Take any morphism and let be its categorical cokernel. Let be the categorical kernel of . Since , kernel universality gives with . The map is a monomorphism. We must establish the pseudo-epimorphism property of , rather than merely name an image.
Suppose satisfies . Let be its categorical kernel. Then for some . The composite is a monomorphism, so the hypothesis that every monomorphism is normal provides a map of which is a categorical kernel, up to its unique compatible isomorphism. Now . The categorical cokernel property of gives , and therefore .
Since is the categorical kernel of , the map factors through it: for some . Cancel the monomorphism to get . As is also a monomorphism, , so is an isomorphism. Finally , proving that is a pseudo-epimorphism.
ThusThe pseudo-epimorphism factorization through the kernel of a cokernel proof uses a pointed category, existence of categorical kernels and categorical cokernels, and normality of every monic map. It does not silently replace pseudo-epic cancellation against zero by the stronger epimorphism condition.
In an abelian category, for set and . The canonical map from coimage to image is an isomorphism. Thus the image factorization in an abelian category isunique up to a unique compatible isomorphism.
The short five lemma states that in a commuting diagram of short exact sequences and in an abelian category, if the vertical maps and are isomorphisms, then so is .
For the Five lemma, consider commuting exact rowswith vertical maps . We prove the usual stronger version: epic, and invertible, and monic imply that is invertible. In particular the conclusion holds if all four outer vertical maps are isomorphisms.
Put and , and define similarly. Exactness gives short exact sequencesThe functoriality of abelian image factorization supplies their vertical outer maps. We show that both are isomorphisms.
First, and : exactness at identifies with , and the image-coimage isomorphism for gives the claimed cokernel. Write and for these categorical cokernels. To construct an inverse to the induced map , observeCancellation of the epimorphism gives , so factors uniquely as with . From , composing with the epic gives and .
Second, and by exactness at the fourth objects. Write their inclusions as and . The equality gives the induced map . Sinceand is a monomorphism, . Kernel universality gives with . Cancelling the monic proves and .
Apply the short five lemma to the two short exact sequences: their outer maps are invertible, so is an isomorphism. This five lemma via image factorization argument uses only universal properties and therefore works in any abelian category, without treating its objects as literal sets of elements.
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