open import Cat.Functor.Adjoint.Monadic
open import Cat.Diagram.Coequaliser
open import Cat.Prelude

import Cat.Functor.Reasoning as F-r
import Cat.Reasoning as C-r

module
{o ℓ o′ ℓ′} {C : Precategory o ℓ} {D : Precategory o′ ℓ′}
{F : Functor C D} {G : Functor D C}
(F⊣G : F ⊣ G)
where


# Beck’s coequaliser🔗

Let $F : \ca{C} \to \ca{D}$ be a functor admitting a right adjoint $U : \ca{D} \to \ca{C}$. Recall that every adjunction induces a monad $UF$ (which we will call $T$ for short) on the category $\ca{C}$, and a “comparison” functor $K : \ca{D} \to \ca{C}^{T}$ into the Eilenberg-Moore category of $T$. In this module we will lay out a sufficient condition for the functor $K$ to have a left adjoint, which we call $K^{-1}$ (Comparison⁻¹). Let us first establish a result about the presentation of $T$-algebras by “generators and relations”.

Suppose that we are given a $T$-algebra $(A, \nu)$. Note that $(TA, \mu)$ is also a $T$-algebra, namely the free $T$-algebra on the object $A$. Let us illustrate this with a concrete example: Suppose we have the cyclic group $\bb{Z}/n\bb{Z}$, for some natural number $n$, which we regard as a subgroup of $\bb{Z}$. The corresponding algebra $(TA, \mu)$ would be the free group on one generator1 whence2 we conclude that, in general, this “free-on-underlying” $(TA, \mu)$ algebra is very far from being the $(A, \nu)$ algebra we started with.

Still, motivated by our $\bb{Z}/n\bb{Z}$ example, it feels like we should be able to quotient the algebra $(TA, \mu)$ by some set of relations and get back the algebra $(A, \nu)$ we started with. This is absolutely true, and it’s true even when the category of $T$-algebras is lacking in quotients! In particular, we have the following theorem:

Theorem. Every $T$-algebra $(A, \nu)$ is the coequaliser of the diagram

called the Beck coequaliser (of $A$). Furthermore, this coequaliser is reflexive in $\ca{C}^T$, meaning that the arrows $\mu$ and $T\nu$ have a common right inverse. Elaborating a bit, this theorem lets us decompose any $T$-algebra $(A, \nu)$ into a canonical presentation of $A$ by generators and relations, as a quotient of a free algebra by a relation (induced by) the fold $\nu$.

Proof. The proof is by calculation, applying the monad laws where applicable, so we present it without much further commentary. Observe that $\nu$ coequalises $\mu$ and $T\nu$ essentially by the definition of being an algebra map. Note that we do not make use of the fact that the monad $T$ was given by a specified adjunction $F \dashv U$ in the proof, and any adjunction presenting $T$ will do.

  open is-coequaliser
algebra-is-coequaliser
: is-coequaliser C^T {A = TTA} {B = TA} {E = Aalg}
mult fold (record { morphism = A.ν ; commutes = sym A.ν-mult })
algebra-is-coequaliser .coequal = Algebra-hom-path C $A.ν C.∘ T.mult .η _ ≡˘⟨ A.ν-mult ⟩≡˘ A.ν C.∘ T.M₁ A.ν ∎  The colimiting map from $(A, \nu)$ to some other algebra $(F, \nu')$ which admits a map $e' : (TA, \mu) \to (F, \nu')$ is given by the composite $A \xto{\eta} TA \xto{e'} F\text{,}$ which is a map of algebras by a long computation, and satisfies the properties of a coequalising map by slightly shorter computations, which can be seen below. Uniqueness of this map follows almost immediately from the algebra laws.  algebra-is-coequaliser .coequalise {F = F} {e'} p .morphism = e' .morphism C.∘ T.unit .η A algebra-is-coequaliser .coequalise {F = F} {e'} p .commutes = (e' .morphism C.∘ unit.η A) C.∘ A.ν ≡⟨ C.extendr (unit.is-natural _ _ _) ⟩≡ (e' .morphism C.∘ T.M₁ A.ν) C.∘ unit.η (T.M₀ A) ≡˘⟨ ap morphism p C.⟩∘⟨refl ⟩≡˘ (e' .morphism C.∘ T.mult .η A) C.∘ unit.η (T.M₀ A) ≡⟨ C.cancelr T.right-ident ⟩≡ e' .morphism ≡⟨ C.intror (sym (T.M-∘ _ _) ∙ ap T.M₁ A.ν-unit ∙ T.M-id) ⟩≡ e' .morphism C.∘ T.M₁ A.ν C.∘ T.M₁ (unit.η A) ≡⟨ C.pulll (sym (ap morphism p)) ⟩≡ (e' .morphism C.∘ T.mult .η A) C.∘ T.M₁ (unit.η A) ≡⟨ C.pushl (e' .commutes) ⟩≡ F .snd .ν C.∘ T.M₁ (e' .morphism) C.∘ T.M₁ (unit.η A) ≡˘⟨ C.refl⟩∘⟨ T.M-∘ _ _ ⟩≡˘ F .snd .ν C.∘ T.M₁ (e' .morphism C.∘ unit.η A) ∎ algebra-is-coequaliser .universal {F = F} {e'} {p} = Algebra-hom-path C$
(e' .morphism C.∘ unit.η _) C.∘ A.ν          ≡⟨ C.extendr (unit.is-natural _ _ _) ⟩≡
(e' .morphism C.∘ T.M₁ A.ν) C.∘ unit.η  _    ≡˘⟨ ap morphism p C.⟩∘⟨refl ⟩≡˘
(e' .morphism C.∘ T.mult .η _) C.∘ unit.η  _ ≡⟨ C.cancelr T.right-ident ⟩≡
e' .morphism                                 ∎
algebra-is-coequaliser .unique {F = F} {e'} {p} {colim} q =
Algebra-hom-path C $sym$
e' .morphism C.∘ unit.η A              ≡⟨ ap morphism q C.⟩∘⟨refl ⟩≡
(colim .morphism C.∘ A.ν) C.∘ unit.η A ≡⟨ C.cancelr A.ν-unit ⟩≡
colim .morphism                        ∎


# Presented algebras🔗

The lemma above says that every algebra which exists can be presented by generators and relations; The relations are encoded by the pair of maps $T^2A \tto TA$ in Beck’s coequaliser, above. But what about the converse? The following lemma says that, if every algebra presented by generators-and-relations (encoded by a parallel pair $T^2A \tto TA$) has a coequaliser in $\ca{D}$, then the comparison functor $\ca{D} \to \ca{C}^T$ has a left adjoint.

On objects, this left adjoint acts by sending each algebra $M$ to the coequaliser of (the diagram underlying) its Beck coequaliser. In a sense, this is “the best approximation in $\ca{D}$ of the algebra”. The action on morphisms is uniquely determined since it’s a map out of a coequaliser.

If we consider the comparison functor as being “the $T$-algebra underlying an object of $\ca{D}$”, then the functor we construct below is the “free object of $\ca{D}$ on a $T$-algebra”. Why is this adjunction relevant, though? Its failure to be an equivalence measures just how far our original adjunction is from being monadic, that is, how far $\ca{D}$ is from being the category of $T$-algebras.

  Comparison⁻¹ : Functor (Eilenberg-Moore C T) D
Comparison⁻¹ .F₀ = coapex ⊙ has-coeq
Comparison⁻¹ .F₁ {X} {Y} alg-map =
has-coeq X .coequalise {e′ = e′} path where
e′ : D.Hom (F.F₀ (X .fst)) (Comparison⁻¹ .F₀ Y)
e′ = has-coeq Y .coeq D.∘ F.₁ (alg-map .morphism)


The adjunction unit and counit are again very simple, and it’s evident to a human looking at them that they satisfy the triangle identities (and are algebra homomorphisms). Agda is not as easy to convince, though, so we leave the computation folded up for the bravest of readers.

  Comparison⁻¹⊣Comparison : Comparison⁻¹ ⊣ Comparison F⊣G
Comparison⁻¹⊣Comparison .unit .η x .morphism =
G.₁ (has-coeq _ .coeq) C.∘ T.unit .η _
Comparison⁻¹⊣Comparison .counit .η x =
has-coeq _ .coequalise (F⊣G .counit.is-natural _ _ _)

Nah, really, it’s quite ugly.
  Comparison⁻¹⊣Comparison .unit .η x .commutes =
C.pullr (T.unit .is-natural _ _ _)
∙ G.extendl (has-coeq _ .coequal)
∙ C.elimr (F⊣G .zag)
∙ G.intror (F⊣G .zig)
∙ G.weave (D.pulll (sym (F⊣G .counit.is-natural _ _ _)) ∙ D.pullr (sym (F.F-∘ _ _)))
Comparison⁻¹⊣Comparison .unit .is-natural x y f = Algebra-hom-path C $(G.₁ (has-coeq y .coeq) C.∘ T.unit.η _) C.∘ f .morphism ≡⟨ C.pullr (T.unit.is-natural _ _ _) ⟩≡ G.₁ (has-coeq y .coeq) C.∘ T.M₁ (f .morphism) C.∘ T.unit .η (x .fst) ≡⟨ C.pulll (sym (G.F-∘ _ _)) ⟩≡ G.₁ (has-coeq y .coeq D.∘ F.₁ (f .morphism)) C.∘ T.unit .η (x .fst) ≡⟨ ap G.₁ (sym (has-coeq _ .universal)) C.⟩∘⟨refl ⟩≡ G.₁ (has-coeq x .coequalise _ D.∘ has-coeq x .coeq) C.∘ T.unit .η (x .fst) ≡⟨ C.pushl (G.F-∘ _ _) ⟩≡ G.₁ (has-coeq x .coequalise _) C.∘ G.₁ (has-coeq x .coeq) C.∘ T.unit.η _ ∎ Comparison⁻¹⊣Comparison .counit .is-natural x y f = has-coeq (F₀ (Comparison F⊣G) x) .unique {p = ap₂ D._∘_ (F⊣G .counit.is-natural _ _ _) refl ·· D.pullr (F⊣G .counit.is-natural _ _ _) ·· D.pulll (sym (F⊣G .counit.is-natural _ _ _))} (sym (D.pullr (has-coeq _ .universal) ∙ D.pulll (has-coeq _ .universal))) ∙ sym (has-coeq _ .unique (F⊣G .counit.is-natural _ _ _ ∙ D.pushr (sym (has-coeq _ .universal)))) Comparison⁻¹⊣Comparison .zig = has-coeq _ .unique {e′ = has-coeq _ .coeq} {p = has-coeq _ .coequal} (sym (D.pullr (has-coeq _ .universal) ∙ D.pulll (has-coeq _ .universal) ∙ ap₂ D._∘_ refl (F.F-∘ _ _) ∙ D.pulll (F⊣G .counit.is-natural _ _ _) ∙ D.cancelr (F⊣G .zig))) ∙ sym (has-coeq _ .unique (D.introl refl)) Comparison⁻¹⊣Comparison .zag = Algebra-hom-path C$
G.pulll (has-coeq _ .universal) ∙ F⊣G .zag


1. the group of integers↩︎

2. I was feeling pretentious when I wrote this sentence↩︎