Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 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.

A quasi-isomorphism that is not an isomorphism of complexes

Statement refuted

Every quasi-isomorphism of complexes is an isomorphism of complexes.

Facts & Assumptions

Given: The inclusion of the zero complex into 0→Z→1ZZ→0.

[L1]

Ab is an abelian category (Abelian groups form an abelian category).

[L2]

The A-page false statement is indeed false (FALSE: every quasi-isomorphism is an isomorphism of complexes).

Counterexample

technique · direct
1.1

The target complex is acyclic, because both the kernel and cokernel of 1Z are zero. The source zero complex is acyclic as well, so the inclusion induces isomorphisms on all homology groups.

L1givenalgebra
2.1

The target complex is nonzero while the source is zero, so the inclusion is not invertible. Thus this is a concrete counterexample, as asserted by [L2].

L2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

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