Abelian Categories

Lemma. In an Ab-enriched category with zero object every finite product is coproduct and conversely.

π1 Proof. Suppose A × B //A, B is a product. Define ι1 : A → A × B and π2 ι2 : B → A × B by

π1ι1 = id, π2ι1 = 0, π1ι2 = 0, π2ι2 = id.

It follows that ι1π1+ι2π2 = id (both sides are equal upon applying π1 and π2). To show that ι1, ι2 are a coproduct suppose given ϕ : A → C, ψ : B → C. It φ : A × B → C has the properties φι1 = ϕ and φι2 = ψ then we must have

φ = φid = φ(ι1π1 + ι2π2) = ϕπ1 + ψπ2.

Conversely, the formula ϕπ1 + ψπ2 yields the desired map on A × B. An additive category is an Ab-enriched category with a zero object and finite products (or coproducts). In such a category, a kernel of a morphism f : A → B is an equalizer k in the diagram

k f ker(f) / A / B. 0

Dually, a cokernel of f is a coequalizer c in the diagram

f c A / B / coker(f). 0

An Abelian category is an additive category such that

1. every map has a kernel and a cokernel,

2. every mono is a kernel, and every epi is a cokernel.

In fact, it then follows immediatly that a mono is the kernel of its cokernel, while an epi is the cokernel of its kernel.

1 Proof of last statement. Suppose f : B → C is epi and the cokernel of some g : A → B. Write k : ker(f) → B for the kernel of f. Since f ◦ g = 0 the map g¯ indicated in the diagram exists. We claim that f is the cokernel of k.

ker(f) O k f g¯ ( 5 B / C A g Thus suppose that we are given ϕ : B → Z with ϕ ◦ k = 0. We wish to prove that ϕ has a unique extension ψ : C → Z with f ◦ ψ = ϕ. But this follows since ϕ ◦ g = ϕ ◦ k ◦ g¯ = 0 and since f is the cokernel of g. Moreover, from the definitions it is immediate that every kernel is mono while every cokernel is epi. Also, f is mono ⇔ kerf = 0 and f is epi ⇔ cokerf = 0. From the above it follows that f is mono and epi precisely when f is an isomorphism.

The image of a map f : A → B is the kernel of the cokernel of f. We then have a factorization

f c A / B / coker(f). O f¯

& ? ker(c) = im(f)

The coimage of f is the cokernel of the kernel of f for which we have a factorization k f ker(f) / A 7/ B. f 0

 coker(k) = coim(f)

Lemma. The induced map f¯ : A → im(f) is an epi. Similarly, the induced map f 0 : coim(f) → B is a mono. In particular, the map coim(f) → im(f) parallel to f is an isomorphism.

Proof. We prove only the first statement. The second assertion becomes the first in the opposite category. Consider x : im(f) → D with x ◦ f¯ = 0 and

2 let l be the kernel of x. Then we have a diagram

f c A / B / coker(f) O f¯ f 0 k   l $ ? x ker(x) / im(f) / D.

If the kernel l is an isomorphism, then x = 0, so we have to show l is epi. Since k ◦l is mono, it is the kernel of h : B → coker(k ◦l). h may be extended to h0 : coker(f) → coker(k ◦ l) since h ◦ f = hklf 0 = 0. Then hk = h0ck = 0 so k factors through kl: klk0 = k and since k is mono, lk0 = id, so l is epi. We conclude that in Abelian categories, we have epi-mono factorizations: f : A  im(f) ,→ B.

Example. The categories Ab, R−Mod, and Ch are Abelian, for any ring R. More generally, for any Abelian category A we have a category of chain complexes Ch(A ) with chain maps as morphisms. It is again Abelian with monos and epis being the levelwise monos and epis. Kernels and Cokernels are given levelwise as well.

For any category C the functor category A C is again abelian will all of the relevant structure determined pointwise. In particular, this applies to the category of presheaves of Abelian groups on C .

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