TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A morphism factors uniquely through its coimage

Statement

Let $f : A \to B$ be a morphism in a category with kernels and cokernels. If $k : K \to A$ is a kernel of $f$ and $q : A \to coim (f)$ is a cokernel of $k$ , then there exists a unique morphism $\tilde{f} : coim (f) \to B$ with

$\tilde{f} \circ q = f .$

Facts & Assumptions

Given: A morphism $f : A \to B$ , a kernel $k : K \to A$ of $f$ , and a cokernel $q : A \to coim (f)$ of $k$ .

[L1]

The coimage of $f$ is the cokernel of a kernel of $f$ (Image and coimage in a category with kernels and cokernels).

Proof

technique · direct

1.1L1given

Because $k$ is a kernel of $f$ , one has $f \circ k = 0$ .

2.1L1step 1.1∎

The morphism $q$ is a cokernel of $k$ , so step 1.1 gives a unique $\tilde{f} : coim (f) \to B$ with $\tilde{f} q = f$ . That is exactly the claimed factorization through the coimage.

A morphism factors uniquely through its coimage

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources