LemmaStatement: AI-generatedProof: AI-generated

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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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Finite derived products and motivic coefficient operations

PDF p. 8: Computations after the splitting principle

Statement

For a finite family $C_{α} \in D (Z)$ and $i \in Z$ , $H^{i} ⁣ (\prod_{α} C_{α}) ≅ \prod_{α} H^{i} (C_{α}) .$ For every positive integer $n$ , writing $C / n = C \otimes^{L} Z / n$ , one also has $(\prod_{α} C_{α}) / n ≃ \prod_{α} (C_{α} / n) .$ These equivalences are induced by the coordinate projections. Consequently a finite-product decomposition of derived complexes induces the corresponding decomposition on integral and finite-coefficient cohomology, and each such joint restriction map is injective. The empty product is zero throughout.

Facts & Assumptions

Given: A finite family of derived complexes, an integer $i$ , and $n > 0$ .

Goal: Prove compatibility with cohomology and derived finite coefficients.

[F1]

Cohomology is functorial on $D (Z)$ . Cohomology descends to the derived category

[F2]

Derived tensor with $Z / n$ is the mapping cone of multiplication by $n$ and is functorial. Functorial finite coefficients for derived complexes

Proof

1.1F1constructalgebra

Represent each $C_{α}$ by a cochain complex. Their finite degreewise direct sum is also their finite product. Products of finitely many abelian groups are exact: a coordinatewise surjection lifts a tuple by finitely many lifts, and kernels are computed coordinatewise. Hence this degreewise product preserves quasi-isomorphisms and represents the derived product. Its cycles and boundaries in every degree are the respective finite products of the coordinate cycles and boundaries, proving the first isomorphism.

2.1F2step 1.1algebra

Multiplication by $n$ on this product is coordinatewise. Its cone has in degree $d$ the sum of the product in degree $d$ with that in degree $d + 1$ , with the usual coordinatewise cone differential. Finite sums commute with finite products, so this cone is canonically the finite product of the individual cones. By [F2] this is exactly the second displayed equivalence.

3.1F1step 1.1step 2.1∎

Apply [F1] and step 1.1 to step 2.1. An element whose every product coordinate is zero is zero, yielding the claimed detection for both coefficient choices. With no indices every complex and group displayed is zero, and the same conclusions hold.

Depends on

Used by · 3

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