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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The singular augmentation commutes with the boundary

Statement

For every topological space X and every abelian group G, εX1=0:C1(X;G)G. Consequently im1kerεX, 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
1.1

Let σ:Δ1X be a singular 1-simplex and let gG. By [L2], 1(σg)=(σδ0)g(σδ1)g. Applying [L1] gives εX1(σg)=gg=0.

L1L2givenalgebra
2.1

The generators σg span C1(X;G), so step 1.1 proves εX1=0 on all of C1(X;G). Therefore every 1-boundary lies in kerεX, which is exactly the compatibility needed in degree 0 for the reduced chain complex.

step 1.1

Depends on

Used by

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