module 1Lab.Equiv.HalfAdjoint where
Adjoint equivalences🔗
An adjoint equivalence is an isomorphism where the homotopies ( satisfy the triangle identities, thus witnessing and as adjoint functors. In Homotopy Type Theory, we can use a half adjoint equivalence - satisfying only one of the triangle identities - as a good notion of equivalence.
is-half-adjoint-equiv : ∀ {ℓ₁ ℓ₂} {A : Type ℓ₁} {B : Type ℓ₂} (f : A → B) → Type _ is-half-adjoint-equiv {A = A} {B = B} f = Σ[ g ∈ (B → A) ] Σ[ η ∈ ((x : A) → g (f x) ≡ x) ] Σ[ ε ∈ ((y : B) → f (g y) ≡ y) ] ((x : A) → ap f (η x) ≡ ε (f x))
The argument is an adaptation of a standard result of both category theory and homotopy theory - where we can “improve” an equivalence of categories into an adjoint equivalence, or a homotopy equivalence into a strong homotopy equivalence (Vogt’s lemma). In HoTT, we show this synthetically for equivalences between
is-iso→is-half-adjoint-equiv : ∀ {ℓ₁ ℓ₂} {A : Type ℓ₁} {B : Type ℓ₂} {f : A → B} → is-iso f → is-half-adjoint-equiv f is-iso→is-half-adjoint-equiv {A = A} {B} {f} iiso = g , η , ε' , λ x → sym (zig x) where open is-iso iiso renaming (inv to g ; linv to η ; rinv to ε)
For
and
we can take the values provided by is-iso. However, if we want
to satisfy the triangle identities, we can not in general take
We can, however, alter it like thus:
ε' : (y : B) → f (g y) ≡ y ε' y = sym (ε (f (g y))) ∙ ap f (η (g y)) ∙ ε y
Drawn as a diagram, the path above factors like:
There is a great deal of redundancy in this definition, given that and have the same boundary! The point is that while the definition of is entirely opaque to us, is written in such a way that we can use properties of paths to make the and cancel:
zig : (x : A) → ε' (f x) ≡ ap f (η x) zig x = ε' (f x) ≡⟨⟩ sym (ε (f (g (f x)))) ∙ ap f ⌜ (η (g (f x))) ⌝ ∙ ε (f x) ≡⟨ ap (λ e → sym (ε _) ∙ ap f e ∙ ε _) (homotopy-invert η) ⟩≡ sym (ε (f (g (f x)))) ∙ ⌜ ap (f ∘ g ∘ f) (η x) ∙ ε (f x) ⌝ ≡˘⟨ ap¡ (homotopy-natural ε _) ⟩≡˘ sym (ε (f (g (f x)))) ∙ ε (f (g (f x))) ∙ ap f (η x) ≡⟨ ∙-cancell (ε (f (g (f x)))) (ap f (η x)) ⟩≡ ap f (η x) ∎
The notion of half-adjoint equivalence is
a useful stepping stone in writing a more comprehensible proof that
isomorphisms are equivalences.
Since this result is fundamental, the proof we actually use is written
with efficiency of computation in mind - hence, cubically. The proof
here is intended to be more educational.
First, we give an equivalent characterisation of paths in fibres, which will be used
in proving that half adjoint equivalences are equivalences.
fibre-paths : ∀ {ℓ₁ ℓ₂} {A : Type ℓ₁} {B : Type ℓ₂} {f : A → B} {y : B} → {f1 f2 : fibre f y} → (f1 ≡ f2) ≃ (Σ[ γ ∈ f1 .fst ≡ f2 .fst ] (ap f γ ∙ f2 .snd ≡ f1 .snd))
The proof of this is not very enlightening, but it’s included here (rather than being completely invisible) for completeness:
fibre-paths {f = f} {y} {f1} {f2} = Path (fibre f y) f1 f2 ≃⟨ Iso→Equiv Σ-path-iso e⁻¹ ⟩≃ (Σ[ γ ∈ f1 .fst ≡ f2 .fst ] (subst (λ x₁ → f x₁ ≡ _) γ (f1 .snd) ≡ f2 .snd)) ≃⟨ Σ-ap-snd (λ x → path→equiv (lemma x)) ⟩≃ (Σ[ γ ∈ f1 .fst ≡ f2 .fst ] (ap f γ ∙ f2 .snd ≡ f1 .snd)) ≃∎ where helper : (p' : f (f1 .fst) ≡ y) → (subst (λ x → f x ≡ y) refl (f1 .snd) ≡ p') ≡ (ap f refl ∙ p' ≡ f1 .snd) helper p' = subst (λ x → f x ≡ y) refl (f1 .snd) ≡ p' ≡⟨ ap₂ _≡_ (transport-refl _) refl ⟩≡ (f1 .snd) ≡ p' ≡⟨ Iso→Path (sym , iso sym (λ x → refl) (λ x → refl)) ⟩≡ ⌜ p' ⌝ ≡ f1 .snd ≡˘⟨ ap¡ (∙-idl _) ⟩≡˘ refl ∙ p' ≡ f1 .snd ≡⟨⟩ ap f refl ∙ p' ≡ f1 .snd ∎ lemma : ∀ {x'} {p'} → (γ : f1 .fst ≡ x') → (subst (λ x → f x ≡ _) γ (f1 .snd) ≡ p') ≡ (ap f γ ∙ p' ≡ f1 .snd) lemma {x'} {p'} p = J (λ x' γ → ∀ p' → (subst (λ x → f x ≡ _) γ (f1 .snd) ≡ p') ≡ (ap f γ ∙ p' ≡ f1 .snd)) helper p p'
Then, given an element we can construct a fibre of of and, using the above characterisation of paths, prove that this fibre is a centre of contraction:
is-half-adjoint-equiv→is-equiv : ∀ {ℓ₁ ℓ₂} {A : Type ℓ₁} {B : Type ℓ₂} {f : A → B} → is-half-adjoint-equiv f → is-equiv f is-half-adjoint-equiv→is-equiv {A = A} {B} {f} (g , η , ε , zig) .is-eqv y = contr fib contract where fib : fibre f y fib = g y , ε y
The fibre is given by which we can prove identical to another using a very boring calculation:
contract : (fib₂ : fibre f y) → fib ≡ fib₂ contract (x , p) = (fibre-paths e⁻¹) .fst (x≡gy , path) where x≡gy = ap g (sym p) ∙ η x path : ap f (ap g (sym p) ∙ η x) ∙ p ≡ ε y path = ap f (ap g (sym p) ∙ η x) ∙ p ≡⟨ ap₂ _∙_ (ap-∙ f (ap g (sym p)) (η x)) refl ∙ sym (∙-assoc _ _ _) ⟩≡ ap (λ x → f (g x)) (sym p) ∙ ⌜ ap f (η x) ⌝ ∙ p ≡⟨ ap! (zig _) ⟩≡ -- by the triangle identity ap (f ∘ g) (sym p) ∙ ⌜ ε (f x) ∙ p ⌝ ≡⟨ ap! (homotopy-natural ε p) ⟩≡ -- by naturality of ε
The calculation of path factors as a bunch of
boring adjustments to paths using the groupoid structure of types, and
the two interesting steps above: The triangle identity says that
and naturality of
lets us “push it past
”
to get something we can cancel:
ap (f ∘ g) (sym p) ∙ ap (f ∘ g) p ∙ ε y ≡⟨ ∙-assoc _ _ _ ⟩≡ ⌜ ap (f ∘ g) (sym p) ∙ ap (f ∘ g) p ⌝ ∙ ε y ≡˘⟨ ap¡ (ap-∙ (f ∘ g) (sym p) p) ⟩≡˘ ap (f ∘ g) ⌜ sym p ∙ p ⌝ ∙ ε y ≡⟨ ap! (∙-invr _) ⟩≡ ap (f ∘ g) refl ∙ ε y ≡⟨⟩ refl ∙ ε y ≡⟨ ∙-idl (ε y) ⟩≡ ε y ∎
Putting these together, we get an alternative definition of is-iso→is-equiv:
is-iso→is-equiv' : ∀ {ℓ₁ ℓ₂} {A : Type ℓ₁} {B : Type ℓ₂} {f : A → B} → is-iso f → is-equiv f is-iso→is-equiv' = is-half-adjoint-equiv→is-equiv ∘ is-iso→is-half-adjoint-equiv