open import Cat.Prelude

import Cat.Diagram.Equaliser
import Cat.Diagram.Zero

module Cat.Diagram.Equaliser.Kernel {o β} (C : Precategory o β) where

open import Cat.Reasoning C

open Cat.Diagram.Equaliser C
open Cat.Diagram.Zero C


# Kernelsπ

In a category with equalisers and a zero object $0$, a kernel of a morphism $f : a \to b$ is an equaliser of $f$ and the zero morphism $0 : a \to b$, hence a subobject $\mathrm{ker}(f) \hookrightarrow a$ of the domain of $f$.

module _ (β : Zero) where
open Zero β

is-kernel : β {a b ker} (f : Hom a b) (k : Hom ker a) β Type _
is-kernel f = is-equaliser f zeroβ

kernels-are-subobjects
: β {a b ker} {f : Hom a b} (k : Hom ker a)
β is-kernel f k β is-monic k
kernels-are-subobjects = is-equaliserβis-monic

record Kernel {a b} (f : Hom a b) : Type (o β β) where
field
{ker} : Ob
kernel : Hom ker a
has-is-kernel : is-kernel f kernel
open is-equaliser has-is-kernel public


Lemma: Let $\mathcal{C}$ be a category with equalisers and a zero object. In $\mathcal{C}$, the kernel of a kernel is zero. First, note that if a category has a choice of zero object and a choice of equaliser for any pair of morphisms, then it has canonically-defined choices of kernels:

module Canonical-kernels
(zero : Zero) (eqs : β {a b} (f g : Hom a b) β Equaliser f g) where
open Zero zero
open Kernel

Ker : β {a b} (f : Hom a b) β Kernel zero f
Ker f .ker           = _
Ker f .kernel        = eqs f zeroβ .Equaliser.equ
Ker f .has-is-kernel = eqs _ _ .Equaliser.has-is-eq


We now show that the maps $! : \ker\ker f \to \emptyset$ and $Β‘ : \emptyset \to \ker\ker f$ are inverses. In one direction the composite is in $\emptyset \to \emptyset$, so it is trivially unique. In the other direction, we have maps $\ker\ker f \to \ker\ker f$, which, since $\ker$ is a limit, are uniquely determined if they are cone homomorphisms.

  Ker-of-kerββ : β {a b} (f : Hom a b) β Ker (Ker f .kernel) .ker β β
Ker-of-kerββ f = make-iso ! Β‘ (!-uniqueβ _ _) p where
module Kf = Kernel (Ker f)
module KKf = Kernel (Ker (Ker f .kernel))


The calculation is straightforward enough: The hardest part is showing that $\ker f \circ \ker \ker f = 0$ (here we are talking about the inclusion maps, not the objects) β but recall that $\ker \ker f$ equalises $\ker f$ and $0$, so we have

$\ker f \circ \ker \ker f = 0 \circ \ker \ker f = 0$

    p : Β‘ β ! β‘ id
p = KKf.uniqueβ (zero-comm _ _) (zero-βl _)
(Kf.uniqueβ (extendl (zero-comm _ _))
(pulll KKf.equal β idr _)
(zero-comm _ _))