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

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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.

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The singular augmentation commutes with the boundary

Statement

For every topological space $X$ and every abelian group $G$ , $ε_{X} \partial_{1} = 0 : C_{1} (X; G) \to G .$ Consequently $im \partial_{1} \subseteq \ker ε_{X}$ , so the reduced singular chain complex of Augmentation at 0-simplices and reduced singular homology is well defined.

Facts & Assumptions

Given: A topological space $X$ and an abelian group $G$ .

[L1]

The augmentation sends every $0$ -simplex tensor $g$ to $g$ (Augmentation at 0-simplices and reduced singular homology).

[L2]

The singular boundary of a $1$ -simplex is the terminal $0$ -face minus the initial $0$ -face (The singular boundary operator).

Proof

technique · direct

1.1L1L2givenalgebra

Let $σ : Δ^{1} \to X$ be a singular $1$ -simplex and let $g \in G$ . By [L2], $\partial_{1} (σ \otimes g) = (σ δ_{0}) \otimes g - (σ δ_{1}) \otimes g .$ Applying [L1] gives $ε_{X} \partial_{1} (σ \otimes g) = g - g = 0.$

2.1step 1.1∎

The generators $σ \otimes g$ span $C_{1} (X; G)$ , so step 1.1 proves $ε_{X} \partial_{1} = 0$ on all of $C_{1} (X; G)$ . Therefore every $1$ -boundary lies in $\ker ε_{X}$ , which is exactly the compatibility needed in degree $0$ for the reduced chain complex.

Depends on

Dependency tree · 2 levels

Sources

Allen Hatcher, Algebraic Topology