open import 1Lab.Path hiding (_∙_)
open import 1Lab.Type

module 1Lab.Path.Groupoid where

_ = Path
_ = hfill
_ = ap-∙

Types are groupoids🔗

The Path types equip every Type with the structure of an $\infty \text{-groupoid}$. The higher structure of a type begins with its inhabitants (the 0-cells); Then, there are the paths between inhabitants - these are inhabitants of the type Path A x y, which are the 1-cells in A. Then, we can consider the inhabitants of Path (Path A x y) p q, which are homotopies between paths.

This construction iterates forever - and, at every stage, we have that the $n\text{-cells}$ in A are the 0-cells of some other type (e.g.: The 1-cells of A are the 0-cells of Path A ...). Furthermore, this structure is weak: The laws, e.g. associativity of composition, only hold up to a homotopy. These laws satisfy their own laws — again using associativity as an example, the associator satisfies the pentagon identity.

These laws can be proved in two ways: Using path induction, or directly with a cubical argument. Here, we do both.

Book-style🔗

This is the approach taken in the HoTT book. We fix a type, and some variables of that type, and some paths between variables of that type, so that each definition doesn’t start with 12 parameters.

module WithJ where
  private variable
    ℓ : Level
    A : Type ℓ
    w x y z : A

First, we (re)define the operations using J. These will be closer to the structure given in the book.

  _∙_ : x ≡ y → y ≡ z → x ≡ z
  _∙_ {x = x} {y} {z} = J (λ y _ → y ≡ z → x ≡ z) (λ x → x)

First we define path composition. Then, we can prove that the identity path - refl - acts as an identity for path composition.

  ∙-idr : (p : x ≡ y) → p ∙ refl ≡ p
  ∙-idr {x = x} {y = y} p =
    J (λ _ p → p ∙ refl ≡ p)
      (happly (J-refl (λ y _ → y ≡ y → x ≡ y) (λ x → x)) _)
      p

This isn’t as straightforward as it would be in “Book HoTT” because - remember - J doesn’t compute definitionally, only up to the path J-refl. Now the other identity law:

  ∙-idl : (p : y ≡ z) → refl ∙ p ≡ p
  ∙-idl {y = y} {z = z} p = happly (J-refl (λ y _ → y ≡ z → y ≡ z) (λ x → x)) p

This case we get for less since it’s essentially the computation rule for J.

  ∙-assoc : (p : w ≡ x) (q : x ≡ y) (r : y ≡ z)
          → p ∙ (q ∙ r) ≡ (p ∙ q) ∙ r
  ∙-assoc {w = w} {x = x} {y = y} {z = z} p q r =
    J (λ x p → (q : x ≡ y) (r : y ≡ z) → p ∙ (q ∙ r) ≡ (p ∙ q) ∙ r) lemma p q r
    where
      lemma : (q : w ≡ y) (r : y ≡ z)
            → (refl ∙ (q ∙ r)) ≡ ((refl ∙ q) ∙ r)
      lemma q r =
        (refl ∙ (q ∙ r)) ≡⟨ ∙-idl (q ∙ r) ⟩≡
        q ∙ r            ≡⟨ sym (ap (λ e → e ∙ r) (∙-idl q)) ⟩≡
        (refl ∙ q) ∙ r   ∎

The associativity rule is trickier to prove, since we do inductive where the motive is a dependent product. What we’re doing can be summarised using words: By induction, it suffices to assume p is refl. Then, what we want to show is (refl ∙ (q ∙ r)) ≡ ((refl ∙ q) ∙ r). But both of those compute to q ∙ r, so we are done. This computation isn’t automatic - it’s expressed by the lemma.

This expresses that the paths behave like morphisms in a category. For a groupoid, we also need inverses and cancellation:

  inv : x ≡ y → y ≡ x
  inv {x = x} = J (λ y _ → y ≡ x) refl

The operation which assigns inverses has to be involutive, which follows from two computations.

  inv-inv : (p : x ≡ y) → inv (inv p) ≡ p
  inv-inv {x = x} =
    J (λ y p → inv (inv p) ≡ p)
      (ap inv (J-refl (λ y _ → y ≡ x) refl) ∙ J-refl (λ y _ → y ≡ x) refl)

And we have to prove that composing with an inverse gives the reflexivity path.

  ∙-invr : (p : x ≡ y) → p ∙ inv p ≡ refl
  ∙-invr {x = x} = J (λ y p → p ∙ inv p ≡ refl)
                     (∙-idl (inv refl) ∙ J-refl (λ y _ → y ≡ x) refl)

  ∙-invl : (p : x ≡ y) → inv p ∙ p ≡ refl
  ∙-invl {x = x} = J (λ y p → inv p ∙ p ≡ refl)
                     (∙-idr (inv refl) ∙ J-refl (λ y _ → y ≡ x) refl)

Cubically🔗

Now we do the same using hfill instead of path induction.

module _ where
  private variable
    ℓ : Level
    A B : Type ℓ
    w x y z : A
    a b c d : A

  open 1Lab.Path

  ∙-filler₂ : ∀ {ℓ} {A : Type ℓ} {x y z : A} (q : x ≡ y) (r : y ≡ z)
            → Square q (q ∙ r) r refl
  ∙-filler₂ q r k i = hcomp (k ∨ ∂ i) λ where
    l (l = i0) → q (i ∨ k)
    l (k = i1) → r (l ∧ i)
    l (i = i0) → q k
    l (i = i1) → r l

The left and right identity laws follow directly from the two fillers for the composition operation.

  ∙-idr : (p : x ≡ y) → p ∙ refl ≡ p
  ∙-idr p = sym (∙-filler p refl)

  ∙-idl : (p : x ≡ y) → refl ∙ p ≡ p
  ∙-idl p = sym (∙-filler' refl p)

For associativity, we use both:

  ∙-assoc : (p : w ≡ x) (q : x ≡ y) (r : y ≡ z)
          → p ∙ (q ∙ r) ≡ (p ∙ q) ∙ r
  ∙-assoc p q r i = ∙-filler p q i ∙ ∙-filler' q r (~ i)

For cancellation, we need to sketch an open cube where the missing square expresses the equation we’re looking for. Thankfully, we only have to do this once!

  ∙-invr : ∀ {x y : A} (p : x ≡ y) → p ∙ sym p ≡ refl
  ∙-invr {x = x} p i j = hcomp (∂ j ∨ i) λ where
    k (k = i0) → p (j ∧ ~ i)
    k (i = i1) → x
    k (j = i0) → x
    k (j = i1) → p (~ k ∧ ~ i)

For the other direction, we use the fact that p is definitionally equal to sym (sym p). In that case, we show that sym p ∙ sym (sym p) ≡ refl - which computes to the thing we want!

  ∙-invl : (p : x ≡ y) → sym p ∙ p ≡ refl
  ∙-invl p = ∙-invr (sym p)

In addition to the groupoid identities for paths in a type, it has been established that functions behave like functors: Other than the lemma ap-∙, preservation of reflexivity and of inverses is definitional.

Convenient helpers🔗

Since a lot of Homotopy Type Theory is dealing with paths, this section introduces useful helpers for dealing with $n\text{-ary}$ compositions. For instance, we know that $p^{-1}\bullet p\bullet q$ is $q\text{,}$ but this involves more than a handful of intermediate steps:

  ∙-cancell
    : ∀ {ℓ} {A : Type ℓ} {x y z : A} (p : x ≡ y) (q : y ≡ z)
    → (sym p ∙ p ∙ q) ≡ q
  ∙-cancell {y = y} p q i j = hcomp (∂ i ∨ ∂ j) λ where
    k (k = i0) → p (i ∨ ~ j)
    k (i = i0) → ∙-filler (sym p) (p ∙ q) k j
    k (i = i1) → q (j ∧ k)
    k (j = i0) → y
    k (j = i1) → ∙-filler₂ p q i k

  ∙-cancelr
    : ∀ {ℓ} {A : Type ℓ} {x y z : A} (p : x ≡ y) (q : z ≡ y)
    → ((p ∙ sym q) ∙ q) ≡ p
  ∙-cancelr {x = x} {y = y} q p = sym $ ∙-unique _ λ i j →
    ∙-filler q (sym p) (~ i) j

  commutes→square
    : {p : w ≡ x} {q : w ≡ y} {s : x ≡ z} {r : y ≡ z}
    → p ∙ s ≡ q ∙ r
    → Square p q s r
  commutes→square {p = p} {q} {s} {r} fill i j =
    hcomp (∂ i ∨ ∂ j) λ where
      k (k = i0) → fill j i
      k (i = i0) → q (k ∧ j)
      k (i = i1) → s (~ k ∨ j)
      k (j = i0) → ∙-filler p s (~ k) i
      k (j = i1) → ∙-filler₂ q r k i

  square→commutes
    : {p : w ≡ x} {q : w ≡ y} {s : x ≡ z} {r : y ≡ z}
    → Square p q s r → p ∙ s ≡ q ∙ r
  square→commutes {p = p} {q} {s} {r} fill i j = hcomp (∂ i ∨ ∂ j) λ where
    k (k = i0) → fill j i
    k (i = i0) → ∙-filler p s k j
    k (i = i1) → ∙-filler₂ q r (~ k) j
    k (j = i0) → q (~ k ∧ i)
    k (j = i1) → s (k ∨ i)

  ∙-cancel'-l : {x y z : A} (p : x ≡ y) (q r : y ≡ z)
              → p ∙ q ≡ p ∙ r → q ≡ r
  ∙-cancel'-l p q r sq = sym (∙-cancell p q) ·· ap (sym p ∙_) sq ·· ∙-cancell p r

  ∙-cancel'-r : {x y z : A} (p : y ≡ z) (q r : x ≡ y)
              → q ∙ p ≡ r ∙ p → q ≡ r
  ∙-cancel'-r p q r sq =
       sym (∙-cancelr q (sym p))
    ·· ap (_∙ sym p) sq
    ·· ∙-cancelr r (sym p)

While connections give us degenerate squares where two adjacent faces are some path and the other two are refl, it is sometimes also useful to have a degenerate square with two pairs of equal adjacent faces. We can build this using hcomp and more connections:

  double-connection
    : (p : a ≡ b) (q : b ≡ c)
    → Square p p q q
  double-connection {b = b} p q i j = hcomp (∂ i ∨ ∂ j) λ where
    k (k = i0) → b
    k (i = i0) → p (j ∨ ~ k)
    k (i = i1) → q (j ∧ k)
    k (j = i0) → p (i ∨ ~ k)
    k (j = i1) → q (i ∧ k)

This corresponds to the following diagram, which expresses the trivial equation $p\bullet q\equiv p\bullet q\text{:}$

Groupoid structure of types (cont.)🔗

A useful fact is that if $H$ is a homotopy $f\sim \mathrm{id}\text{,}$ then it “commutes with $f$”, in the following sense:

open 1Lab.Path

homotopy-invert
  : ∀ {a} {A : Type a} {f : A → A}
  → (H : (x : A) → f x ≡ x) {x : A}
  → H (f x) ≡ ap f (H x)

The proof proceeds entirely using $H$ itself, together with the De Morgan algebra structure on the interval.

homotopy-invert {f = f} H {x} i j = hcomp (∂ i ∨ ∂ j) λ where
  k (k = i0) → H x       (j ∧ i)
  k (j = i0) → H (f x)   (~ k)
  k (j = i1) → H x       (~ k ∧ i)
  k (i = i0) → H (f x)   (~ k ∨ j)
  k (i = i1) → H (H x j) (~ k)