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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The tensor-total differential is balanced, well defined, and squares to zero

Statement

Let $P$ be a chain complex of right $R$ -modules and $Q$ a chain complex of left $R$ -modules. On the finite-diagonal total module $Tot (P \otimes_{R} Q)$ , the formula $d (p \otimes q) = d_{P} p \otimes q + (- 1)^{p} p \otimes d_{Q} q$ is balanced and satisfies $d^{2} = 0$ .

Proof

Given: homogeneous $p \in P_{p}$ , $q \in Q_{q}$ , and the tensor-total convention.

1.1

For $r \in R$ , $d ((p r) \otimes q) = d_{P} p r \otimes q + (- 1)^{p} p r \otimes d_{Q} q = d (p \otimes r q)$ , since both differentials are $R$ -linear; thus the formula descends from elementary tensors.

given

2.1

Applying $d$ again gives $d_{P}^{2} p \otimes q + (- 1)^{p - 1} d_{P} p \otimes d_{Q} q + (- 1)^{p} d_{P} p \otimes d_{Q} q + p \otimes d_{Q}^{2} q$ .

step 1.1algebra

3.1

The first and last terms vanish and the two middle terms cancel. Linearity then proves $d^{2} = 0$ on every finite sum in each total degree.

step 2.1algebra∎

Depends on

Used by

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Dependency tree · two levels

7 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 (standard reference, not scraped)