module Cat.Abelian.Images {o ℓ} {C : Precategory o ℓ} (A : is-abelian C) where
open is-abelian A


# Images in abelian categories🔗

Let $f : A \to B$ be a morphism in an abelian category $\mathcal{A}$, which (by definition) admits a canonical decomposition as

$A \xtwoheadrightarrow{p} \operatorname{coker}(\ker f) \cong \ker (\operatorname{coker}f) \xhookrightarrow{i} B\text{,}$

where the map $p$ is epic, $i$ is monic, and the indicated isomorphism arises from $f$ in a canonical way, using the universal properties of kernels and cokernels. What we show in this section is that the arrow $\ker (\operatorname{coker}f) \hookrightarrow B$ is an image for $f$: It is the largest monomorphism through which $f$ factors. Since this construction does not depend on any specificities of $f$, we conclude that every map in an abelian category factors as a regular epi followed by a regular mono.

## The image🔗

images : ∀ {A B} (f : Hom A B) → Image f
images f = im where
the-img : ↓Obj (const! (cut f)) Forget-full-subcat
the-img .x = tt
the-img .y .object = cut (Ker.kernel (Coker.coeq f))
the-img .y .witness {c} = kernels-are-subobjects C ∅ _ (Ker.has-is-kernel _)


Break $f$ down as an epi $p : A \twoheadrightarrow \ker (\operatorname{coker}f)$ followed by a mono $i : \ker (\operatorname{coker}f) \hookrightarrow B$. We can take the map $i$ as the “image” subobject. We must provide a map filling the dotted line in

but this is the epimorphism $p$ in our factorisation. More precisely, it’s the epimorphism $p$ followed by the isomorphism $f'$ in the decomposition

$A \xtwoheadrightarrow{p} \operatorname{coker}(\ker f) \xrightarrow{f'} \ker (\operatorname{coker}f) \xhookrightarrow{i} B\text{.}$

  the-img .map ./-Hom.map = decompose f .fst ∘ Coker.coeq _
the-img .map ./-Hom.commutes = pulll (Ker.factors _) ∙ Coker.factors _


## Universality🔗

Suppose that we’re given another decomposition of $f$ as

$A \xrightarrow{q} X \hookrightarrow{i'} B\times{,}$

and we wish to construct a map $g : i' \le i$1, hence a map $\operatorname*{im}f \to X$ such that the triangle

commutes.

  im : Image f
im .Initial.bot = the-img
im .Initial.has⊥ other = contr factor unique where
factor : ↓Hom (const! (cut f)) Forget-full-subcat the-img other
factor .α = tt
factor .β ./-Hom.map =
Coker.universal (Ker.kernel f) {e′ = other .map .map} path
∘ coker-ker≃ker-coker f .is-invertible.inv


Observe that by the universal property of $\operatorname{coker}(\ker f)$2, if we have a map $q : A \to X$ such that $0 = q\ker f$, then we can obtain a (unique) map $\operatorname{coker}(\ker f) \to X$ s.t. the triangle above commutes!

      where abstract
path : other .map .map ∘ 0m ≡ other .map .map ∘ kernel f .Kernel.kernel
path =
other .y .witness _ _ $sym$
pulll (other .map .commutes)
·· Ker.equal f
·· ∅.zero-∘r _
·· 0m-unique
·· sym (ap₂ _∘_ refl ∘-zero-r ∙ ∘-zero-r)


To satisfy that equation, observe that since $i'$ is monic, it suffices to show that $i'0 = i'q\ker f$, but we have assumed that $i'q = f$, and $f\ker f = 0$ by the definition of kernel. Some tedious isomorphism-algebra later, we have shown that $\ker (\operatorname{coker}f) \hookrightarrow B$ is the image of $f$.

Here’s the tedious isomorphism algebra.
    factor .β ./-Hom.commutes = invertible→epic (coker-ker≃ker-coker f) _ _ $Coker.unique₂ (Ker.kernel f) (sym (Ker.equal f ∙ ∅.zero-∘r _ ∙ 0m-unique ∙ sym ∘-zero-r)) (ap₂ _∘_ ( sym (assoc _ _ _) ∙ ap₂ _∘_ refl (cancelr (coker-ker≃ker-coker f .is-invertible.invr))) refl ∙ pullr (Coker.factors _) ∙ other .map .commutes) (sym (decompose f .snd ∙ assoc _ _ _)) factor .sq = /-Hom-path$ sym $other .y .witness _ _$
pulll (factor .β .commutes)
∙ the-img .map .commutes
·· sym (other .map .commutes)
·· ap (other .y .object .map ∘_) (intror refl)

unique : ∀ x → factor ≡ x
unique x = ↓Hom-path _ _ refl $/-Hom-path$ other .y .witness _ _ \$
sym (x .β .commutes ∙ sym (factor .β .commutes))


1. this is an inequality in the poset of subobjects of $B$, which is a map $i' \to i$ in the slice over $B$.↩︎

2. hence of $\ker (\operatorname{coker}f)$, after adjusting by our old friendly isomorphism↩︎