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 ' : A ! C; : B ! C. It φ : A × B ! C has the properties φι1 = ' and φι2 = then we must have φ = φid = φ(ι1π1 + ι2π2) = ϕπ1 + π2: Conversely, the formula ϕπ1 + π2 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 ' : B ! Z with ' ◦ k = 0. We wish to prove that ' has a unique extension : C ! Z with f ◦ = '. But this follows since ' ◦ g = ' ◦ 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 . 3.