False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

FALSE: a chain map is determined by its maps on homology

Statement

If two chain maps induce the same morphism on every homology object, then the two chain maps are equal.

Facts & Assumptions

Given: The two-term complex $0 \to Z \overset{1_{Z}}{\to} Z \to 0$ and the endomorphisms $f = 1_{C}$ and $g = 0_{C}$ of this complex.

[L1]

$A b$ is an abelian category (Abelian groups form an abelian category).

[L2]

Chain maps induce homology maps (A chain map induces a well-defined map on homology).

Refutation

technique · direct

1.1

The complex is acyclic: in degree $1$ the kernel of $1_{Z}$ is $0$ , and in degree $0$ the cokernel of $1_{Z}$ is also $0$ . Hence all its homology objects are zero. The maps $f$ and $g$ are distinct because $f_{1} = 1_{Z}$ while $g_{1} = 0$ .

L1givenalgebra

2.1

By [L2], both $f$ and $g$ induce the zero endomorphism on every homology object, since those homology objects are zero by step 1.1. Therefore equal homology maps do not force equality of chain maps.

L2step 1.1∎

Depends on

Used by

Two distinct chain maps inducing the same homology map Counterexample

Dependency tree · two levels

11 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

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)