LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-06

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.

Boundary on relative chains

Statement

For $A \subseteq X$ , the singular boundary induces homomorphisms ${\overset{ˉ}{\partial}}_{n} : C_{n} (X, A; G) \to C_{n - 1} (X, A; G)$ , and ${\overset{ˉ}{\partial}}_{n - 1} {\overset{ˉ}{\partial}}_{n} = 0$ .

Facts & Assumptions

Given: A subspace $A \subseteq X$ and a fixed abelian group $G$ .

Proof

technique · direct

1.1

Inclusion sends every singular simplex of $A$ to the same simplex in $X$ ; its faces still have image in $A$ . Hence $\partial C_{n} (A; G) \subseteq C_{n - 1} (A; G)$ , so $[c] \mapsto [\partial c]$ is well defined on the quotient.

givenconstruct

2.1

Applying this map twice gives $[\partial^{2} c] = 0$ , since the singular boundary squares to zero.

step 1.1algebra∎

Depends on

Used by

Relative singular homology Definition

Cited to discharge well-definedness by Relative singular chain complex.

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

Allen Hatcher, Algebraic Topology, §2.1 (standard reference, not scraped)