TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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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

1.1L1L2L3

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

2.1L3L4L5step 1.1

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.

3.1L6step 2.1∎

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.

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources