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 $H_{n} (C [k]) = Z_{n} (C [k]) / B_{n} (C [k]) ≅ Z_{n - k} (C) / B_{n - k} (C) = H_{n - k} (C) .$

Depends on

Used by

A degreewise split cone sequence with no chain splitting Counterexample
Shifting a three-term complex with all signs Example
FALSE: the degreewise splitting of the cone sequence is a chain splitting False statement
Shift preserves homotopy equivalences, contractibility, and quasi-isomorphisms Proposition
The cone connecting map agrees with the shifted identity up to the declared sign Proposition
The cone long exact sequence Theorem

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

Charles A. Weibel, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)
The Stacks Project, Section 12.14: Homotopy and the shift functor (standard reference, not scraped)