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

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The two Hom double-complex differentials commute before signing

Statement

For the Hom double complex $K^{p, q} = {Hom}_{A} (P_{p}, I^{q})$ , the unsigned horizontal and vertical maps commute:

h^{p, q + 1} v^{p, q} = v^{p + 1, q} h^{p, q} : K^{p, q} ⟶ K^{p + 1, q + 1} .

Moreover, $h^{2} = 0 = v^{2}$ . Consequently the signed total differential $D ∣_{K^{p, q}} = h + (- 1)^{p} v$ satisfies $D^{2} = 0$ .

Facts & Assumptions

Given: Projective and injective resolutions $P_{∙} \to M$ and $N \to I^{∙}$ , with the maps $h (f) = f \circ d_{P}$ and $v (f) = d_{I} \circ f$ .

Proof

technique · direct

1.1givenalgebra

For $f \in K^{p, q}$ , both mixed composites equal $d_{I}^{q} \circ f \circ d_{P, p + 1}$ , so $h v (f) = v h (f)$ . The resolution identities also give $h^{2} (f) = f \circ d_{P, p + 1} d_{P, p + 2} = 0$ and $v^{2} (f) = d_{I}^{q + 1} d_{I}^{q} \circ f = 0$ .

2.1step 1.1algebra∎

Since the horizontal degree increases from $p$ to $p + 1$ after applying $h$ , $D^{2} f = h^{2} f + (- 1)^{p + 1} v h f + (- 1)^{p} h v f + v^{2} f = 0.$

The two Hom double-complex differentials commute before signing

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources