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lambeks lemma without math prose
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SirDavid12 committed Feb 11, 2023
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```agda
open import Cat.Prelude
open import Cat.Diagram.Initial
open import Cat.Displayed.Total
open import Cat.Displayed.Base
open Total-hom

import Cat.Reasoning

module Cat.Instances.FAlg where
```
```agda

module _ {o ℓ} {C : Precategory o ℓ} (F : Functor C C) where

open Cat.Reasoning C
open Functor F
open Displayed

FAlg : Displayed C _ _
Ob[ FAlg ] A = Hom (F₀ A) A
Hom[ FAlg ] h α β = h ∘ α ≡ β ∘ F₁ h
Hom[ FAlg ]-set _ _ _ = hlevel!
FAlg .id′ = idl _ ∙ intror F-id
FAlg ._∘′_ p q = pullr q ∙ extendl p ∙ ap (_ ∘_) (sym (F-∘ _ _))
FAlg .idr′ _ = prop!
FAlg .idl′ _ = prop!
FAlg .assoc′ _ _ _ = prop!

```

```agda

FAlgebras : Precategory _ _
FAlgebras = ∫ FAlg

module FAlgebras = Cat.Reasoning FAlgebras

lambek : (i : FAlgebras.Ob) is-initial FAlgebras i is-invertible (i .snd)
lambek (I , i) init = make-invertible (j .hom) p q
where
j : FAlgebras.Hom (I , i) (F₀ I , F₁ i)
j = init (F₀ I , F₁ i) .centre

i' : FAlgebras.Hom (F₀ I , F₁ i) (I , i)
i' .hom = i
i' .preserves = refl

p = ap hom (is-contr→is-prop (init (I , i)) (i' FAlgebras.∘ j) FAlgebras.id)
q = (j .hom ∘ i) ≡⟨ j .preserves ⟩
F₁ i ∘ F₁ (j .hom) ≡˘⟨ F-∘ _ _ ⟩
F₁ (i ∘ j .hom) ≡⟨ ap F₁ p ⟩
F₁ id ≡⟨ F-id ⟩
id ∎
```

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