LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31

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The shifted differential squares to zero

Statement

For every chain complex $C_{∙}$ and every $k \in Z$ , the shifted differential on $C [k]_{∙}$ satisfies $d_{n - 1}^{C [k]} d_{n}^{C [k]} = 0$ for all $n \in Z$ .

Facts & Assumptions

Given: A chain complex $C_{∙}$ , an integer $k$ , and an integer $n$ .

[L1]

The shift satisfies $d_{n}^{C [k]} = (- 1)^{k} d_{n - k}^{C}$ (The shift of a chain complex).

Proof

technique · direct

1.1L1givenalgebra

By [L1], $d_{n - 1}^{C [k]} d_{n}^{C [k]} = (- 1)^{k} d_{n - k - 1}^{C} (- 1)^{k} d_{n - k}^{C} = (- 1)^{2 k} d_{n - k - 1}^{C} d_{n - k}^{C} .$

2.1step 1.1algebra∎

Since $C_{∙}$ is a chain complex, the last composite in step 1.1 is zero. Therefore $d_{n - 1}^{C [k]} d_{n}^{C [k]} = 0$ for all $n$ .

Depends on

The shift of a chain complex

The shifted differential squares to zero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources