Alphabeta Math
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→B be a morphism in a category with kernels and cokernels. If k:K→A is a kernel of f and q:A→coim⁡(f) is a cokernel of k, then there exists a unique morphism f~:coim⁡(f)→B with

f~∘q=f.

Facts & Assumptions

Given: A morphism f:A→B, a kernel k:K→A of f, and a cokernel q:A→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∘k=0.

2.1L1step 1.1∎

The morphism q is a cokernel of k, so step 1.1 gives a unique f~:coim⁡(f)→B with f~q=f. That is exactly the claimed factorization through the coimage.

Depends on

Used by

Dependency tree · two levels

2 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