ExampleConstruction: 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.

Acyclic assembly with exact columns

Example

Let $k = Z / 2$ , put $C_{p, q} = k$ for $p, q \in {0, 1}$ and zero elsewhere, set both vertical maps $C_{p, 1} \to C_{p, 0}$ equal to identity, and set all horizontal maps zero. Every column and the total complex are acyclic.

Facts & Assumptions

[F1]

Acyclic assembly lemma for a first quadrant double complex makes the total complex acyclic when every column of a first-quadrant double complex is acyclic.

[F2]

Abelian-group model for spectral-sequence computations supplies the binary group and coordinate homology quotients. Direct sum total complex of a double complex uses differential $h + v$ .

Verification

Given: The four components and maps in the example. Both vertical squares are zero because all components outside vertical indices zero and one vanish; mixed composites vanish because every horizontal map is zero.

1.1

Each nonzero column is $k \overset{1}{\to} k$ in degrees one and zero. Its degree-one kernel and degree-zero cokernel are zero, and all other columns and homology degrees are zero. Thus every column is acyclic, including its degree-zero homology. The first-quadrant support satisfies [F1], so there is no surviving edge complex and assembly gives an acyclic total complex.

F1F2

2.1

For a direct check order total degree one as $C_{1, 0} \oplus C_{0, 1}$ . The total complex is $k \overset{x \mapsto (x, 0)}{\to} k^{2} \overset{(u, v) \mapsto v}{\to} k$ in degrees $2, 1, 0$ . The first map is injective, the second is surjective, and the kernel of the second is $k \oplus 0$ , exactly the first image. Hence $H_{2} = H_{1} = H_{0} = 0$ , with all other degrees already zero. This checks the two endpoints and the middle image-kernel equality independently of the assembly invocation, using explicit maps and no AC.

F2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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, An Introduction to Homological Algebra, Chapter 5 (standard reference, not scraped)
The Stacks Project, Homological Algebra (standard reference, not scraped)