[Agda] formalization of category theory
Thorsten Altenkirch
txa at Cs.Nott.AC.UK
Sun Jun 20 17:46:21 CEST 2010
P.S.
Actually, to partially answer my own question: This doesn't seem to be
the case since the natural definition of Ar would be Ar = Σ (A B :
Obj) Arrow A B (not official Agda Syntax) and then you can only define
partial composition, if equality of objects is decidable.
Thorsten
On 20 Jun 2010, at 16:34, Thorsten Altenkirch wrote:
> Hi Vag,
>
> I am afraid I agree with Wolfgang. The indexed way is more in the
> spirit of Type Theory while using partial application seems to be
> the natural choice if you work in an algebraic framework.
>
> Btw, can you show that both definitions are equivalent, i.e. give
> translations in both directions and show that there composites lead
> to equivalent categories in the appropriate sense.
>
> Cheers,
> Thorsten
>
> On 17 Jun 2010, at 08:42, Vag Vagoff wrote:
>
>> > record Cat (l : Level) : Set (S l) where
>> > field
>> > Ob : Set l
>> > Arrow : Ob → Ob → Set l
>> > id : ∀ {o1} → (Arrow o1 o1)
>> > comp : ∀ {o1 o2 o3} → (Arrow o2 o3) → (Arrow o1 o2) →
>> (Arrow o1 o3)
>>
>> It is more convenient to define category without mentioning objects:
>>
>> record CategoryWO (Ar : Set) : Set where
>> field
>> _∘_ : Ar → Ar → Maybe Ar
>> infix 90 _∘_
>> field
>> idR : ∀ f → ∃ λ g → f ∘ g ≡ just f
>> idL : ∀ f → ∃ λ g → g ∘ f ≡ just f
>> identity : ∀ {f g} → Ex f ∘ g → un∃ (idL g) ≡
>> un∃ (idR f)
>> composition : ∀ {f g h} → Ex f ∘ g → Ex g ∘ h → Ex
>> f ∘ h
>> associativity : ∀ {f g h fg gh} → f ∘ g ≡ just fg →
>> g ∘ h ≡ just gh → fg ∘ h ≡ f ∘ gh
>>
>> -- or --
>>
>> record CategoryWO' (Ar : Set) : Set where
>> field
>> _∘_ : Maybe Ar → Maybe Ar → Maybe Ar
>> undefR : ∀ {f} → f ∘ nothing ≡ nothing
>> undefL : ∀ {f} → nothing ∘ f ≡ nothing
>> infix 90 _∘_
>> field
>> idR : ∀ f → ∃ λ g → f ∘ g ≡ f
>> idL : ∀ f → ∃ λ g → g ∘ f ≡ f
>> identity : ∀ {f g} → Ex f ∘ g → un∃ (idL g) ≡
>> un∃ (idR f)
>> composition : ∀ {f g h} → Ex f ∘ g → Ex g ∘ h → Ex
>> f ∘ h
>> associativity : ∀ {f g h} → (f ∘ g) ∘ h ≡ f ∘ (g
>> ∘ h)
>>
>>
>> Isn't it?
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