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

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Homology of a shift is shifted homology

Statement

For every chain complex $C_{∙}$ , every integer $k$ , and every degree $n$ , there is a natural isomorphism $H_{n} (C [k]) ≅ H_{n - k} (C) .$

Facts & Assumptions

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

[L1]

The shifted differential is $d_{n}^{C [k]} = (- 1)^{k} d_{n - k}^{C},$ so $C [k]_{n} = C_{n - k}$ (The shift of a chain complex).

[L2]

Homology is the quotient of cycles by boundaries (Homology object of a chain complex).

Proof

technique · direct

1.1L1givenalgebra

Because the differential in [L1] differs from $d_{n - k}^{C}$ only by the unit $(- 1)^{k}$ , its kernel and image are the same subobjects. Hence $Z_{n} (C [k]) = Z_{n - k} (C), B_{n} (C [k]) = B_{n - k} (C) .$

2.1L2step 1.1algebra∎

Applying [L2] to the equalities of step 1.1 yields $H_{n} (C [k]) = Z_{n} (C [k]) / B_{n} (C [k]) ≅ Z_{n - k} (C) / B_{n - k} (C) = H_{n - k} (C) .$

Homology of a shift is shifted homology

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources