open import Cat.Diagram.Coequaliser.RegularEpi
open import Cat.Diagram.Coequaliser
open import Cat.Morphism.Strong.Epi
open import Cat.Prelude

open import Homotopy.Connectedness

import Cat.Reasoning as Cr

module Data.Set.Surjection where

open is-coequaliser
open is-regular-epi
open Coequaliser

Surjections between sets🔗

Here we prove that surjective maps are exactly the regular epimorphisms in the category of sets: Really, we prove that surjections are regular epimorphisms (this is straightforward), then we prove that every epimorphism is a surjection (this takes a detour). Putting these together, we get every epimorphism is regular as a corollary.

Surjections are epic🔗

This is the straightforward direction: We know (since $\mathbf{Sets}$ has pullbacks) that if a morphism is going to be a regular epimorphism, then it must be an effective epimorphism, too: so if it is to coequalise any morphisms, it certainly coequalises its kernel pair.

surjective→regular-epi
  : ∀ {ℓ} (c d : n-Type ℓ ) (f : ∣ c ∣ → ∣ d ∣)
  → is-surjective f
  → is-regular-epi (Sets ℓ) {c} {d} f
surjective→regular-epi c _ f x .r = el! (Σ ∣ c ∣ λ x → Σ ∣ c ∣ λ y → f x ≡ f y)
surjective→regular-epi _ _ f x .arr₁ = λ (y , _ , _) → y
surjective→regular-epi _ _ f x .arr₂ = λ (_ , y , _) → y

That’s our choice of cofork: it remains to show that $f$ coequalises this. Given any map $e^{\prime }:c\to F\text{,}$ element $x:d\text{,}$ mere fibre $f^{\ast }(x)\text{,}$ and proof $p$ that $e^{\prime }$ is constant on the kernel pair of $f$ (i.e. that it also coequalises the kernel pair cofork), we can construct an element of $F\text{:}$ since $e^{\prime }$ is suitably constant, we can use the elimination principle for $\Vert f^{\ast }x\Vert \to F\text{,}$ since $F$ is a set.

surjective→regular-epi c d f surj .has-is-coeq = coeqs where
  go : ∀ {F} (e' : ∣ c ∣ → ∣ F ∣) p (x : ∣ d ∣) → ∥ fibre f x ∥ → ∣ F ∣
  go e' p x =
    ∥-∥-rec-set (hlevel ) (λ x → e' (x .fst))
      (λ x y → p $ₚ (x .fst , y .fst , x .snd ∙ sym (y .snd)))

After a small amount of computation to move the witnesses of surjectivity out of the way, we get what we wanted.

  coeqs : is-coequaliser (Sets _) _ _ _
  coeqs .coequal i (x , y , p) = p i
  coeqs .universal {F} {e'} p x = go {F = F} e' p x (surj x)
  coeqs .factors {F} {e'} {p = p} = funext λ x →
    ∥-∥-elim {P = λ e → go {F} e' p (f x) e ≡ e' x}
      (λ x → hlevel ) (λ e → p $ₚ (e .fst , x , e .snd)) (surj (f x))
  coeqs .unique {F} {e'} {p} {colim} comm = funext λ a →
    ∥-∥-elim {P = λ e → colim a ≡ go {F} e' p a e} (λ x → hlevel )
      (λ x → ap colim (sym (x .snd)) ∙ comm $ₚ x .fst)
      (surj a)

surjective→strong-epi
  : ∀ {ℓ} (c d : n-Type ℓ ) (f : ∣ c ∣ → ∣ d ∣)
  → is-surjective f
  → is-strong-epi (Sets ℓ) {c} {d} f
surjective→strong-epi c d f f-surj =
  is-regular-epi→is-strong-epi (Sets _) f $
  surjective→regular-epi c d f f-surj

surjective→epi
  : ∀ {ℓ} (c d : n-Type ℓ ) (f : ∣ c ∣ → ∣ d ∣)
  → is-surjective f
  → Cr.is-epic (Sets ℓ) {c} {d} f
surjective→epi c d f surj {c = x} =
  is-regular-epi→is-epic (surjective→regular-epi c d f surj) {c = x}

Epis are surjective🔗

Now this is the hard direction. Our proof follows (Rijke and Spitters 2015, sec. 2.9), so if the exposition below doesn’t make a lot of sense, be sure to check their paper, too. We define a higher inductive type standing for the cofibre of $f\text{,}$ also known as the mapping cone. Strictly speaking, this is the homotopy pushout

but Cubical Agda makes it very convenient to define a purpose-built HIT in-line. We use the names “tip”, “base”, and “cone” to evoke the construction of a cone in solid geometry: the tip is.. the tip, the base is the circle, and the cone is the triangular side which we have rotated around the vertical axis.

data Cofibre {ℓ ℓ'} {A : Type ℓ} {B : Type ℓ'} (f : A → B) : Type (ℓ ⊔ ℓ') where
  tip  : Cofibre f
  base : B → Cofibre f
  cone : ∀ a → tip ≡ base (f a)

What’s important here is that if a map $f$ has connected cofibre, then it is a surjection — so our proof that epis are surjective will factor through showing that epis have connected cofibres¹.

connected-cofibre→surjective
  : ∀ {ℓ ℓ'} {A : Type ℓ} {B : Type ℓ'} (f : A → B)
  → is-connected (Cofibre f)
  → is-surjective f
connected-cofibre→surjective {A = A} {B = B} f conn x = transport cen (lift tt) where

We define a family $P$ of propositions over $\mathrm{Cofibre}(f)$ which maps tip to the unit proposition and the base point with coordinate $x$ to the fibre $f^x\text{.}$ Since $\mathrm{Prop}$ is a set, $P$ extends to a map $P^{\prime }:\Vert \mathrm{Cofibre}(f){\Vert }_0\to \mathrm{Prop}\text{.}$

  P : Cofibre f → Prop _
  P tip      = el! (Lift _ ⊤)
  P (base x) = el! ∥ fibre f x ∥
  P (cone a i) =
    n-ua {X = el! (Lift _ ⊤)} {Y = el! ∥ fibre f (f a) ∥}
      (prop-ext! (λ _ → inc (a , refl)) λ _ → lift tt) i

  P' : ∥ Cofibre f ∥₀ → Prop _
  P' = ∥-∥₀-elim (λ _ → hlevel ) P

Letting $x$ be an element of the codomain, and since by assumption $f\text{’s}$ cofibre is connected, we have a path

so since the unit type is trivially inhabited, so is the fibre of $f$ over $x\text{:}$ $f$ is surjective.

  cen : Lift _ ⊤ ≡ ∥ fibre f x ∥
  cen = ap ∣_∣ (ap P' (is-contr→is-prop conn (inc tip) (inc (base x))))

Epis have connected cofibre🔗

Now suppose that $f$ is an epic morphism ^{2 3}. We know that the set-truncation of $f\text{’s}$ cofibre is inhabited (since every cofibre is inhabited), so it remains to show that any two points are merely equal.

epi→connected-cofibre
  : ∀ {ℓ} (c d : n-Type ℓ ) (f : ∣ c ∣ → ∣ d ∣)
  → Cr.is-epic (Sets ℓ) {c} {d} f
  → is-connected (Cofibre f)
epi→connected-cofibre c d f epic = contr (inc tip) $
  ∥-∥₀-elim (λ _ → is-prop→is-set (squash _ _)) λ w →
    ∥-∥₀-path.from (hom w)
  where

Let $x:d$ — we’ll show that $\Vert \mathrm{tip}={\mathrm{base}}_x\Vert \text{.}$ Here’s where we use $f\text{’s}$ epimorphy: we have a homotopy $|\mathrm{tip}|=|{\mathrm{base}}_{(fx)}|\text{,}$ namely the cone — and since we can write its left-hand-side as the composition of $f$ with a constant function, $f$ gives us a path $|\mathrm{tip}|=|{\mathrm{base}}_x|$ — which, by the characterisation of paths in the set truncation, means we merely have $\Vert \mathrm{tip}={\mathrm{base}}_x\Vert \text{.}$

    go : ∀ x → ∥ tip ≡ base x ∥
    go x = ∥-∥₀-path.to (p $ₚ x) where
      p = epic {c = el ∥ Cofibre f ∥₀ squash}
        (λ _ → inc tip)
        (λ x → inc (base x))
        (funext λ x → ap ∥_∥₀.inc (cone x))

An appeal to induction over the cofibre completes the proof: Epimorphisms have connected cofibres. Note that the path case is discharged automatically since we’re mapping into a proposition.

    hom : ∀ w → ∥ tip ≡ w ∥
    hom tip = inc refl
    hom (base x) = go x
    hom (cone a i) = is-prop→pathp (λ i → ∥_∥.squash {A = tip ≡ cone a i})
      (inc refl) (go (f a)) i

And, composing this with the previous step, we have what we originally set out to prove: all $\mathbf{Sets}\text{-epimorphisms}$ are regular, because they’re all surjections!

epi→surjective
  : ∀ {ℓ} (c d : n-Type ℓ ) (f : ∣ c ∣ → ∣ d ∣)
  → Cr.is-epic (Sets ℓ) {c} {d} f
  → is-surjective f
epi→surjective {ℓ} c d f epi x =
  connected-cofibre→surjective f (epi→connected-cofibre c d f (λ {x} → epi {x})) x