Edited 1FundamentalGroup/Quest0.agda

1FundamentalGroup/Quest0.agda

@@ -1,21 +1,209 @@
 module 1FundamentalGroup.Quest0 where
 
-open import Cubical.Core.Everything
 open import Cubical.Data.Empty
+open import Cubical.Data.Unit renaming ( Unit to ⊤ )
+open import Cubical.Data.Bool
 open import Cubical.Foundations.Prelude
+open import Cubical.Foundations.Isomorphism
 open import Cubical.Foundations.Path
-open import Cubical.Foundations.GroupoidLaws
+
+private
+  variable
+    u : Level
 
 data S¹ : Type where
   base : S¹
   loop : base ≡ base
 
-¬ : Type → Type
+-- if you don't know how to input a character
+-- go to evil-mode, put your cursor on the character
+-- and do `SPC h '`
+
+¬ : Type u → Type u
 ¬ A = A → ⊥
 
-doubleCover : type
+_≢_ : {A : Type u} → (x y : A) → Type u
-doubleCover = ?
+x ≢ y = ¬ (x ≡ y)
+
+_≅_ = Iso
+
+{- Bool
+
+data Bool : Type where
+  true : Bool
+  false : Bool
+
+The above definition for the Booleans
+can be interpreted as
+
+- a construction with only two recipes
+  `true` and `false`
+- a space with two points `true` and `false`.
+  This space is discrete in the sense that
+  we haven't specified any paths.
+
+Our goal is to show
+
+  refl ≢ loop (input \nequiv)
+
+that there is path (aka homotopy) from `refl` to `loop`.
+To do so we must assume there is such a path and derive
+a contradiction.
+The contradiction we will try to reach is that `true ≡ false`.
+Indeed it does not hold:
+
+-}
 
 
-refl≢loop : ¬ ( refl ≡ loop )
+{- transport
-refl≢loop h = {!!}
+
+To follow a point in `a : A` along a path `p : A ≡ B`
+we use
+
+  transport : {A B : Type u} → A ≡ B → A → B
+
+Why do we propify? Discuss.
+
+-}
+
+true≢false' : true ≢ false
+true≢false' h = transport ⊤≡⊥ tt where
+
+  propify : Bool → Type
+  propify false = ⊥
+  propify true = ⊤
+
+  ⊤≡⊥ : ⊤ ≡ ⊥
+  ⊤≡⊥ = cong propify h
+
+
+
+Flip : Bool → Bool
+Flip false = true
+Flip true = false
+
+{- Iso
+
+We show that Flip is an isomorphism from Bool → Bool
+with inverse Flip.
+
+A proof of `A ≅ B` (input \cong or write Iso A B) is given by
+
+  iso f i s r
+
+where
+
+  f : A → B and i : B → A
+
+are the map and its inverse,
+here both `f` and `i` are Flip
+
+`s` is a proof that `f` is a section with
+right inverse `i` and
+`r` is a proof that `f` is a retraction
+with left inverse `i`
+
+-}
+
+flipIso : Bool ≅ Bool
+flipIso = iso Flip Flip s r where
+  s : section Flip Flip
+  s false = refl
+  s true = refl
+
+  r : retract Flip Flip
+  r false = refl
+  r true = refl
+
+{- Path ≡
+
+A corollary of univalence is
+`isoToPath` which takes an isomorphism
+`f : A ≅ B` and gives a path
+`fPath : A ≡ B`.
+The resulting path has the important property
+that when you follow (transport/subst)
+a point in `A` along the path
+you will get the point `f(a)` in `B`
+
+-}
+
+flipPath : Bool ≡ Bool
+flipPath = isoToPath flipIso
+
+{-
+
+Try out `transport` on `true : Bool` and
+`flipPath` by doing `C-c C-n`
+and typing in `transport flipPath true`
+
+-}
+
+{- bundle over S¹
+
+We want to construct a bundle over S¹
+that looks like this:
+-- insert image of double cover
+
+to do so we use flipPath
+to specify the fibers of the bundle
+at each point on the `loop`.
+These fibers must coincide at the end-points
+with the fiber we set for `base`, which is `Bool`.
+
+-}
+
+-- the bundle
+doubleCover : S¹ → Type
+doubleCover base = Bool
+doubleCover (loop i) = flipPath i
+
+{- subst
+
+Given a bundle `B : A → Type u`
+over a space `A` and a path `p : x ≡ y`
+between points in `x y : A`,
+
+  subst : (B : A → Type u) (p : x ≡ y) → B x → B y
+
+follows the path _over_ `p`, taking one
+end point of the path in fiber `B x` to
+the other end point in fiber `B y`.
+
+We use `subst` to get a function that takes a path `p : base ≡ base`
+and follows the point `true` in the fiber `doubleCover base`
+along the path over `p` to some point in `doubleCover base`.
+
+Note that `doubleCover base` is just `Bool` (externally).
+
+-}
+
+SubstTrue : (p : base ≡ base) → doubleCover base
+SubstTrue p = subst doubleCover p true
+
+{-
+
+You can check that `SubstTrue refl` and `SubstTrue loop`
+are using `C-c C-n`
+
+-}
+
+{- cong
+
+Given a function `f : A → B`
+and a path `p : x ≡ y` between points `x y : A`
+
+  cong : (f : A → B) → (p : x ≡ y) → f x ≡ f y
+
+gives us a path in `B` from `f x` to `f y`
+
+We can use the above to get the contradiction we want
+by
+
+- assuming `p : refl ≡ loop`
+- deducing `SubstTrue refl ≡ SubstTrue loop` using `cong`
+
+-}
+
+refl≢loop : refl ≢ loop
+refl≢loop p = true≢false (cong SubstTrue p)