CounterexampleConstruction: 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.

A degreewise split cone sequence with no chain splitting

Statement refuted

Every degreewise split cone sequence splits as a sequence of complexes.

Facts & Assumptions

Given: The cone sequence attached to $\times 2 : Z [0] \to Z [0]$ .

[L1]

Every cone sequence is degreewise split short exact (The canonical mapping-cone sequence is degreewise split short exact).

[L2]

Homology of a shift is shifted homology (Homology of a shift is shifted homology).

Counterexample

technique · direct

1.1

By [L1], the cone sequence for $\times 2$ is degreewise split. Its middle term is the complex $0 \to Z \overset{2}{\to} Z \to 0.$

L1givenalgebra

2.1

If the sequence split as complexes, the middle term would be isomorphic to $Z [0] \oplus Z [1]$ , whose homology is $Z$ in degrees $0$ and $1$ by [L2]. But the middle term from step 1.1 has homology $H_{0} ≅ Z / 2$ and $H_{1} = 0$ . Hence no chain splitting exists, so the displayed sequence is a counterexample.

L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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)