blog/content/posts/2023-05-06-equivalences.lagda.md

+++ title = "Equivalences" slug = "equivalences" date = 2023-05-06 tags = ["type-theory", "agda", "hott"] math = true draft = true +++

Imports

{-# OPTIONS --cubical #-}

open import Agda.Primitive.Cubical
open import Cubical.Foundations.Equiv
open import Cubical.Foundations.Prelude
open import Data.Bool

Bool-id : Bool → Bool
Bool-id true = true
Bool-id false = false

unap : {A B : Type} {x y : A} (f : A → B) → f x ≡ f y → x ≡ y
unap p i = ?

-- Need to convert point-wise equality into universally quantified equality?
Bool-id-refl : (x : Bool) → (Bool-id x ≡ x)
Bool-id-refl true = refl
Bool-id-refl false = refl

The equivalence proof below involves the contractibility-of-fibers definition of an equivalence. There are others, but the "default" one used by the Cubical standard library uses this.

Bool-id-is-equiv : isEquiv Bool-id

In the contractibility-of-fibers proof, we must first establish our fibers. If we had (f : A \rightarrow B), then this is saying given any (y : B), we must provide:

• an (x : A) that would have gotten mapped to y (preimage), and
• a proof that f\ x \equiv y

These are the two elements of the pair given below. Since our function is id, we can just give y again, and use the refl function above for the equality proof

Bool-id-is-equiv .equiv-proof y .fst = y , Bool-id-refl y

The next step is to prove that it's contractible. Using the same derivation for y as above, this involves taking in another fiber y_1, and proving that it's equivalent the fiber we've just defined above.

To prove fiber equality, we can just do point-wise equality over both the preimage of y, and then the second-level equality of the proof of $f\ x \equiv y$.

In the first case here, we need to provide something that equals our x above when i = i0, and something that equals the fiber $y_1$'s preimage x_1 when i = i1, aka y \equiv proj_1\ y_1.

Bool-id-is-equiv .equiv-proof y .snd y₁ i .fst =
  let
    eqv = snd y₁
    -- eqv : Bool-id (fst y₁) ≡ y

    eqv2 = eqv ∙ sym (Bool-id-refl y)
    -- eqv2 : Bool-id (fst y₁) ≡ Bool-id y

    -- Ok, unap doesn't actually exist unless f is known to have an inverse.
    -- Fortunately, because we're proving an equivalence, we know that f has an
    -- inverse, in particular going from y to x, which in thise case is also y.
    eqv3 = unap Bool-id eqv2

    Bool-id-inv : Bool → Bool
    Bool-id-inv b = (((Bool-id-is-equiv .equiv-proof) b) .fst) .fst

    eqv3′ = cong Bool-id-inv eqv2
    give-me-info = ?
    -- eqv3 : fst y₁ ≡ y

    eqv4 = sym eqv3
    -- eqv4 : y ≡ fst y₁
  in
  eqv4 i

Now we can prove that the path is the same

\begin{CD} A @> > > B \\ @VVV @VVV \\ C @> > > D \end{CD}

• A \rightarrow B is the path of the original fiber that we've specified, which is f\ x \equiv y
• C \rightarrow D is the path of the other fiber that we're proving, which is proj_2\ y_1

So what we want now is a-b ≡ c-d

Bool-id-is-equiv .equiv-proof y .snd y₁ i .snd j =
  let
    a-b = Bool-id-refl y
    c-d = y₁ .snd
  in
  ?