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Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel

Statement

In an abelian category, every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel.

Facts & Assumptions

Given: An abelian category and a monomorphism $m : A \to B$ with cokernel $c : B \to C$ .

[L1]

The kernel of a monomorphism is zero (The kernel of a monomorphism is zero and the cokernel of an epimorphism is zero).

[L2]

The cokernel of $0 \to A$ is $A$ (The cokernel of the zero map out of the zero object is the target, and dually for kernels).

[L3]

The image and coimage are defined by kernels and cokernels (Image and coimage in a category with kernels and cokernels).

[L4]

A morphism factors through its image, and the canonical $coim \to im$ map exists (A morphism factors uniquely through its image, The canonical morphism from the coimage to the image exists and is unique).

[L5]

In an abelian category the canonical $coim (m) \to im (m)$ map is an isomorphism (Abelian category).

[L6]

The opposite of an abelian category is abelian (The opposite of an abelian category is abelian).

Proof

technique · direct

1.1

By [L1], the kernel of $m$ is zero. Therefore [L2] identifies the coimage projection $q_{m} : A \to coim (m)$ with an isomorphism.

L1L2L3

2.1

By definition, the image inclusion $i_{m} : im (m) \to B$ is a kernel of the cokernel $c$ . Since $m = i_{m} \overline{m} q_{m}$ by [L4], and both $q_{m}$ and $\overline{m}$ are isomorphisms by step 1.1 and [L5], there is an isomorphism $u : A \tilde{⟶} im (m)$ with $i_{m} u = m$ . So $m$ is itself a kernel of $c$ up to the unique compatible isomorphism of kernel objects.

L3L4L5step 1.1

3.1

By [L6], the opposite of an abelian category is again abelian. Applying step 2.1 there to the opposite of an epimorphism gives that every epimorphism in the original category is the cokernel of its kernel.

L6step 2.1∎

Depends on

Used by

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Section 12.5, Lemma 12.5.4 (standard reference, not scraped)
Saunders Mac Lane, Categories for the Working Mathematician, VIII.3 (standard reference, not scraped)