LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07‡ 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.

The cellular boundary squares to zero

Statement

For every $n \geq 1$ , $d_{n - 1} d_{n} = 0$ .

Facts & Assumptions

Given: The three-skeleton definition of $d$ in Cellular boundary from three consecutive skeleta.

Proof

technique · direct

1.1

For $n = 1$ the composite is zero because $d_{0} = 0$ by definition.

given

1.2

Let $n \geq 2$ . Write $d_{n} = j_{n - 1} \partial_{n}, d_{n - 1} = j_{n - 2} \partial_{n - 1},$ where $j_{n - 1} : H_{n - 1} (X^{n - 1}) \to H_{n - 1} (X^{n - 1}, X^{n - 2})$ is the quotient map. In the long exact sequence of the pair $(X^{n - 1}, X^{n - 2})$ , $j_{n - 1}$ is immediately followed by $\partial_{n - 1}$ .

given

2.1

Exactness gives $\partial_{n - 1} j_{n - 1} = 0$ , and hence $d_{n - 1} d_{n} = j_{n - 2} (\partial_{n - 1} j_{n - 1}) \partial_{n} = 0.$ Together with step 1.1 this covers every $n \geq 1$ .

step 1.1step 1.2∎

Depends on

Used by

Cellular homology Definition

Dependency tree · two levels

5 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

J. Peter May, A Concise Course in Algebraic Topology, Chapter 13 (standard reference, not scraped)