open import 1Lab.HLevel.Retracts
open import 1Lab.HLevel
open import 1Lab.Equiv
open import 1Lab.Path
open import 1Lab.Type

module 1Lab.Equiv.Biinv where


# Bi-invertible maps🔗

Recall the three conditions that make up the notion of equivalence.

To be more specific, what we need for a notion of equivalence ${\mathrm{is-equiv}}(f)$ to be “coherent” is:

• Being an isomorphism implies being an equivalence (${\mathrm{is-iso}}(f) \to {\mathrm{is-equiv}}(f)$)

• Being an equivalence implies being an isomorphism (${\mathrm{is-equiv}}(f) \to {\mathrm{is-iso}}(f)$); Taken together with the first point we may summarise: “Being an equivalence and being an isomorphism are logically equivalent.”

• Most importantly, being an equivalence must be a proposition.

The “blessed” definition of equivalence is that of a map with contractible fibres. However, this definition is highly abstract, so it begs the question: Is it possible to describe a simpler notion of equivalence that still satisfies the three conditions? The answer is yes! Paradoxically, adding more data to is-iso leaves us with a good notion of equivalence.

A left inverse to a function $f : A \to B$ is a function $g : B \to A$ equipped with a homotopy $g \circ f \sim {{\mathrm{id}}_{}}$. Symmetrically, a right inverse to $f$ is a function $h : B \to A$ equipped with a homotopy $f \circ h \sim {{\mathrm{id}}_{}}$.

linv : (A → B) → Type _
linv f = Σ[ g ∈ (_ → _) ] (g ∘ f ≡ id)

rinv : (A → B) → Type _
rinv f = Σ[ h ∈ (_ → _) ] (f ∘ h ≡ id)


A map $f$ equipped with a choice of left- and right- inverse is said to be biinvertible. Perhaps surprisingly, is-biinv is a good notion of equivalence.

is-biinv : (A → B) → Type _
is-biinv f = linv f × rinv f


If $f$ is an equivalence, then so are pre- and post- composition with $f$. This can be proven using equivalence induction, but a more elementary argument — directly constructing quasi-inverses — suffices, and doesn’t use the sledgehammer that is univalence.

The proof is as follows: If $f : A \to B$ has inverse $f^{-1} : B → A$, then $(f^{-1} \circ -)$ and $(- \circ f^{-1})$ are inverses to $(f \circ -)$ and $(- \circ f)$.

is-equiv→pre-is-equiv  : {f : A → B} → is-equiv f → is-equiv {A = C → A} (f ∘_)
is-equiv→post-is-equiv : {f : A → B} → is-equiv f → is-equiv {A = B → C} (_∘ f)

The proof is by is-equiv→is-iso and is-iso→is-equiv. Nothing too clever.
is-equiv→pre-is-equiv {f = f} f-eqv = is-iso→is-equiv isiso where
f-iso : is-iso f
f-iso = is-equiv→is-iso f-eqv

f⁻¹ : _
f⁻¹ = f-iso .is-iso.inv

isiso : is-iso (_∘_ f)
isiso .is-iso.inv f x = f⁻¹ (f x)
isiso .is-iso.rinv f = funext λ x → f-iso .is-iso.rinv _
isiso .is-iso.linv f = funext λ x → f-iso .is-iso.linv _

is-equiv→post-is-equiv {f = f} f-eqv = is-iso→is-equiv isiso where
f-iso : is-iso f
f-iso = is-equiv→is-iso f-eqv

f⁻¹ : _
f⁻¹ = f-iso .is-iso.inv

isiso : is-iso _
isiso .is-iso.inv f x = f (f⁻¹ x)
isiso .is-iso.rinv f = funext λ x → ap f (f-iso .is-iso.linv _)
isiso .is-iso.linv f = funext λ x → ap f (f-iso .is-iso.rinv _)


With this lemma, it can be shown that if $f$ is an isomorphism, then linv(f) and rinv(f) are both contractible.

is-iso→is-contr-linv : {f : A → B} → is-iso f → is-contr (linv f)
is-iso→is-contr-linv isiso =
is-equiv→post-is-equiv (is-iso→is-equiv isiso) .is-eqv id

is-iso→is-contr-rinv : {f : A → B} → is-iso f → is-contr (rinv f)
is-iso→is-contr-rinv isiso =
is-equiv→pre-is-equiv (is-iso→is-equiv isiso) .is-eqv id


This is because linv(f) is the fibre of $(- \circ f)$ over id, and the fibres of an equivalence are contractible. Dually, rinv(f) is the fibre of $(f \circ -)$ over id.

_ : {f : A → B} → linv f ≡ fibre (_∘ f) id
_ = refl

_ : {f : A → B} → rinv f ≡ fibre (f ∘_) id
_ = refl


We show that if a map is biinvertible, then it is invertible. This is because if a function has two inverses, they coincide:

is-biinv→is-iso : {f : A → B} → is-biinv f → is-iso f
is-biinv→is-iso {f = f} ((g , g∘f≡id) , h , h∘f≡id) = iso h (happly h∘f≡id) beta
where
beta : (x : _) → h (f x) ≡ x
beta x =
h (f x)         ≡⟨ happly (sym g∘f≡id) _ ⟩≡
g (f (h (f x))) ≡⟨ ap g (happly h∘f≡id _) ⟩≡
g (f x)         ≡⟨ happly g∘f≡id _ ⟩≡
x               ∎


Finally, we can show that being biinvertible is a proposition. Since propositions are those types which are contractible if inhabited suffices to show that is-biinv is contractible when it is inhabited:

is-biinv-is-prop : {f : A → B} → is-prop (is-biinv f)
is-biinv-is-prop {f = f} = contractible-if-inhabited contract where
contract : is-biinv f → is-contr (is-biinv f)
contract ibiinv =
×-is-hlevel 0 (is-iso→is-contr-linv iiso)
(is-iso→is-contr-rinv iiso)
where
iiso = is-biinv→is-iso ibiinv


Since is-biinv is a product of contractibles whenever it is inhabited, then it is contractible. Finally, we have that ${\mathrm{is-iso}}(f) \to {\mathrm{is-biinv}}(f)$: pick the given inverse as both a left- and right- inverse.

is-iso→is-biinv : {f : A → B} → is-iso f → is-biinv f
is-iso→is-biinv iiso .fst = iiso .is-iso.inv , funext (iiso .is-iso.linv)
is-iso→is-biinv iiso .snd = iiso .is-iso.inv , funext (iiso .is-iso.rinv)