Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31‡ 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: every quasi-isomorphism is a chain homotopy equivalence

Statement

Every quasi-isomorphism is a chain homotopy equivalence.

Facts & Assumptions

Given: The acyclic noncontractible complex 0→Z→2Z→ mod 2Z/2→0, and the zero map from it to the zero complex.

[A1]

The statement refuted is: every quasi-isomorphism is a chain homotopy equivalence.

[L1]

A quasi-isomorphism is a chain map inducing isomorphisms on all homology objects (Quasi-isomorphism).

[L2]

Every chain homotopy equivalence is a quasi-isomorphism (A chain homotopy equivalence is a quasi-isomorphism).

[L3]

A complex is zero in the homotopy category exactly when it is contractible (Zero homology does not make an object zero in the homotopy category).

Refutation

technique · direct
1.1

Both the source complex and the zero complex have zero homology in every degree, so the zero map between them is a quasi-isomorphism by [L1].

L1givenalgebra
2.1

If that map were a chain homotopy equivalence, its source would be isomorphic to the zero object in the homotopy category. By [L3], the source would then be contractible, contrary to the explicit nonsplit example. Hence [A1] is false. This does not contradict [L2], which gives only the forward implication.

A1L2L3step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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