-∥-out! do
    (x* , i) ← eso x
    (f* , q) ← ff→full {F = F} ff (f 𝒟.∘ 𝒟.to i)
    (g* , r) ← ff→full {F = F} ff (g 𝒟.∘ 𝒟.to i)

Next, note that $a\circ f^{\prime }=a\circ g^{\prime }\text{:}$ this follows from faithfulness of $F\text{,}$ and our hypothesis that $F(a)\circ f=F(a)\circ g\text{.}$

    let
      s =
        ff→faithful {F = F} ff $
        F₁ (a 𝒞.∘ f*)           ≡⟨ F-∘ _ _ ∙ 𝒟.pushr q ⟩≡
        (F₁ a 𝒟.∘ f) 𝒟.∘ 𝒟.to i ≡⟨ ap₂ 𝒟._∘_ p refl ⟩≡
        (F₁ a 𝒟.∘ g) 𝒟.∘ 𝒟.to i ≡⟨ 𝒟.pullr (sym r) ∙ sym (F-∘ _ _) ⟩≡
        F₁ (a 𝒞.∘ g*)           ∎

To wrap things up, recall that $a$ is monic, so $f^{\prime }=g^{\prime }\text{,}$ and $F(f^{\prime })=F(g^{\prime })\text{.}$ However, $F(f^{\prime })=f\circ i$ and $F(g^{\prime })=g\circ i$ by definition, so we can deduce that $f\circ i=g\circ i\text{.}$ Finally, isomorphisms are epic, so we can cancel $i$ on the left, concluding that $f=g\text{.}$

    pure $ 𝒟.iso→epic i f g $
      f 𝒟.∘ 𝒟.to i ≡˘⟨ q ⟩≡˘
      F₁ f*        ≡⟨ ap F₁ (a-monic f* g* s) ⟩≡
      F₁ g*        ≡⟨ r ⟩≡
      g 𝒟.∘ 𝒟.to i ∎

  ff+eso→preserves-epi : 𝒞.is-epic a → 𝒟.is-epic (F₁ a)
  ff+eso→preserves-epi {a = a} a-epic {x} f g p = ∥-∥-out! do
    (x* , i) ← eso x
    (f* , q) ← ff→full {F = F} ff (𝒟.from i 𝒟.∘ f)
    (g* , r) ← ff→full {F = F} ff (𝒟.from i 𝒟.∘ g)
    let s = F-∘ _ _ ∙ 𝒟.pushl q ∙ ap₂ 𝒟._∘_ refl p ∙ 𝒟.pulll (sym r) ∙ sym (F-∘ _ _)
    pure $ 𝒟.iso→monic (i 𝒟.Iso⁻¹) f g $
      sym q
      ·· ap F₁ (a-epic f* g* (ff→faithful {F = F} ff s))
      ·· r

Left and right adjoints🔗

If we are given an adjunction $L⊣F\text{,}$ then the right adjoint $F$ preserves monomorphisms. Fix a mono $a:\mathcal{C}(A,B)\text{,}$ and let $f,g:\mathcal{D}(X,FA)$ satisfy $F(a)f=F(a)g\text{.}$ We want to show $f=g\text{,}$ and, by the adjunction, it will suffice to show that $\epsilon L(f)=\epsilon L(g)\text{.}$ Since $a$ is a monomorphism, we can again reduce this to showing

which follows by a quick calculation.

module _ {L : Functor 𝒟 𝒞} (L⊣F : L ⊣ F) where
  private
    module L = Cat.Functor.Reasoning L
  open _⊣_ L⊣F

  right-adjoint→preserves-monos : 𝒞.is-monic a → 𝒟.is-monic (F₁ a)
  right-adjoint→preserves-monos {a = a} a-monic f g p =
    R-adjunct.injective L⊣F $
    a-monic _ _ $
    a 𝒞.∘ ε _ 𝒞.∘ L.₁ f            ≡⟨ 𝒞.pulll (sym (counit.is-natural _ _ _)) ⟩≡
    (ε _ 𝒞.∘ L.₁ (F₁ a)) 𝒞.∘ L.₁ f ≡⟨ L.extendr p ⟩≡
    (ε _ 𝒞.∘ L.₁ (F₁ a)) 𝒞.∘ L.₁ g ≡⟨ 𝒞.pushl (counit.is-natural _ _ _) ⟩≡
    a 𝒞.∘ ε _ 𝒞.∘ L.₁ g            ∎

module _ {R : Functor 𝒟 𝒞} (F⊣R : F ⊣ R) where
  private
    module R = Cat.Functor.Reasoning R
  open _⊣_ F⊣R

  left-adjoint→preserves-epis : 𝒞.is-epic a → 𝒟.is-epic (F₁ a)
  left-adjoint→preserves-epis {a = a} a-epic f g p =
    L-adjunct.injective F⊣R $
    a-epic _ _ $
    𝒞.pullr (unit.is-natural _ _ _)
    ∙ R.extendl p
    ∙ 𝒞.pushr (sym (unit.is-natural _ _ _))