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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The cokernel of the zero map out of the zero object is the target, and dually for kernels

Statement

Let $0$ be a zero object in a category with zero morphisms and kernels and cokernels. For every object $A$ , the identity $1_{A} : A \to A$ is a cokernel of the zero morphism $0 \to A$ . Dually, $1_{A}$ is a kernel of the zero morphism $A \to 0$ .

So $coker (0 \to A) ≅ A$ and $\ker (A \to 0) ≅ A$ .

Facts & Assumptions

Given: A zero object $0$ and an object $A$ .

[L1]

There is a unique morphism $0 \to A$ and a unique morphism $A \to 0$ (Initial object, terminal object, and zero object).

[L2]

A cokernel of $u$ is a morphism $q$ with $q u = 0$ through which every morphism annihilating $u$ factors uniquely, and a kernel is dual (Kernels and cokernels in a category with zero morphisms as equalizers and coequalizers).

Proof

technique · direct

1.1L1L2

Let $u : 0 \to A$ be the unique map from [L1]. Since every composite $h \circ u : 0 \to X$ equals the unique map $0 \to X$ , every morphism $h : A \to X$ annihilates $u$ . Each such $h$ factors uniquely through $1_{A}$ , namely as $h = h \circ 1_{A}$ . Therefore $1_{A}$ is a cokernel of $u$ .

2.1L1L2step 1.1∎

The dual argument with the unique map $A \to 0$ shows that $1_{A}$ is also a kernel of $A \to 0$ . So both displayed identifications hold up to the unique compatible isomorphism of kernels and cokernels.

Depends on

Used by

Dependency tree · two levels

6 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)