open import 1Lab.Path.Reasoning
open import 1Lab.Path.Groupoid
open import 1Lab.HLevel
open import 1Lab.Path
open import 1Lab.Type

open is-contr

module 1Lab.Equiv where

Equivalences🔗

The key principle that distinguishes HoTT from other type theories is the univalence principle: isomorphic types can be identified. In cubical type theory, this turns out to be a theorem; but from a neutral point of view, it should be read as the definition of the identity types of universes — which are otherwise left unspecified by MLTT. However, the notion of isomorphism is not well-behaved when higher dimensional types are involved, so we can’t actually use it to formulate univalence.

The key issue is that any identity system is equipped with a path induction principle, which, if extended to the type of isomorphisms, would let us conclude that being an isomorphism is a proposition: but this is provably not the case. The result of formulating univalence using isomorphisms is sometimes called isovalence — that page has a formalisation of its failure.

We must therefore define a property of functions which refines being an isomorphism but is an isomorphism: functions satisfying this property will be called equivalences. More specifically, we’re looking for a family $is-equiv(f)$ which

• is implied by $is-iso(f)\text{,}$ so that any function with a two-sided inverse is an equivalence. Apart from preserving the connection with the intuitive motivation behind univalence, exhibiting a two-sided inverse to a given map is simply easier than proving that it’s a coherent equivalence.

• implies $is-iso(f)\text{,}$ so that knowing that a map is an equivalence also lets you extract a two-sided inverse. Most notions of equivalence will also tell you additional properties about the connection between $f$ and its inverse map.

• and finally, is an actual proposition.

Put concisely, we’re looking to define a propositional truncation of $is-iso(f)$ — one which allows us to conveniently project out the underlying data. Since propositional truncations are unique, this means that all type families satisfying the three bullet points above are themselves equivalent; so any choice made must be meta-mathematical, driven by concerns like ease of use, and efficiency of the formalisation.

Here in the 1Lab, we formalise three acceptable notions of equivalence:

• biinvertible maps, which store two inverses to the given function;
• half-adjoint equivalences, which, in addition to the inverse map and the witnesses of invertibility, have an extra coherence datum; and
• the notion defined in this file, contractible maps, which repackage the notion of equivalence in a homotopically-inspired way.

private variable
  ℓ₁ ℓ₂ : Level
  A B C : Type ℓ₁

Isomorphisms🔗

Before we set about defining and working with equivalences, we’ll warm up by defining, and proving basic things about, isomorphisms. First, we define what it means for functions to be inverses of eachother, on both the left and the right.

is-left-inverse : (B → A) → (A → B) → Type _
is-left-inverse g f = (x : _) → g (f x) ≡ x

is-right-inverse : (B → A) → (A → B) → Type _
is-right-inverse g f = (x : _) → f (g x) ≡ x

A function $g$ which is both a left- and right inverse to $f$ is called a two-sided inverse (to $f\text{).}$ To say that a function is an isomorphism is to equip it with a two-sided inverse.

record is-iso (f : A → B) : Type (level-of A ⊔ level-of B) where
  no-eta-equality
  constructor iso
  field
    inv  : B → A
    rinv : is-right-inverse inv f
    linv : is-left-inverse inv f

It’s immediate from the symmetry of the definition that if $g$ is a two-sided inverse to $f\text{,}$ then $f$ also inverts $g\text{:}$ an isomorphism’s inverse is again an isomorphism.

  inverse : is-iso inv
  inverse .inv  = f
  inverse .rinv = linv
  inverse .linv = rinv

Iso : ∀ {ℓ₁ ℓ₂} → Type ℓ₁ → Type ℓ₂ → Type _
Iso A B = Σ (A → B) is-iso

module _ where
  open is-iso

Second, a composite of isomorphisms is an isomorphism: if we can undo $f$ and $g$ individually, we can undo them sequentially, by first undoing $g\text{,}$ then undoing $f\text{.}$ The observation that the inverse is in the opposite order is sometimes humorously referred to as the socks-and-shoes principle.

  ∘-is-iso : {f : B → C} {g : A → B} → is-iso f → is-iso g → is-iso (f ∘ g)
  ∘-is-iso f-im g-im .inv x = g-im .inv (f-im .inv x)
  ∘-is-iso {f = f} {g = g} f-im g-im .rinv x =
    f (g (g-im .inv (f-im .inv x))) ≡⟨ ap f (g-im .rinv _) ⟩≡
    f (f-im .inv x)                 ≡⟨ f-im .rinv _ ⟩≡
    x                               ∎
  ∘-is-iso {g = g} f-im g-im .linv x =
    ap (g-im .inv) (f-im .linv (g x)) ∙ g-im .linv x

Finally, the identity map is its own two-sided inverse, so it is an isomorphism. Keep in mind that, again, there are multiple ways for a map to be an isomorphism. It would be more correct to say that there is a parametric way to make the identity map into an isomorphism: if we know more about the specific type, there might be other ways.

  private
    id-iso : is-iso {A = A} id
    id-iso .inv    = id
    id-iso .rinv x = refl
    id-iso .linv x = refl

Equivalences proper🔗

With that out of the way, we can actually define the property of functions that we’ll adopt under the name equivalence. The notion of contractible maps is due to Voevodsky, and it very elegantly packages the data of an inverse in a way that is immediately seen to be coherent. However, it turns out to have a very simple motivating explanation at the level of sets:

Recall that the preimage of a function $f:A\to B$ over a point $y:B\text{,}$ written $f^{\ast }(y)\text{,}$ is the set $\{x:A|f(x)=y\}\text{.}$ The properties of $f$ being injective and surjective can be profitably rephrased in terms of its preimages:

• a function is injective if each of its preimages has at most one element. For if we have $x,y:A$ and $p:f(x)=f(y)\text{,}$ then we can look at the set $f^{\ast }(f(x))\text{:}$ it has elements given by $(x,\mathrm{refl})\text{,}$ and $(y,p)\text{.}$ For these to be the same, we must have $x=y\text{,}$ so $f$ is injective.

• a function is surjective if each of its preimages has at least one element, in the sense of propositional truncations. This is the actual definition: for the preimage $f^{\ast }(y)$ to be inhabited means that there exists $x:A$ with $f(x)=y\text{.}$

Putting these together, a function is a bijection if each of its fibres has exactly one element. Phrased as above, we’ve actually arrived exactly at the definition of equivalence proposed by Voevodsky: we just have to rephrase a few things.

fibre : (A → B) → B → Type _
fibre {A = A} f y = Σ[ x ∈ A ] (f x ≡ y)

Finally, rather than talking of types with exactly one element, we already have a rephrasing which works at the level of homotopy type theory: contractibility. This is exactly the conjunction of the assertions above, just packaged in a way that does not require us to first define propositional truncations. A function is an equivalence if all of its fibres are contractible:

record is-equiv (f : A → B) : Type (level-of A ⊔ level-of B) where
  no-eta-equality
  field
    is-eqv : (y : B) → is-contr (fibre f y)

open is-equiv public

_≃_ : ∀ {ℓ₁ ℓ₂} → Type ℓ₁ → Type ℓ₂ → Type _
_≃_ A B = Σ (A → B) is-equiv

To warm up, let’s show that the identity map is an equivalence, working with the definition directly. First, we have to show that the fibre over $y$ has an element, for which we can take $(y,\mathrm{refl})\text{,}$ since what we’re looking for boils down to an element $y:A$ which is mapped to $y$ by the identity function.

id-equiv : is-equiv {A = A} id
id-equiv .is-eqv y .centre = y , refl

Then, given any pair $(x,p)$ with $x:A$ and $p:x=y\text{,}$ we must show that we have $(y,\mathrm{refl})=(x,p)\text{.}$ But this is exactly what path induction promises us! Rather than an appeal to transport, we can directly use a connection:

id-equiv .is-eqv y .paths (x , p) i = p (~ i) , λ j → p (~ i ∨ j)

-- This helper is for functions f, g that cancel eachother, up to
-- definitional equality, without any case analysis on the argument:

strict-fibres : ∀ {ℓ ℓ'} {A : Type ℓ} {B : Type ℓ'} {f : A → B} (g : B → A) (b : B)
  → Σ[ t ∈ fibre f (f (g b)) ]
    ((t' : fibre f b) → Path (fibre f (f (g b))) t
                          (g (f (t' .fst)) , ap (f ∘ g) (t' .snd)))
strict-fibres {f = f} g b .fst = (g b , refl)
strict-fibres {f = f} g b .snd (a , p) i = (g (p (~ i)) , λ j → f (g (p (~ i ∨ j))))

-- This is more efficient than using Iso→Equiv. When f (g x) is definitionally x,
-- the type reduces to essentially is-contr (fibre f b).

Since Cubical Agda has primitives that refer to equivalences, we have to take some time to teach the system about the type we just defined above. In addition to teaching it how to project the underlying function of an equivalence, we must prove that an equivalence has contractible fibres — but what the system asks of us is that each partial fibre be extensible: the cubical phrasing of contractibility.

{-# BUILTIN EQUIV _≃_ #-}
{-# BUILTIN EQUIVFUN fst #-}

is-eqv' : ∀ {a b} (A : Type a) (B : Type b)
        → (w : A ≃ B) (a : B)
        → (ψ : I)
        → (p : Partial ψ (fibre (w .fst) a))
        → fibre (w .fst) a [ ψ ↦ p ]
is-eqv' A B (f , is-equiv) a ψ u0 = inS (
  hcomp (∂ ψ) λ where
    i (ψ = i0) → c .centre
    i (ψ = i1) → c .paths (u0 1=1) i
    i (i = i0) → c .centre)
  where c = is-equiv .is-eqv a

{-# BUILTIN EQUIVPROOF is-eqv' #-}

equiv-centre : (e : A ≃ B) (y : B) → fibre (e .fst) y
equiv-centre e y = e .snd .is-eqv y .centre

equiv-path : (e : A ≃ B) (y : B) →
  (v : fibre (e .fst) y) → Path _ (equiv-centre e y) v
equiv-path e y = e .snd .is-eqv y .paths

is-equiv is a proposition🔗

Since being contractible is a proposition, and being an equivalence simply quantifies contractibility over each fibre, we can directly conclude that being an equivalence is a proposition.

is-equiv-is-prop : (f : A → B) → is-prop (is-equiv f)
is-equiv-is-prop f x y i .is-eqv p = is-contr-is-prop (x .is-eqv p) (y .is-eqv p) i

Equivalences are invertible🔗

We can now show that equivalences are invertible. This will amount to taking apart the proof that each fibre is contractible. If we have $y:B\text{,}$ then the centre of contraction of $f^{\ast }(y)$ gives us a distinguished $x:A$ and a path $f(x)=y\text{:}$

equiv→inverse : {f : A → B} → is-equiv f → B → A
equiv→inverse eqv y = eqv .is-eqv y .centre .fst

equiv→counit
  : ∀ {f : A → B} (eqv : is-equiv f) (y : B) → f (equiv→inverse eqv y) ≡ y
equiv→counit eqv y = eqv .is-eqv y .centre .snd

Now we have to show that applying the inverse to $f(x)$ gives us back $x\text{.}$ But note that $f^{\ast }(fx)$ has two inhabitants: the centre of contraction, which is projected by the inverse, but also the simpler $(x,\mathrm{refl})\text{.}$ Since the fibre is contractible, we have a path $f^{-1}(fx)=x\text{.}$

equiv→unit
  : ∀ {f : A → B} (eqv : is-equiv f) x → equiv→inverse eqv (f x) ≡ x
equiv→unit {f = f} eqv x i = eqv .is-eqv (f x) .paths (x , refl) i .fst

Contractibility gives us, in addition to a path between the points $f^{-1}(fx)=x\text{,}$ a particular higher-dimensional square. Note that we have two ways of proving that $f{f}^{-1}fx=fx\text{,}$ corresponding to either cancelling the outer $f{f}^{-1}$ or the inner $f^{-1}f\text{:}$ $f(\eta x)$ and $\epsilon fx\text{.}$ The square we obtain tells us that these are actually the same:

equiv→square
  : ∀ {f : A → B} (eqv : is-equiv f) (x : A)
  → Square (ap f (equiv→unit eqv x)) (equiv→counit eqv (f x)) refl refl
equiv→square {f = f} eqv x i j = eqv .is-eqv (f x) .paths (x , refl) i .snd j

However, the square above is slightly skewed: we would have preferred for the two non-trivial paths to be in opposite sides, so that it would correspond to a Path between the paths. We can use hcomp to slide one of the faces around to get us the more useful direct proof that $f(\eta x)=\epsilon fx\text{.}$

equiv→zig
  : ∀ {f : A → B} (eqv : is-equiv f) x
  → ap f (equiv→unit eqv x) ≡ equiv→counit eqv (f x)
equiv→zig {f = f} eqv x i j = hcomp (∂ i ∨ ∂ j) λ where
  k (k = i0) → equiv→square eqv x j i
  k (i = i0) → f (equiv→unit eqv x j)
  k (i = i1) → equiv→counit eqv (f x) (j ∨ ~ k)
  k (j = i0) → equiv→counit eqv (f x) (i ∧ ~ k)
  k (j = i1) → f x

equiv→zag
  : ∀ {f : A → B} (eqv : is-equiv f) x
  → ap (equiv→inverse eqv) (equiv→counit eqv x)
  ≡ equiv→unit eqv (equiv→inverse eqv x)
equiv→zag {f = f} eqv b =
  subst (λ b → ap g (ε b) ≡ η (g b)) (ε b) (helper (g b)) where
  g = equiv→inverse eqv
  ε = equiv→counit eqv
  η = equiv→unit eqv

  helper : ∀ a → ap g (ε (f a)) ≡ η (g (f a))
  helper a i j = hcomp (∂ i ∨ ∂ j) λ where
    k (i = i0) → g (ε (f a) (j ∨ ~ k))
    k (i = i1) → η (η a (~ k)) j
    k (j = i0) → g (equiv→zig eqv a (~ i) (~ k))
    k (j = i1) → η a (i ∧ ~ k)
    k (k = i0) → η a (i ∧ j)

Finally, it’ll be convenient for us to package some of the theorems above into a proof that every equivalence is an isomorphism:

is-equiv→is-iso : {f : A → B} → is-equiv f → is-iso f
is-equiv→is-iso eqv .is-iso.inv  = equiv→inverse eqv
is-equiv→is-iso eqv .is-iso.rinv = equiv→counit eqv
is-equiv→is-iso eqv .is-iso.linv = equiv→unit eqv

Improving isomorphisms🔗

Having shown that being an equivalence is a proposition and implies being an isomorphism, we’re left with the trickiest part: showing that it’s implied by being an isomorphism. The argument presented here is natively cubical; we do everything in terms of lifting properties. There is a more “algebraic” version, factoring through the intermediate notion of half-adjoint equivalence, available on that page.

We’ll write $(f,g,s,t)$ for the the function $f:A\to B\text{,}$ its inverse $g:B\to A\text{,}$ and the two homotopies. Our construction definitionally preserves $f\text{,}$ $g$ and the homotopy $s:fg\sim \mathrm{id}\text{.}$ However, it can not preserve the proof $t:gf\sim \mathrm{id}\text{,}$ since that would imply that is-iso is a proposition, which, again, is provably not the case.

module _ {f : A → B} (i : is-iso f) where
  open is-iso i renaming (inv to g ; rinv to s ; linv to t)

We want to show that, for any $y\text{,}$ the fibre of $f$ over $y$ is contractible. According to our decomposition above, it will suffice to show that the fibre is propositional, and that it is inhabited. Inhabitation is easy — $(gy,sy)$ belongs to the fibre $f^{\ast }(y)\text{.}$ Let’s focus on propositionality.

  _ = PathP

Suppose that we have $y\text{,}$ $x_0\text{,}$ $x_1\text{,}$ $p_0$ and $p_1$ as below; Note that $(x_0,p_0)$ and $(x_1,p_1)$ are fibres of $f$ over $y\text{.}$ What we need to show is that we have some $\pi :x_0\equiv x_1$ and $p_0\equiv p_1$ over $\pi \text{.}$

  private module _ (y : B) (x0 x1 : A) (p0 : f x0 ≡ y) (p1 : f x1 ≡ y) where

As an intermediate step in proving that $p_0\equiv p_1\text{,}$ we must show that $x_0\equiv x_1$ — without this, we can’t even state that $p_0$ and $p_1$ are identified, since they live in different types! To this end, we will build $\pi :x_0\equiv x_1\text{,}$ parts of which will be required to assemble the overall proof that $p_0\equiv p_1\text{.}$

We’ll detail the construction of ${\pi }_0\text{;}$ for ${\pi }_1\text{,}$ the same method is used. We want to construct a line, which we can do by exhibiting that line as the missing face in a square. We have equations $gy\equiv gy$ (refl), $g(f{x}_0)\equiv gy$ (the action of g on p0), and $g(f{x}_0)=x_0$ by the assumption that $g$ is a right inverse to $f\text{.}$ Diagrammatically, these fit together into a square:

The missing line in this square is ${\pi }_0\text{.}$ Since the inside (the filler) will be useful to us later, we also give it a name: $\theta \text{.}$

    π₀ : g y ≡ x0
    π₀ i = hcomp (∂ i) λ where
      k (i = i0) → g y
      k (i = i1) → t x0 k
      k (k = i0) → g (p0 (~ i))

    θ₀ : Square (ap g (sym p0)) refl (t x0) π₀
    θ₀ i j = hfill (∂ i) j λ where
      k (i = i0) → g y
      k (i = i1) → t x0 k
      k (k = i0) → g (p0 (~ i))

Since the construction of ${\pi }_1$ is analogous, I’ll simply present the square. We correspondingly name the missing face ${\pi }_1$ and the filler ${\theta }_1\text{.}$

    π₁ : g y ≡ x1
    π₁ i = hcomp (∂ i) λ where
      j (i = i0) → g y
      j (i = i1) → t x1 j
      j (j = i0) → g (p1 (~ i))

    θ₁ : Square (ap g (sym p1)) refl (t x1) π₁
    θ₁ i j = hfill (∂ i) j λ where
      j (i = i0) → g y
      j (i = i1) → t x1 j
      j (j = i0) → g (p1 (~ i))

Joining these paths by their common $gy$ face, we obtain $\pi :x_0\equiv x_1\text{.}$ This square also has a filler, connecting ${\pi }_0$ and ${\pi }_1$ over the line $gy\equiv \pi i\text{.}$

    π : x0 ≡ x1
    π i = hcomp (∂ i) λ where
      j (j = i0) → g y
      j (i = i0) → π₀ j
      j (i = i1) → π₁ j

This concludes the construction of $\pi \text{,}$ and thus, the 2D part of the proof. Now, we want to show that $p_0\equiv p_1$ over a path induced by $\pi \text{.}$ This is a square with a specific boundary, which can be built by constructing an appropriate open cube, where the missing face is that square. As an intermediate step, we define $\theta$ to be the filler for the square above.

    θ : Square refl π₀ π₁ π
    θ i j = hfill (∂ i) j λ where
      k (i = i1) → π₁ k
      k (i = i0) → π₀ k
      k (k = i0) → g y

Observe that we can coherently alter $\theta$ to get $\iota$ below, which expresses that $\mathrm{ap}g{p}_0$ and $\mathrm{ap}g{p}_1$ are identified.

    ι : Square (ap (g ∘ f) π) (ap g p0) (ap g p1) refl
    ι i j = hcomp (∂ i ∨ ∂ j) λ where
      k (k = i0) → θ i (~ j)
      k (i = i0) → θ₀ (~ j) (~ k)
      k (i = i1) → θ₁ (~ j) (~ k)
      k (j = i0) → t (π i) (~ k)
      k (j = i1) → g y

This composition can be visualised as the red (front) face in the diagram below. The back face is $\theta i(\neg j)\text{,}$ i.e. (θ i (~ j)) in the code. Similarly, the $j=i1$ (bottom) face is g y, the $j=i0$ (top) face is t (π i) (~ k), and similarly for $i=i0$ (left) and $i=i1$ (right).

The fact that $j$ only appears as $\neg j$ can be understood as the diagram above being upside-down. Indeed, ${\pi }_0$ and ${\pi }_1$ in the boundary of $\theta$ (the inner, blue face) are inverted when their types are considered. We’re in the home stretch: Using our assumption $s:f(gx)=x\text{,}$ we can cancel all of the $f\circ g\text{s}$ in the diagram above to get what we wanted: $p_0\equiv p_1\text{.}$

    sq1 : Square (ap f π) p0 p1 refl
    sq1 i j = hcomp (∂ i ∨ ∂ j) λ where
      k (i = i0) → s (p0 j) k
      k (i = i1) → s (p1 j) k
      k (j = i0) → s (f (π i)) k
      k (j = i1) → s y k
      k (k = i0) → f (ι i j)

The composition above can be visualised as the front (red) face in the cubical diagram below. Once more, left is $i=i0\text{,}$ right is $i=i1\text{,}$ up is $j=i0\text{,}$ and down is $j=i1\text{.}$

Putting all of this together, we get that $(x_0,p_0)\equiv (x_1,p_1)\text{.}$ Since there were no assumptions on any of the variables under consideration, this indeed says that the fibre over $y$ is a proposition for any choice of $y\text{.}$

    is-iso→fibre-is-prop : (x0 , p0) ≡ (x1 , p1)
    is-iso→fibre-is-prop i .fst = π i
    is-iso→fibre-is-prop i .snd = sq1 i

Since the fibre over $y$ is inhabited by $(gy,sy)\text{,}$ we get that any isomorphism has contractible fibres:

  is-iso→is-equiv : is-equiv f
  is-iso→is-equiv .is-eqv y .centre .fst = g y
  is-iso→is-equiv .is-eqv y .centre .snd = s y
  is-iso→is-equiv .is-eqv y .paths z =
    is-iso→fibre-is-prop y (g y) (fst z) (s y) (snd z)

If we package this differently, then we can present it as a map between the types of isomorphisms $A\to B$ and equivalences $A\simeq B\text{.}$

Iso→Equiv : Iso A B → A ≃ B
Iso→Equiv (f , is-iso) = f , is-iso→is-equiv is-iso

inverse-is-equiv : {f : A → B} (eqv : is-equiv f) → is-equiv (equiv→inverse eqv)
inverse-is-equiv {f = f} eqv .is-eqv x .centre = record
  { fst = f x ; snd = equiv→unit eqv x }
inverse-is-equiv {A = A} {B = B} {f = f} eqv .is-eqv x .paths (y , p) = q where
  g = equiv→inverse eqv
  η = equiv→unit eqv
  ε = equiv→counit eqv
  zag = equiv→zag eqv

  q : (f x , η x) ≡ (y , p)
  q i .fst = (ap f (sym p) ∙ ε y) i
  q i .snd j = hcomp (∂ i ∨ ∂ j) λ where
    k (k = i0) → zag y j i
    k (i = i0) → η (p k) (j ∧ k)
    k (i = i1) → p (k ∧ j)
    k (j = i0) → g (∙-filler' (ap f (sym p)) (ε y) k i)
    k (j = i1) → η (p k) (i ∨ k)

module Equiv {ℓ ℓ'} {A : Type ℓ} {B : Type ℓ'} (f : A ≃ B) where
  to = f .fst
  from = equiv→inverse (f .snd)
  η    = equiv→unit (f .snd)
  ε    = equiv→counit (f .snd)
  zig  = equiv→zig (f .snd)
  zag  = equiv→zag (f .snd)

  injective : ∀ {x y} → to x ≡ to y → x ≡ y
  injective p = ap fst $ is-contr→is-prop (f .snd .is-eqv _) (_ , refl) (_ , sym p)

  injective₂ : ∀ {x y z} → to x ≡ z → to y ≡ z → x ≡ y
  injective₂ p q = ap fst $ is-contr→is-prop (f .snd .is-eqv _)
    (_ , p) (_ , q)

  inverse : B ≃ A
  inverse = from , inverse-is-equiv (f .snd)

  adjunctl : ∀ {x y} → to x ≡ y → x ≡ from y
  adjunctl p = sym (η _) ∙ ap from p

  adjunctr : ∀ {x y} → x ≡ from y → to x ≡ y
  adjunctr p = ap to p ∙ ε _

  open is-iso

  adjunct : ∀ {x y} → (to x ≡ y) ≃ (x ≡ from y)
  adjunct {x} {y} .fst = adjunctl
  adjunct {x} {y} .snd = is-iso→is-equiv λ where
    .inv    → adjunctr
    .rinv p → J (λ _ p → sym (η _) ∙ ap from (ap to (sym p) ∙ ε _) ≡ sym p)
      (sym (∙-swapl (∙-idr _ ∙ sym (zag _) ∙ sym (∙-idl _) ∙ sym (ap-∙ from _ _))))
      (sym p)
    .linv → J (λ _ p → ap to (sym (η _) ∙ ap from p) ∙ ε _ ≡ p)
      (sym (∙-swapr (∙-idl _ ∙ ap sym (sym (zig _)) ∙ sym (∙-idr _) ∙ sym (ap-∙ to _ _))))

module Iso {ℓ ℓ'} {A : Type ℓ} {B : Type ℓ'} ((f , f-iso) : Iso A B) where
  open is-iso f-iso renaming (inverse to inverse-iso)

  injective : ∀ {x y} → f x ≡ f y → x ≡ y
  injective p = sym (linv _) ·· ap inv p ·· linv _

  inverse : Iso B A
  inverse = inv , inverse-iso

Simple constructions🔗

Having established the three desiderata for a notion of equivalence, we will spend the rest of this module constructing readily-available equivalences, and establishing basic facts about them.

Contractible types🔗

If $A$ and $B$ are contractible types, then any function $f:A\to B$ must be homotopic to the function which sends everything in $A$ to the centre of $B\text{;}$ This function is invertible by sending everything in $B$ to the centre of $A\text{.}$ Therefore, any function between contractible types is an equivalence.

is-contr→is-equiv : is-contr A → is-contr B → {f : A → B} → is-equiv f
is-contr→is-equiv cA cB = is-iso→is-equiv f-is-iso where
  f-is-iso : is-iso _
  f-is-iso .is-iso.inv  _ = cA .centre
  f-is-iso .is-iso.rinv _ = is-contr→is-prop cB _ _
  f-is-iso .is-iso.linv _ = is-contr→is-prop cA _ _

Pairing this with the “canonical” function, we obtain an equivalence between any contractible types. Since the unit type is contractible, any contractible type is equivalent to the unit.

is-contr→≃ : is-contr A → is-contr B → A ≃ B
is-contr→≃ cA cB = (λ _ → cB .centre) , is-contr→is-equiv cA cB

is-contr→≃⊤ : is-contr A → A ≃ ⊤
is-contr→≃⊤ c = is-contr→≃ c ⊤-is-contr

Strictness of the empty type🔗

We say that an initial object is strict if every map into it is an equivalence. This holds of the empty type, and moreover, the proof is delightfully simple! Since showing that $f:A\to \perp$ is an equivalence boils down to showing something about $f$ given a point $y:\perp$, it’s immediate that any map into the empty type is an equivalence.

¬-is-equiv : (f : A → ⊥) → is-equiv f
¬-is-equiv f .is-eqv ()

is-empty→≃⊥ : ¬ A → A ≃ ⊥
is-empty→≃⊥ ¬a = _ , ¬-is-equiv ¬a

Propositional extensionality (redux)🔗

module _ {ℓ ℓ'} {P : Type ℓ} {Q : Type ℓ'} (pprop : is-prop P) (qprop : is-prop Q) where

Following the trend set by contractible types, above, another h-level for which constructing equivalences is easy are the propositions. If $P$ and $Q$ are propositions, then any map $P\to P$ (resp. $Q\to Q\text{)}$ must be homotopic to the identity, and consequently any pair of functions $P\to Q$ and $Q\to P$ is a pair of inverses. Put another way, any biimplication between propositions is an equivalence.

  biimp-is-equiv : (f : P → Q) → (Q → P) → is-equiv f
  biimp-is-equiv f g .is-eqv y .centre .fst = g y
  biimp-is-equiv f g .is-eqv y .centre .snd = qprop (f (g y)) y
  biimp-is-equiv f g .is-eqv y .paths (p' , path) = Σ-pathp
    (pprop (g y) p')
    (is-prop→squarep (λ _ _ → qprop) _ _ _ _)

  prop-ext : (P → Q) → (Q → P) → P ≃ Q
  prop-ext p→q q→p .fst = p→q
  prop-ext p→q q→p .snd = biimp-is-equiv p→q q→p

Groupoid operations🔗

module _ where
  open is-iso

Since types are higher groupoids, we have certain algebraic laws regarding the behaviour of path operations which can be expressed as saying that they form equivalences. First, the inverse path operation is definitionally involutive, so it’s its own two-sided inverse:

  sym-equiv : ∀ {ℓ} {A : Type ℓ} {x y : A} → (x ≡ y) ≃ (y ≡ x)
  sym-equiv .fst = sym
  sym-equiv .snd = is-iso→is-equiv (iso sym (λ _ → refl) (λ _ → refl))

If we have a path $p:x=y\text{,}$ then $p^{-1}pq=q\text{,}$ and $p{p}^{-1}r=r\text{.}$ Viewing this as a function of $q\text{,}$ it says that the operation precompose with $p$ is inverted on both sides by precomposition with $p^{-1}\text{.}$

  ∙-pre-equiv : ∀ {ℓ} {A : Type ℓ} {x y z : A} → x ≡ y → (y ≡ z) ≃ (x ≡ z)
  ∙-pre-equiv p .fst q = p ∙ q
  ∙-pre-equiv p .snd = is-iso→is-equiv λ where
    .inv q  → sym p ∙ q
    .rinv q → ∙-assoc p _ _       ·· ap (_∙ q) (∙-invr p) ·· ∙-idl q
    .linv q → ∙-assoc (sym p) _ _ ·· ap (_∙ q) (∙-invl p) ·· ∙-idl q

Similarly, postcomposition with $p$ is inverted on both sides by postcomposition with $p^{-1}\text{,}$ so it too is an equivalence.

  ∙-post-equiv : ∀ {ℓ} {A : Type ℓ} {x y z : A} → y ≡ z → (x ≡ y) ≃ (x ≡ z)
  ∙-post-equiv p .fst q = q ∙ p
  ∙-post-equiv p .snd = is-iso→is-equiv λ where
    .inv q  → q ∙ sym p
    .rinv q → sym (∙-assoc q _ _) ·· ap (q ∙_) (∙-invl p) ·· ∙-idr q
    .linv q → sym (∙-assoc q _ _) ·· ap (q ∙_) (∙-invr p) ·· ∙-idr q

The Lift type🔗

Because Agda’s universes are not cumulative, we can not freely move a type $A:{\mathrm{Type}}_0$ to conclude that $A:{\mathrm{Type}}_1\text{,}$ or to higher universes. To work around this, we have a Lift type, which, given a small type $A:{\mathrm{Type}}_i$ and some universe $j>i\text{,}$ gives us a name for $A$ in ${\mathrm{Type}}_j\text{.}$ To know that this operation is coherent, we can prove that the lifting map

is an equivalence: the name of $A$ in ${\mathrm{Type}}_j$ really is equivalent to the type we started with. Because Lift is a very well-behaved record type, the proof that this is an equivalence looks very similar to the proof that the identity function is an equivalence:

Lift-≃ : ∀ {a ℓ} {A : Type a} → Lift ℓ A ≃ A
Lift-≃ .fst (lift a) = a
Lift-≃ .snd .is-eqv a .centre = lift a , refl
Lift-≃ .snd .is-eqv a .paths (x , p) i .fst = lift (p (~ i))
Lift-≃ .snd .is-eqv a .paths (x , p) i .snd j = p (~ i ∨ j)

Finally, while we’re here, we can easily prove that the Lift type respects equivalences, in that if we have small $A\simeq B\text{,}$ then their liftings ${\mathrm{Lift}}_{j}A$ and ${\mathrm{Lift}}_{j}B$ will still be equivalent.

Lift-ap
  : ∀ {a b ℓ ℓ'} {A : Type a} {B : Type b}
  → A ≃ B → Lift ℓ A ≃ Lift ℓ' B
Lift-ap (f , eqv) .fst (lift x) = lift (f x)
Lift-ap f .snd .is-eqv (lift y) .centre = lift (Equiv.from f y) , ap lift (Equiv.ε f y)
Lift-ap f .snd .is-eqv (lift y) .paths (lift x , q) i = lift (p i .fst) , λ j → lift (p i .snd j)
  where p = f .snd .is-eqv y .paths (x , ap Lift.lower q)

Involutions🔗

Finally, we can show here that any involution — a function $A\to A\text{,}$ which inverts itself — is an equivalence.

is-involutive→is-equiv : {f : A → A} → (∀ a → f (f a) ≡ a) → is-equiv f
is-involutive→is-equiv inv = is-iso→is-equiv (iso _ inv inv)

Closure properties🔗

We will now show a rather fundamental property of equivalences: they are closed under two-out-of-three. This means that, considering $f:A\to B\text{,}$ $g:B\to C\text{,}$ and $(g\circ f):A\to C$ as a set of three things, if any two are an equivalence, then so is the third:

module _ {ℓ ℓ₁ ℓ₂} {A : Type ℓ} {B : Type ℓ₁} {C : Type ℓ₂} {f : A → B} {g : B → C} where

  ∙-is-equiv    : is-equiv f → is-equiv g → is-equiv (g ∘ f)
  equiv-cancell : is-equiv g → is-equiv (g ∘ f) → is-equiv f
  equiv-cancelr : is-equiv f → is-equiv (g ∘ f) → is-equiv g

We have already shown the first of these, when the individual functions are equivalences, since being an equivalence is logically equivalent to being an isomorphism. Supposing that $g$ and $g\circ f$ are, then we can show that

is an inverse to $f\text{;}$ the other case is symmetric. Since both of the proofs are just calculations, we will not comment on them.

  ∙-is-equiv ef eg = is-iso→is-equiv (∘-is-iso (is-equiv→is-iso eg) (is-equiv→is-iso ef))

  equiv-cancell eg egf = is-iso→is-equiv (iso inv right left) where
    inv : B → A
    inv x = equiv→inverse egf (g x)
    opaque
      right : is-right-inverse inv f
      right x =
        f (equiv→inverse egf (g x))                        ≡˘⟨ equiv→unit eg _ ⟩≡˘
        equiv→inverse eg (g (f (equiv→inverse egf (g x)))) ≡⟨ ap (equiv→inverse eg) (equiv→counit egf _) ⟩≡
        equiv→inverse eg (g x)                             ≡⟨ equiv→unit eg _ ⟩≡
        x                                                  ∎
      left : is-left-inverse inv f
      left x = equiv→unit egf x

  equiv-cancelr ef egf = is-iso→is-equiv (iso inv right left) where
    inv : C → B
    inv x = f (equiv→inverse egf x)
    right : is-right-inverse inv g
    right x = equiv→counit egf x
    left : is-left-inverse inv g
    left x =
      f (equiv→inverse egf (g x))                        ≡˘⟨ ap (f ∘ equiv→inverse egf ∘ g) (equiv→counit ef _) ⟩≡˘
      f (equiv→inverse egf (g (f (equiv→inverse ef x)))) ≡⟨ ap f (equiv→unit egf _) ⟩≡
      f (equiv→inverse ef x)                             ≡⟨ equiv→counit ef _ ⟩≡
      x                                                  ∎

Specialising these, any left- or right- inverse of an equivalence must be homotopic to the specified one, so that it too is an equivalence.

left-inverse→equiv
  : {f : A → B} {g : B → A}
  → is-left-inverse g f → is-equiv f → is-equiv g
left-inverse→equiv linv ef = equiv-cancelr ef
  (subst is-equiv (sym (funext linv)) id-equiv)

right-inverse→equiv
  : {f : A → B} {g : B → A}
  → is-right-inverse g f → is-equiv f → is-equiv g
right-inverse→equiv rinv ef = equiv-cancell ef
  (subst is-equiv (sym (funext rinv)) id-equiv)

Equivalence reasoning🔗

To simplify working with chains of equivalences, we can re-package their closure operations as mixfix operators, so that we have a shorthand notation for working with composites of pairs of equivalences and long chains of equivalences, where making the intermediate steps explicit is more important.

id≃ : ∀ {ℓ} {A : Type ℓ} → A ≃ A
id≃ = id , id-equiv

_∙e_ : A ≃ B → B ≃ C → A ≃ C
_∙e_ (f , ef) (g , eg) = g ∘ f , ∙-is-equiv ef eg

_e⁻¹ : A ≃ B → B ≃ A
((f , ef) e⁻¹) = equiv→inverse ef , inverse-is-equiv ef

≃⟨⟩-syntax : ∀ {ℓ ℓ₁ ℓ₂} (A : Type ℓ) {B : Type ℓ₁} {C : Type ℓ₂}
           → B ≃ C → A ≃ B → A ≃ C
≃⟨⟩-syntax A g f = f ∙e g

_≃˘⟨_⟩_ : ∀ {ℓ ℓ₁ ℓ₂} (A : Type ℓ) {B : Type ℓ₁} {C : Type ℓ₂}
        → B ≃ A → B ≃ C → A ≃ C
A ≃˘⟨ f ⟩≃˘ g = f e⁻¹ ∙e g

_≃⟨⟩_ : ∀ {ℓ ℓ₁} (A : Type ℓ) {B : Type ℓ₁} → A ≃ B → A ≃ B
x ≃⟨⟩ x≡y = x≡y

_≃∎ : ∀ {ℓ} (A : Type ℓ) → A ≃ A
x ≃∎ = id≃

infixr  _∙e_
infix  _e⁻¹

infixr  ≃⟨⟩-syntax _≃⟨⟩_ _≃˘⟨_⟩_
infix   _≃∎
infix  _≃_

syntax ≃⟨⟩-syntax x q p = x ≃⟨ p ⟩ q

lift-inj
  : ∀ {ℓ ℓ'} {A : Type ℓ} {a b : A}
  → lift {ℓ = ℓ'} a ≡ lift {ℓ = ℓ'} b → a ≡ b
lift-inj p = ap Lift.lower p

-- Fibres of composites are given by pairs of fibres.
fibre-∘-≃
  : ∀ {ℓ ℓ' ℓ''} {A : Type ℓ} {B : Type ℓ'} {C : Type ℓ''}
  → {f : B → C} {g : A → B}
  → ∀ c → fibre (f ∘ g) c ≃ (Σ[ (b , _) ∈ fibre f c ] (fibre g b))
fibre-∘-≃ {f = f} {g = g} c = Iso→Equiv (fwd , iso bwd invl invr)
    where
      fwd : fibre (f ∘ g) c → Σ[ (b , _) ∈ fibre f c ] (fibre g b)
      fwd (a , p) = ((g a) , p) , (a , refl)

      bwd : Σ[ (b , _) ∈ fibre f c ] (fibre g b) → fibre (f ∘ g) c
      bwd ((b , p) , (a , q)) = a , ap f q ∙ p

      invl : ∀ x → fwd (bwd x) ≡ x
      invl ((b , p) , (a , q)) i .fst .fst = q i
      invl ((b , p) , (a , q)) i .fst .snd j =
        hcomp (∂ i ∨ ∂ j) λ where
          k (i = i0) → ∙-filler (ap f q) p k j
          k (i = i1) → p (j ∧ k)
          k (j = i0) → f (q i)
          k (j = i1) → p k
          k (k = i0) → f (q (i ∨ j))
      invl ((b , p) , a , q) i .snd .fst = a
      invl ((b , p) , a , q) i .snd .snd j = q (i ∧ j)

      invr : ∀ x → bwd (fwd x) ≡ x
      invr (a , p) i .fst = a
      invr (a , p) i .snd = ∙-idl p i