module Cat.Diagram.Pullback.Properties where

Properties of pullbacks🔗

This module chronicles some general properties of pullbacks.

Identity🔗

Degenerate squares where two opposite sides are identities are pullbacks.

  id-is-pullback : ∀ {a b : Ob} {f : Hom a b} → is-pullback C id f f id
  id-is-pullback .square = id-comm
  id-is-pullback .universal {p₁' = p₁'} _ = p₁'
  id-is-pullback .p₁∘universal = idl _
  id-is-pullback .p₂∘universal {p = p} = p ∙ idl _
  id-is-pullback .unique q r = sym (idl _) ∙ q

The pasting law for pullbacks says that, if in the commutative diagram below the the right square is a pullback, then the left square is universal if, and only if, the outer rectangle is, too. Note the emphasis on the word commutative: if we don’t know that both squares (and the outer rectangle) all commute, the pasting law does not go through.

  module _ {a b c d e f : Ob}
           {a→d : Hom a d} {a→b : Hom a b} {b→c : Hom b c}
           {d→e : Hom d e} {b→e : Hom b e} {e→f : Hom e f}
           {c→f : Hom c f}
           (right-pullback : is-pullback C b→c c→f b→e e→f)
    where

    private module right = is-pullback right-pullback

Let’s start with proving that, if the outer rectangle is a pullback, then so is the left square. Assume, then, that we have some other object $x,$ which fits into a cone, like in the diagram below. I’ve coloured the two arrows we assume commutative.

    pasting-outer→left-is-pullback
      : is-pullback C (b→c ∘ a→b) c→f a→d (e→f ∘ d→e)
      → (square : b→e ∘ a→b ≡ d→e ∘ a→d)
      → is-pullback C a→b b→e a→d d→e
    pasting-outer→left-is-pullback outer square = pb where
      module outer = is-pullback outer

To appeal to the universal property of the outer pullback, we must somehow extend our red cone over $b \to e \leftarrow d$ to one over $c \to f \leftarrow e .$ Can we do this? Yes! By assumption, the square on the right commutes, which lets us apply commutativity of the red diagram (the assumption $p$ in the code). Check out the calculation below:

      abstract
        path : ∀ {P} {P→b : Hom P b} {P→d : Hom P d} (p : b→e ∘ P→b ≡ d→e ∘ P→d)
             → c→f ∘ b→c ∘ P→b ≡ (e→f ∘ d→e) ∘ P→d
        path {_} {P→b} {P→d} p =
          c→f ∘ b→c ∘ P→b   ≡⟨ extendl right.square ⟩
          e→f ∘ b→e ∘ P→b   ≡⟨ ap (e→f ∘_) p ⟩
          e→f ∘ d→e ∘ P→d   ≡⟨ cat! C ⟩
          (e→f ∘ d→e) ∘ P→d ∎

      pb : is-pullback C _ _ _ _
      pb .is-pullback.square =
        b→e ∘ a→b ≡⟨ square ⟩
        d→e ∘ a→d ∎

We thus have an induced map $x \to a,$ which, since $a$ is a pullback, makes everything in sight commute, and does so uniquely.

      pb .universal {p₁' = P→b} {p₂' = P→d} p =
        outer.universal {p₁' = b→c ∘ P→b} {p₂' = P→d} (path p)

      pb .p₁∘universal {p₁' = P→b} {p₂' = P→d} {p = p} =
        right.unique₂ {p = pulll right.square ∙ pullr p}
          (assoc _ _ _ ∙ outer.p₁∘universal)
          (pulll square ∙ pullr outer.p₂∘universal)
          refl p

      pb .p₂∘universal {p₁' = P→b} {p₂' = P→d} {p = p} = outer.p₂∘universal

      pb .unique {p = p} q r =
        outer.unique (sym (ap (_ ∘_) (sym q) ∙ assoc _ _ _)) r

For the converse, suppose that both small squares are a pullback, and we have a cone over $c \to f \leftarrow d .$ By the universal property of the right pullback, we can find a map $x \to b$ forming the left leg of a cone over $b \to e \leftarrow d;$ By the universal property of the left square, we then have a map $x \to a,$ as we wanted.

    pasting-left→outer-is-pullback
      : is-pullback C a→b b→e a→d d→e
      → (square : c→f ∘ b→c ∘ a→b ≡ (e→f ∘ d→e) ∘ a→d)
      → is-pullback C (b→c ∘ a→b) c→f a→d (e→f ∘ d→e)
    pasting-left→outer-is-pullback left square = pb where
      module left = is-pullback left

      pb : is-pullback C (b→c ∘ a→b) c→f a→d (e→f ∘ d→e)
      pb .is-pullback.square =
        c→f ∘ b→c ∘ a→b   ≡⟨ square ⟩
        (e→f ∘ d→e) ∘ a→d ∎
      pb .universal {p₁' = P→c} {p₂' = P→d} x =
        left.universal {p₁' = right.universal (x ∙ sym (assoc _ _ _))} {p₂' = P→d}
          right.p₂∘universal
      pb .p₁∘universal = pullr left.p₁∘universal ∙ right.p₁∘universal
      pb .p₂∘universal = left.p₂∘universal
      pb .unique {p₁' = P→c} {P→d} {p = p} {lim'} q r =
        left.unique (right.unique (assoc _ _ _ ∙ q) s) r
        where
          s : b→e ∘ a→b ∘ lim' ≡ d→e ∘ P→d
          s =
            b→e ∘ a→b ∘ lim'   ≡⟨ pulll left.square ⟩
            (d→e ∘ a→d) ∘ lim' ≡⟨ pullr r ⟩
            d→e ∘ P→d          ∎

Monomorphisms🔗

Being a monomorphism is a “limit property”. Specifically, $f : A \to B$ is a monomorphism iff. the square below is a pullback.

  module _ {a b} {f : Hom a b} where
    is-monic→is-pullback : is-monic f → is-pullback C id f id f
    is-monic→is-pullback mono .square = refl
    is-monic→is-pullback mono .universal {p₁' = p₁'} p = p₁'
    is-monic→is-pullback mono .p₁∘universal = idl _
    is-monic→is-pullback mono .p₂∘universal {p = p} = idl _ ∙ mono _ _ p
    is-monic→is-pullback mono .unique p q = introl refl ∙ p

    is-pullback→is-monic : is-pullback C id f id f → is-monic f
    is-pullback→is-monic pb f g p = sym (pb .p₁∘universal {p = p}) ∙ pb .p₂∘universal

Pullbacks additionally preserve monomorphisms, as shown below:

  is-monic→pullback-is-monic
    : ∀ {x y z} {f : Hom x z} {g : Hom y z} {p} {p1 : Hom p x} {p2 : Hom p y}
    → is-monic f
    → is-pullback C p1 f p2 g
    → is-monic p2
  is-monic→pullback-is-monic {f = f} {g} {p1 = p1} {p2} mono pb h j p = eq
    where
      module pb = is-pullback pb
      q : f ∘ p1 ∘ h ≡ f ∘ p1 ∘ j
      q =
        f ∘ p1 ∘ h ≡⟨ extendl pb.square ⟩
        g ∘ p2 ∘ h ≡⟨ ap (g ∘_) p ⟩
        g ∘ p2 ∘ j ≡˘⟨ extendl pb.square ⟩
        f ∘ p1 ∘ j ∎

      r : p1 ∘ h ≡ p1 ∘ j
      r = mono _ _ q

      eq : h ≡ j
      eq = pb.unique₂ {p = extendl pb.square} r p refl refl

Properties of pullbacks🔗

Identity🔗

Pasting law🔗

Monomorphisms🔗