LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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.

Induced singular chain maps commute with boundaries

Statement

If $f : X \to Y$ is continuous, then for every $n \geq 1$ and every abelian group $G$ , $\partial_{n} \circ f_{#, n} = f_{#, n - 1} \circ \partial_{n} : C_{n} (X; G) \to C_{n - 1} (Y; G) .$ In degree $0$ , both composites from $C_{0} (X; G)$ to $0$ are the zero map.

Facts & Assumptions

Given: A continuous map $f : X \to Y$ , an abelian group $G$ , and an integer $n \geq 0$ .

[L1]

The induced map sends a singular simplex $σ$ to the composite $f \circ σ$ (The induced singular chain map of a continuous map).

[L2]

The singular boundary is the alternating sum of the affine face restrictions (The singular boundary operator).

Proof

technique · direct

1.1L2given

If $n = 0$ , then $\partial_{0} = 0$ by [L2], so both composites from $C_{0} (X; G)$ to $0$ are the zero map.

1.2L1L2givenalgebra

Assume $n \geq 1$ and let $σ : Δ^{n} \to X$ be a singular $n$ -simplex. By [L1] and [L2], $\partial_{n} f_{#, n} (σ) = \sum_{i = 0}^{n} (- 1)^{i} (f \circ σ) \circ δ_{i} = \sum_{i = 0}^{n} (- 1)^{i} f \circ (σ \circ δ_{i}) .$ The same formulas give $f_{#, n - 1} \partial_{n} (σ) = f_{#, n - 1} (\sum_{i = 0}^{n} (- 1)^{i} σ \circ δ_{i}) = \sum_{i = 0}^{n} (- 1)^{i} f \circ (σ \circ δ_{i}),$ so the two values agree on every singular simplex.

2.1step 1.1step 1.2∎

Singular simplices generate $C_{n} (X; Z)$ , and the coefficient- $G$ map is the tensor extension of the integer-coefficient map. Therefore step 1.2 proves the identity for every $n \geq 1$ , while step 1.1 handles degree $0$ .

Depends on

Used by

Functoriality of relative homology Proposition
Singular chains and singular homology are covariantly functorial Proposition
Singular cohomology is contravariantly functorial Proposition

Dependency tree · two levels

4 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 (standard reference, not scraped)