open import Cat.Diagram.Coproduct.Indexed open import Cat.Instances.Sets.Complete open import Cat.Diagram.Colimit.Base open import Cat.Diagram.Initial open import Cat.Prelude module Cat.Instances.Sets.Cocomplete where
Sets is cocomplete🔗
Before proving that the category of sets is cocomplete, as a warm-up exercise, we prove that the category of sets admits indexed coproducts, and furthermore, that these are disjoint: all of the coproduct inclusions are monomorphisms, and distinct inclusions have initial images. This will be illustrative of a minor sticking point that will come up in the construction of arbitrary colimits.
Sets-has-coproducts : ∀ {κ ℓ} → has-indexed-coproducts (Sets (κ ⊔ ℓ)) κ Sets-has-coproducts {κ} {ℓ} {I = I} F = coprod where
The coproduct of the family is given by the type . However, this type is in general not a set! Consider a family of sets indexed by the circle. Its total space will, by necessity, be a groupoid rather than a set.
However, we can always truncate the sum down to a set, and it turns out that this truncation does serve as a coproduct of the family in the category of sets. The point here is that, since the objects of are.. well, sets, they can’t have any interesting paths, by definition. A grim slogan: In the category of Sets, nobody can hear your paths scream.
sum : Type (κ ⊔ ℓ) sum = Σ[ i ∈ I ] ∣ F i ∣ open Indexed-coproduct open is-indexed-coproduct coprod : Indexed-coproduct (Sets _) F coprod .ΣF = ∥ sum ∥₀ , squash coprod .ι i x = inc (i , x) coprod .has-is-ic .match {Y = Y} f = ∥-∥₀-elim (λ _ → Y .is-tr) λ { (i , x) → f i x } {- 1 -} coprod .has-is-ic .commute = refl coprod .has-is-ic .unique {Y = Y} f p = funext (∥-∥₀-elim (λ _ → is-prop→is-set (Y .is-tr _ _)) λ x → happly (p _) _)
Note that, in the construction of match above, we used the fact that (the common codomain of all the ) is a set to eliminate from the truncation — by definition, can’t tell that might have had some extra paths we squashed away.
Colimits🔗
Perfectly dually to the construction of limits in , rather than taking the equaliser of a product, we take the coequaliser of a sum. The same considerations about truncation level that apply for arbitrary coproducts apply to arbitrary colimits: fortunately, the construction of set-coequalisers already includes a truncation.
Sets-is-cocomplete : ∀ {ι κ o} → is-cocomplete ι κ (Sets (ι ⊔ κ ⊔ o)) Sets-is-cocomplete {D = D} F = colim where module D = Precategory D module F = Functor F sum : Type _ sum = Σ[ d ∈ D.Ob ] ∣ F.₀ d ∣ rel : sum → sum → Type _ rel (X , x) (Y , y) = Σ[ f ∈ D.Hom X Y ] (F.₁ f x ≡ y)
The precise coequaliser we take is the quotient of by the relation (generated by) identifying together all those points and whenever there exists a map such that .
apex : Cocone F apex .coapex = (sum / rel) , squash apex .ψ x p = inc (x , p) apex .commutes f = funext (λ i → sym (quot (f , refl)))
By the same truncation nonsense as above, we can apply Coeq-rec to eliminate from our quotient to the coapex of any other cocone over ; The family of maps respects the quotient essentially by definition.
colim : Initial _ colim .bot = apex colim .has⊥ other = contr map unique where map : Cocone-hom F apex other map .hom = Coeq-rec (other .coapex .is-tr) (λ { (x , p) → other .ψ x p }) λ { ((X , x) , (Y , y) , f , p) → sym (happly (other .commutes f) x) ∙ ap (other .ψ Y) p } map .commutes o = refl unique : ∀ x → map ≡ x unique hom′ = Cocone-hom-path _ (funext (Coeq-elim-prop (λ x → other .coapex .is-tr _ _) λ y → sym (happly (hom′ .commutes _) _)))
Coproducts are disjoint🔗
As a final lemma, we prove that coproducts in , as constructed above, are disjoint. However, this does not apply to arbitrary coproducts; To prove that the injections are monomorphisms, we require that the indexing type be a set.
module _ {κ} {I : Set κ} {F : ∣ I ∣ → Set κ} where private module coprod = Indexed-coproduct (Sets-has-coproducts {ℓ = κ} F) Set-disjoint-coprods : is-disjoint-coproduct (Sets κ) {S = coprod.ΣF} F coprod.ι Set-disjoint-coprods = coprod where open is-disjoint-coproduct open is-indexed-coproduct
We already know that the coproduct is a coproduct (who would have guessed, honestly) — so it remains to show that the injections are monic, the summands intersect, and the intersections of different summands are empty. The intersections are cheap: Sets is finitely complete, so all pullbacks exist, in particular the pullback of .
coprod : is-disjoint-coproduct _ _ _ coprod .is-coproduct = coprod.has-is-ic coprod .summands-intersect i j = Sets-pullbacks _ _
To prove that the injections are monic, we use our assumption that the family was indexed by a set: The sum then is also a set, so we can get it out from under the truncation in the definition of coproduct.
coprod .injections-are-monic _ g h path = funext go where abstract path′ : Path (∀ c → Σ (λ x → ∣ F x ∣)) (λ c → _ , g c) (λ c → _ , h c) path′ i c = ∥-∥₀-elim {B = λ _ → Σ (∣_∣ ⊙ F)} (λ x → Σ-is-hlevel 2 (I .is-tr) (λ x → F x .is-tr)) (λ x → x) (path i c) q : ∀ {c} → ap fst (happly path′ c) ≡ refl q = I .is-tr _ _ _ _ go : ∀ c → g c ≡ h c go c = subst (λ e → PathP (λ i → ∣ F (e i) ∣) (g c) (h c)) q (ap snd (happly path′ c))
The same thing happens in proving that different injections have disjoint images: We must project out a path from a path — using that they are in a set to eliminate from the truncation — to prove using the assumption that .
coprod .different-images-are-disjoint i j i≠j os = contr map uniq where map : Σ[ i ∈ ∣ F i ∣ ] Σ (λ x → _) → ∣ os ∣ map (i , j , p) = absurd (i≠j (ap (∥-∥₀-elim (λ _ → I .is-tr) fst) p)) uniq : ∀ x → map ≡ x uniq _ = funext λ where (_ , _ , p) → absurd (i≠j (ap (∥-∥₀-elim (λ _ → I .is-tr) fst) p))