Add some notes
This commit is contained in:
parent
8466f79ecc
commit
565a7b19d9
3 changed files with 305 additions and 19 deletions
|
|
@ -1,8 +1,19 @@
|
|||
-- {{{
|
||||
|
||||
open import Agda.Builtin.Equality
|
||||
open import Agda.Builtin.Sigma
|
||||
open import Agda.Primitive using (Level; lzero; lsuc; _⊔_) renaming (Set to Type)
|
||||
open import Data.Nat
|
||||
open import Data.Product
|
||||
open import Data.Product.Base
|
||||
open import Data.Product.Properties
|
||||
open import Data.Bool
|
||||
|
||||
open import Data.Product using (_,_)
|
||||
open import Function.Base using (_∘_)
|
||||
open import Relation.Binary.Core
|
||||
open import Relation.Binary.Definitions
|
||||
open import Relation.Nullary.Negation.Core using (¬_)
|
||||
|
||||
import Relation.Binary.PropositionalEquality as Eq
|
||||
open Eq using (_≡_; refl; trans; sym; cong; cong-app; subst)
|
||||
|
|
@ -71,6 +82,23 @@ apd {A} {B} f {x} {y} p =
|
|||
in
|
||||
path-ind D d x y p
|
||||
|
||||
-- Fibers
|
||||
fiber : {A B : Type} → (f : A → B) → (y : B) → Type
|
||||
fiber {A} {B} f y = Σ[ x ∈ A ] f x ≡ y
|
||||
|
||||
-- Contractibility
|
||||
isContr : Type → Type
|
||||
isContr A = Σ[ x ∈ A ] (∀ y → x ≡ y)
|
||||
|
||||
-- Equivalence
|
||||
record isEquiv {A B : Type} (f : A → B) : Type where
|
||||
field
|
||||
equiv-proof : (y : B) → isContr (fiber f y)
|
||||
open isEquiv
|
||||
|
||||
_≃_ : (A B : Type) → Type
|
||||
A ≃ B = Σ[ f ∈ (A → B) ] (isEquiv f)
|
||||
|
||||
-- }}}
|
||||
|
||||
-------------------------------------------------------------------------------
|
||||
|
|
@ -101,20 +129,13 @@ l212-obv-proof-2 {A} x y z p q =
|
|||
in
|
||||
r x p
|
||||
|
||||
l212-obv-proof-3 : {A : Type} → lemma-212
|
||||
l212-obv-proof-3 {A} x y z p q =
|
||||
l212-pairwise-1-2 : {A : Type} → (x y z : A) → (x ≡ y) → (y ≡ z)
|
||||
→ l212-obv-proof-1 ≡ l212-obv-proof-2
|
||||
l212-pairwise-1-2 {A} x y z p q =
|
||||
let
|
||||
f : (y z : A) → (q : y ≡ z) → (x : A) → x ≡ y → x ≡ z
|
||||
f = ?
|
||||
r = f y z q
|
||||
|
||||
g : (x y : A) → (p : x ≡ y) → (x ≡ z)
|
||||
g = path-ind
|
||||
(λ a b a≡b → a ≡ z)
|
||||
(λ a → r a ?)
|
||||
s = g x y p
|
||||
in
|
||||
?
|
||||
left-proof = l212-obv-proof-1 x y z p q
|
||||
right-proof = l212-obv-proof-2 x y z p q
|
||||
in ?
|
||||
|
||||
-- Exercise 2.3: Give a fourth, different proof of Lemma 2.1.2, and prove it
|
||||
-- equal to the others
|
||||
|
|
@ -132,9 +153,44 @@ l212-obv-proof-4 {A} x y z p q =
|
|||
-- Exercise 2.4: Define, by induction on n, a general notion of n-dimensional
|
||||
-- path in a type A, simultaneously with the type of boundaries of such paths
|
||||
|
||||
data nPath {A : Type} : ℕ → Type where
|
||||
zeroPath : (x y : A) → (p : x ≡ y) → nPath zero
|
||||
sucPath : {n : ℕ} → (nPath n)
|
||||
-- p : (2 ≡ 1 + 1)
|
||||
-- p = refl
|
||||
--
|
||||
-- q : (2 ≡ 1 + 1)
|
||||
-- q = ?
|
||||
--
|
||||
-- p2 : p ≡ q
|
||||
-- q2 : p ≡ q
|
||||
--
|
||||
-- p3 : p2 ≡ q2
|
||||
--
|
||||
-- nSuc- npath 5
|
||||
|
||||
-- data nPath {A : Type} : ℕ → Type where
|
||||
-- zeroPath : (x y : A) → (p : x ≡ y) → nPath zero
|
||||
-- sucPath : {n : ℕ}
|
||||
-- → (left : nPath n)
|
||||
-- → (right : nPath n)
|
||||
-- → (p : left ≡ right)
|
||||
-- → nPath (suc n)
|
||||
|
||||
-- data nPath : (A : Type) → (x y : A) → Type where
|
||||
-- path-zero : (A : Type) → (x y : A) → (p : x ≡ y) → nPath A x y
|
||||
-- path-suc : (A : Type) → (x y : A)
|
||||
-- → (left : nPath A x y)
|
||||
-- → (right : nPath A x y)
|
||||
-- → (p : left ≡ right)
|
||||
-- → nPath (nPath A x y) left right
|
||||
|
||||
data nPath : Type where
|
||||
|
||||
-- p : nPath ℕ (1 + 1) 2
|
||||
-- p = path-zero ℕ (1 + 1) 2 refl
|
||||
--
|
||||
-- q : nPath ℕ (1 + 1) 2
|
||||
-- q = path-zero ℕ (1 + 1) 2 refl
|
||||
--
|
||||
-- p2 : nPath (nPath ℕ (1 + 1) 2) p q
|
||||
|
||||
-- Exercise 2.5: Prove that the functions (2.3.6) and (2.3.7) are
|
||||
-- inverse-equivalences
|
||||
|
|
@ -153,7 +209,88 @@ f237 {p = p} {f = f} _ = ap f p
|
|||
-- Exercise 2.6: Prove that if p : x = y, then the function (p ∙ –) : (y = z) →
|
||||
-- (x = z) is an equivalence.
|
||||
|
||||
e26 : {A : Type} {x y : A} → (p : x ≡ y) → ?
|
||||
|
||||
-- Exercise 2.9. Prove that coproducts have the expected universal property,
|
||||
-- (A + B → X) ≃ (A → X) × (B → X).
|
||||
|
||||
-- Exercise 2.13
|
||||
-- Show that (2 ≃ 2) ≃ 2
|
||||
|
||||
-- The fiber of id for the Bool y (aka all x's such that id x ≡ y, which is just
|
||||
-- y)
|
||||
id-fiber-base : (y : Bool) → Σ[ x ∈ Bool ] id x ≡ y
|
||||
proj₁ (id-fiber-base y) = y
|
||||
proj₂ (id-fiber-base y) = refl
|
||||
|
||||
id-fiber-is-contr : (y : Bool) → (y₁ : Σ[ x ∈ Bool ] id x ≡ y) → id-fiber-base y ≡ y₁
|
||||
id-fiber-is-contr y y₁ =
|
||||
let
|
||||
y₁-Bool-equiv = snd y₁
|
||||
|
||||
pair₁ : y ≡ fst y₁
|
||||
pair₁ = sym y₁-Bool-equiv
|
||||
|
||||
snd-fiber : id (fst (id-fiber-base y)) ≡ y
|
||||
snd-fiber = snd (id-fiber-base y)
|
||||
|
||||
pair₂ : subst (λ x → id x ≡ y) pair₁ (snd (id-fiber-base y)) ≡ snd y₁
|
||||
pair₂ =
|
||||
begin
|
||||
subst (λ x → id x ≡ y) pair₁ snd-fiber
|
||||
≡⟨ ? ⟩
|
||||
-- subst : Substitutive {A = A} _≡_ ℓ
|
||||
-- subst P refl p = p
|
||||
-- snd-fiber
|
||||
?
|
||||
≡⟨ ? ⟩
|
||||
snd y₁
|
||||
∎
|
||||
|
||||
pair = pair₁ , pair₂
|
||||
in Σ-≡,≡→≡ pair
|
||||
|
||||
x : Substitutive ? ?
|
||||
x = ?
|
||||
|
||||
-- Σ-≡,≡←≡ : p₁ ≡ p₂ → Σ (a₁ ≡ a₂) (λ p → subst B p b₁ ≡ b₂)
|
||||
-- Σ-≡,≡←≡ refl = refl , refl
|
||||
|
||||
-- A proof that for every y, the fiber of id for that y (which is just going to
|
||||
-- be y itself) is contractible, which means that for every pair of (a, path),
|
||||
-- there needs to be a path between that pair and the pair i gave above in id-fiber-base
|
||||
id-is-contr : (y : Bool) → isContr (fiber id y)
|
||||
proj₁ (id-is-contr y) = id-fiber-base y
|
||||
proj₂ (id-is-contr y) y₁ = id-fiber-is-contr y y₁
|
||||
|
||||
id-is-eqv : isEquiv id
|
||||
id-is-eqv .equiv-proof y = ?
|
||||
|
||||
id-eqv : Bool ≃ Bool
|
||||
proj₁ id-eqv = id
|
||||
proj₂ id-eqv = id-is-eqv
|
||||
|
||||
-- proj₁ (proj₁ (proj₂ id-eqv .equiv-proof y)) = y
|
||||
-- proj₂ (proj₁ (proj₂ id-eqv .equiv-proof y)) = refl
|
||||
-- proj₂ (proj₂ id-eqv .equiv-proof y) y₁ = {! !}
|
||||
|
||||
e213 : (Bool ≃ Bool) ≃ Bool
|
||||
|
||||
e213-helper-1 : (y : Bool) → Σ (Bool ≃ Bool) (λ x → proj₁ e213 x ≡ y)
|
||||
proj₁ (e213-helper-1 y) = {! !}
|
||||
proj₂ (e213-helper-1 y) = {! !}
|
||||
|
||||
proj₁ e213 bool-equiv = (proj₁ bool-equiv) true
|
||||
proj₁ (proj₂ e213 .equiv-proof y) = e213-helper-1 y
|
||||
proj₂ (proj₂ e213 .equiv-proof y) y₁ = {! !}
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
|
|
|||
130
src/2023-05-06-equiv.lagda.md
Normal file
130
src/2023-05-06-equiv.lagda.md
Normal file
|
|
@ -0,0 +1,130 @@
|
|||
+++
|
||||
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
|
||||
|
||||
-- https://git.mzhang.io/school/cubical/issues/1
|
||||
eqv4′ = ?
|
||||
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
|
||||
|
||||
-- https://git.mzhang.io/school/cubical/issues/2
|
||||
?
|
||||
```
|
||||
|
||||
## Other Equivalences
|
||||
|
||||
There are 2 other ways we can define equivalences:
|
||||
|
||||
TODO: Talk about them being equivalent to each other
|
||||
|
|
@ -1,13 +1,15 @@
|
|||
{-# OPTIONS --cubical #-}
|
||||
|
||||
open import Cubical.Foundations.Prelude
|
||||
using (_≡_; refl; _∙_; _≡⟨_⟩_; _∎; cong; sym; fst; snd; _,_; ~_)
|
||||
using (_≡_; refl; _∙_; _∎; cong; sym; fst; snd; _,_; ~_)
|
||||
open import Cubical.Data.Empty as ⊥
|
||||
open import Cubical.Foundations.Equiv using (isEquiv; equivProof; equiv-proof)
|
||||
open import Relation.Nullary using (¬_)
|
||||
open import Relation.Binary.Core using (Rel)
|
||||
open import Data.Nat
|
||||
|
||||
open import Data.Bool
|
||||
|
||||
sanity : 1 + 1 ≡ 2
|
||||
sanity = refl
|
||||
|
||||
|
|
@ -38,3 +40,20 @@ sanity = refl
|
|||
-- _[_;_]′
|
||||
-- a [ 34 ; b ]
|
||||
--
|
||||
|
||||
|
||||
x : Bool
|
||||
x = true
|
||||
|
||||
-- _&&_ : Bool → Bool → Bool
|
||||
|
||||
-- (x && true) : Bool
|
||||
|
||||
|
||||
p : 2 + 3 ≡ 3 + 2
|
||||
p = refl
|
||||
|
||||
-- _+_ : Nat → Nat → Nat
|
||||
-- zero + m = m
|
||||
-- suc n + m = suc (n + m)
|
||||
p2 : (x y : ℕ) → x + y ≡ y + x
|
||||
|
|
|
|||
Loading…
Reference in a new issue