Agda中证明自由回路的依赖递归性质蕴含圆的泛性质
圆𝕊¹的自由回路归纳原理到泛性质的证明困境
背景
我正依据HoTT书籍第6章定义圆𝕊¹的归纳原理并证明其相关性质。核心难点在于HoTT用判断相等定义递归/归纳原理,这无法在Agda中表述,因此参考unimath采用**自由回路(free-loop)**来定义圆𝕊¹的递归/归纳原理,但在证明目标命题时陷入困境。
目标命题
需要证明的命题如下(含义:若ev-free-loop-∏是截面,则它同时也是收缩):
free-loop-∏-induction→universal-property-circle : ∀ {n m : Level} {X : Set n} {A : Set m} → (S : free-loop X) → (∀ P → free-loop-∏-induction {n} {m} {X} S P) → universal-property-circle {n} {m} {X} {A}
相关定义
用到的核心定义:
free-loop : ∀ {n : Level} → Set n → Set n free-loop X = Σ[ x ∈ X ] (x ≡ x) free-dependent-loop : ∀ {n m : Level} {X : Set n} → free-loop X → (X → Set m) → Set m free-dependent-loop ⟨ x , l ⟩ P = Σ[ a ∈ P x ] (transport l P a ≡ a) ev-free-loop-∏ : ∀ {n m : Level} {X : Set n} → (S : free-loop X) → (P : X → Set m) → (∀ (x : X) → P x) → free-dependent-loop S P ev-free-loop-∏ ⟨ b , l ⟩ P h = ⟨ h b , congd h l ⟩ free-loop-∏-induction : ∀ {n m : Level} {X : Set n} (S : free-loop X) (P : X → Set m) → Set (n ⊔ m) free-loop-∏-induction S P = section (ev-free-loop-∏ S P) universal-property-circle : ∀ {n m : Level} {X : Set n} {A : Set m} → Set (n ⊔ m) universal-property-circle {_} {_} {X} {A} = (X → A) ≃ (Σ[ a ∈ A ] a ≡ a)
不完整的证明草稿
目前写出的不完整证明如下,其中包含多个未验证的公设:
free-loop-∏-induction→universal-property-circle {_} {_} {X} {A} ⟨ b , l ⟩ φ = ⟨ f , ⟨ ⟨ g , f∘g∼id ⟩ , ⟨ g , g∘f∼id ⟩ ⟩ ⟩ where postulate ∏-induction→induction : ∀ {n m : Level} {X : Set n} {A : Set m} → (S : free-loop X) → (∀ P → free-loop-∏-induction {n} {m} {X} S P) → free-loop-induction {n} {m} {X} {A} S S = ⟨ b , l ⟩ φ' = (∏-recursion→recursion S φ) f = ev-free-loop S g : (Σ[ a ∈ A ] a ≡ a) → (X → A) g = proj₁ φ' f∘g∼id : (f ∘ g) ∼ id f∘g∼id t = proj₂ φ' t g∘f∼id : (g ∘ f) ∼ id g∘f∼id h = funext (g (f h)) h ξ where ε : ∀ h → g (f h) b ≡ h b ε h = cong proj₁ ( proj₂ φ' (f h)) postulate μ : transport (ε h) (λ z → z ≡ z) (cong (g (f h)) l) ≡ cong h l ξ : ∀ (x : X) → g (f h) x ≡ h x ξ = unicity-circle-∏ {_} {_} {X} {A} S φ (g (f h)) h (ε h) μ
待解决问题
- 依赖自由回路归纳(
free-loop-∏-induction)是否蕴含自由回路归纳(free-loop-induction)? - 公设
μ是否成立? - 如何补全完整的证明,消除这些未验证的公设?
内容的提问来源于stack exchange,提问作者Werner Germán Busch
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