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PropositionStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passaudited 2026-08-31
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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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 Hn(C[k])Hnk(C).

Facts & Assumptions

Given: A chain complex C and integers n,k.

[L1]

The shifted differential is dnC[k]=(1)kdnkC, so C[k]n=Cnk (The shift of a chain complex).

[L2]

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

Proof

technique · direct
1.1

Because the differential in [L1] differs from dnkC only by the unit (1)k, its kernel and image are the same subobjects. Hence Zn(C[k])=Znk(C),Bn(C[k])=Bnk(C).

L1givenalgebra
2.1

Applying [L2] to the equalities of step 1.1 yields Hn(C[k])=Zn(C[k])/Bn(C[k])Znk(C)/Bnk(C)=Hnk(C).

L2step 1.1algebra

Depends on

Used by

Dependency tree · two levels

6 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