False statementConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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fs-every-complex-of-projectives-is-homotopically-projective.md

Statement

Every complex of projective modules, without any boundedness hypothesis, is K-projective.

Facts & Assumptions

Given: Every complex of projective modules, without any boundedness hypothesis, is K-projective.

[F1]

K-projectivity requires vanishing of Hom into every acyclic shift (Homotopically projective bounded above complex).

[F2]

A projective module lifts maps through epimorphisms (Projective modules and the lifting property).

Refutation

1.1

Let $R = Z / 4$ and $P^{i} = R$ for every integer $i$ , with all differentials multiplication by two. Then $d^{2} = 4 = 0$ and $\ker (2) = 2 R = im (2)$ , so $P$ is acyclic. Each term is projective: a map out of $R$ lifts through any epimorphism by choosing a preimage of the value at $1$ .

F2algebra

2.1

Every $R$ -linear $h^{i} : R \to R$ is multiplication by some $a_{i}$ . If $1_{P} = d h + h d$ , degree $i$ gives $1 = 2 a_{i} + 2 a_{i + 1}$ in $R$ , impossible after reduction modulo two. Thus ${Hom}_{K} (P, P)$ contains a nonzero identity although its target $P$ is acyclic. This violates K-projectivity.

F1step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

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