CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)

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.

Sum and product totalisations can differ on infinite diagonals

Statement refuted

Whenever both totalisations of a homological double complex exist, they are isomorphic as chain complexes.

Facts & Assumptions

[F1]

Direct sum total complex of a double complex and The total differential squares to zero give the direct-sum total complex.

[F2]

Product total complex of a double complex gives the product total complex.

[F3]

Countable sequence groups and tail filtrations constructs $S = (Z / 2)^{(N)}$ and $P = (Z / 2)^{N}$ and proves that $S$ is countably infinite while $P$ is uncountable.

Counterexample

Given: The category of abelian groups, and $C_{j, - j} = Z / 2$ for $j \geq 0$ , with every other component zero and every horizontal and vertical arrow zero.

1.1

Each individual square and each mixed composite is zero, so these data are an anticommuting double complex. Its only nonzero diagonal is total degree zero. The two total objects there are respectively $S$ and $P$ by their universal properties; all other total degrees are zero. The total differentials are zero by their defining formulas, so both constructions exist as chain complexes.

F1F2F3given

2.1

Any chain-complex isomorphism between them would have an isomorphism $S \to P$ in degree zero, hence a bijection of the underlying sets. Composing it with the enumeration of $S$ would enumerate $P$ , contrary to its proved uncountability. Thus even an abstract chain isomorphism is impossible. In particular the canonical comparison is the finite-support inclusion, which misses the constant-one sequence. The infinitely many nonzero components in degree zero are essential to this witness; the zero groups in other degrees cause no exception.

F3step 1.1∎

Depends on

Used by

Sum and product totalisations on an infinite diagonal Counterexample
Direct sum and product totalisations are always isomorphic False statement

Dependency tree · two levels

8 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

Weibel, Chapter 5; explicit infinite-diagonal binary witness (standard reference, not scraped)