False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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FALSE: every acyclic complex of projective objects is contractible

Statement

False. Every acyclic complex of projective objects is contractible.

Facts & Assumptions

Given: The bi-infinite chain complex over $R = Z / 4 Z$ with $C_{n} = R$ for every $n$ and differential $d_{n}$ equal to multiplication by $2$ .

[L1]

Contractibility is the existence of a homotopy from the identity to zero (A contractible complex).

[L2]

A bounded-below acyclic complex of projectives is contractible once its cycle epimorphisms split (A bounded below acyclic complex of projective objects is contractible when its cycle epimorphisms split).

Refutation

technique · direct

1.1

Since $2^{2} = 4 = 0$ in $R$ , the displayed differentials satisfy $d_{n - 1} d_{n} = 0$ . Also $\ker (d_{n}) = 2 R = im (d_{n + 1}),$ so the complex is acyclic. Each term $R$ is a free rank-one $R$ -module and hence projective.

givenalgebra

2.1

If a contracting homotopy existed, then for each $n$ one would have $1_{R} = d_{n + 1} s_{n} + s_{n - 1} d_{n} = 2 s_{n} + 2 s_{n - 1} .$ But every endomorphism of the free rank-one module $R$ is multiplication by an element of $R$ , and the right-hand side is always even while $1_{R}$ is not. Contradiction.

L1step 1.1algebra

3.1

Therefore the complex is acyclic and degreewise projective but not contractible. The positive theorem [L2] does not apply because this standard counterexample is not bounded below with the required splitting data.

L2step 2.1∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Charles A. Weibel, An Introduction to Homological Algebra (standard reference, not scraped)
Romyar Sharifi, Homological Algebra (standard reference, not scraped)