LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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Acyclic assembly by exact columns

Statement

Let $K^{p, q}$ be a first-quadrant double cochain complex whose signed total complex uses finite direct sums on every diagonal. Suppose a cochain complex $C^{∙}$ maps to the bottom edge so that, for every $p$ , the augmented column $0 ⟶ C^{p} ⟶ K^{p, 0} \overset{v}{\to} K^{p, 1} \overset{v}{\to} K^{p, 2} ⟶ \dots$ is exact and the augmentations commute with the horizontal maps. Then the induced cochain map $C^{∙} \to Tot K$ is a quasi-isomorphism.

Facts & Assumptions

Given: The first-quadrant double complex, compatible column augmentations, and exact augmented columns stated above.

Proof

technique · direct

1.1

Adjoin $C^{p}$ in vertical degree $- 1$ . Compatibility makes this an augmented double complex, and the cone of $C^{∙} \to Tot K$ is its signed total complex up to shift. In total degree $n$ , only the finitely many columns $0 \leq p \leq n + 1$ occur; filtering by the largest horizontal degree has successive quotients equal to shifts of the exact augmented columns.

givenalgebra

2.1

Starting with one column and adjoining the others, the short exact sequences of successive filtered complexes show inductively that every finite truncation is acyclic. These truncations stabilize degreewise, so the full cone is acyclic. Hence $C^{∙} \to Tot K$ is a quasi-isomorphism.

step 1.1algebra∎

Depends on

The direct-sum total complex on finite diagonals

Used by

FALSE: balance of Ext requires spectral-sequence pages False statement
Acyclic assembly by exact rows Lemma
Projective and injective constructions of Ext agree for supplied resolutions Theorem
The bar cochain complex computes group cohomology Theorem

Dependency tree · two levels

3 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

Charles A. Weibel, An Introduction to Homological Algebra, Chapter 2 (standard reference, not scraped)