121 lines
3.1 KiB
Markdown
121 lines
3.1 KiB
Markdown
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title = "Equivalences"
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slug = "equivalences"
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date = 2023-05-06
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tags = ["type-theory", "agda", "hott"]
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math = true
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draft = true
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+++
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<details>
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<summary>Imports</summary>
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```
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{-# OPTIONS --cubical #-}
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open import Agda.Primitive.Cubical
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open import Cubical.Foundations.Equiv
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open import Cubical.Foundations.Prelude
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open import Data.Bool
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```
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</details>
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```
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Bool-id : Bool → Bool
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Bool-id true = true
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Bool-id false = false
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unap : {A B : Type} {x y : A} (f : A → B) → f x ≡ f y → x ≡ y
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unap p i = ?
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-- Need to convert point-wise equality into universally quantified equality?
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Bool-id-refl : (x : Bool) → (Bool-id x ≡ x)
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Bool-id-refl true = refl
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Bool-id-refl false = refl
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```
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The equivalence proof below involves the contractibility-of-fibers definition of
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an equivalence. There are others, but the "default" one used by the Cubical
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standard library uses this.
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```
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Bool-id-is-equiv : isEquiv Bool-id
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```
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In the contractibility-of-fibers proof, we must first establish our fibers. If
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we had $(f : A \rightarrow B)$, then this is saying given any $(y : B)$, we must
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provide:
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- an $(x : A)$ that would have gotten mapped to $y$ (preimage), and
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- a proof that $f\ x \equiv y$
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These are the two elements of the pair given below. Since our function is `id`,
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we can just give $y$ again, and use the `refl` function above for the equality
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proof
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```
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Bool-id-is-equiv .equiv-proof y .fst = y , Bool-id-refl y
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```
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The next step is to prove that it's contractible. Using the same derivation for
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$y$ as above, this involves taking in another fiber $y_1$, and proving that it's
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equivalent the fiber we've just defined above.
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To prove fiber equality, we can just do point-wise equality over both the
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preimage of $y$, and then the second-level equality of the proof of $f\ x \equiv
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y$.
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In the first case here, we need to provide something that equals our $x$ above
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when $i = i0$, and something that equals the fiber $y_1$'s preimage $x_1$ when
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$i = i1$, aka $y \equiv proj_1\ y_1$.
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```
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Bool-id-is-equiv .equiv-proof y .snd y₁ i .fst =
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let
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eqv = snd y₁
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-- eqv : Bool-id (fst y₁) ≡ y
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eqv2 = eqv ∙ sym (Bool-id-refl y)
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-- eqv2 : Bool-id (fst y₁) ≡ Bool-id y
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-- Ok, unap doesn't actually exist unless f is known to have an inverse.
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-- Fortunately, because we're proving an equivalence, we know that f has an
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-- inverse, in particular going from y to x, which in thise case is also y.
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eqv3 = unap Bool-id eqv2
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Bool-id-inv : Bool → Bool
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Bool-id-inv b = (((Bool-id-is-equiv .equiv-proof) b) .fst) .fst
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eqv3′ = cong Bool-id-inv eqv2
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give-me-info = ?
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-- eqv3 : fst y₁ ≡ y
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eqv4 = sym eqv3
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-- eqv4 : y ≡ fst y₁
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in
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eqv4 i
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```
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Now we can prove that the path is the same
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\begin{CD}
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A @> > > B \\\
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@VVV @VVV \\\
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C @> > > D
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\end{CD}
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- $A \rightarrow B$ is the path of the original fiber that we've specified, which is $f\ x \equiv y$
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- $C \rightarrow D$ is the path of the other fiber that we're proving, which is $proj_2\ y_1$
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So what we want now is `a-b ≡ c-d`
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```
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Bool-id-is-equiv .equiv-proof y .snd y₁ i .snd j =
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let
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a-b = Bool-id-refl y
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c-d = y₁ .snd
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in
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?
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```
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