Michael Zhang
28025a8c66
All checks were successful
ci/woodpecker/push/woodpecker Pipeline was successful
121 lines
3.1 KiB
Markdown
121 lines
3.1 KiB
Markdown
+++
|
||
title = "Equivalences"
|
||
slug = "equivalences"
|
||
date = 2023-05-06
|
||
tags = ["type-theory", "agda", "hott"]
|
||
math = true
|
||
draft = true
|
||
+++
|
||
|
||
<details>
|
||
<summary>Imports</summary>
|
||
|
||
```
|
||
{-# OPTIONS --cubical #-}
|
||
|
||
open import Agda.Primitive.Cubical
|
||
open import Cubical.Foundations.Equiv
|
||
open import Cubical.Foundations.Prelude
|
||
open import Data.Bool
|
||
```
|
||
|
||
</details>
|
||
|
||
```
|
||
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
|
||
?
|
||
```
|