PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The kernel of a monomorphism is zero and the cokernel of an epimorphism is zero

Statement

Let $C$ be a category with a zero object and the needed kernels and cokernels. If $m : A \to B$ is monic and $k : K \to A$ is a kernel of $m$ , then $K$ is a zero object and $k$ is the zero morphism into $A$ .

Dually, if $e : A \to B$ is epic and $q : B \to Q$ is a cokernel of $e$ , then $Q$ is a zero object and $q$ is the zero morphism out of $B$ .

Facts & Assumptions

Given: A zero object $0$ , a monomorphism $m : A \to B$ with kernel $k : K \to A$ , and an epimorphism $e : A \to B$ with cokernel $q : B \to Q$ .

[L1]

A zero object is both initial and terminal, so there are unique morphisms $0 \to X$ and $X \to 0$ for every object $X$ (Initial object, terminal object, and zero object).

[L2]

Monomorphisms are left-cancellable and epimorphisms are right-cancellable (Monomorphism and epimorphism by left and right cancellation).

[L3]

A kernel of $m$ is a morphism $k$ with $m k = 0$ through which every morphism $h$ with $m h = 0$ factors uniquely; a cokernel is dual (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1L1L2L3

If $h : X \to A$ satisfies $m h = 0$ , then also $m 0_{X, A} = 0$ , so [L2] gives $h = 0_{X, A}$ . Therefore the unique map $0 \to A$ from [L1] has the kernel universal property for $m$ , because every morphism killed by $m$ factors uniquely through $0$ .

2.1L1L2L3step 1.1∎

Kernels are unique up to a unique compatible isomorphism, so the displayed kernel $k : K \to A$ is isomorphic to the zero morphism $0 \to A$ from step 1.1. Hence $K$ is a zero object and $k$ is the zero map into $A$ . The cokernel claim is the formal dual of the same argument with right cancellation in [L2].

Depends on

Used by

Dependency tree · two levels

7 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

Saunders Mac Lane, Categories for the Working Mathematician, VIII.1 (standard reference, not scraped)
Gautam Tamme, Algebra II Lecture 9 (standard reference, not scraped)