False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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FALSE: a chain map is determined by its maps on homology

Statement

If two chain maps induce the same morphism on every homology object, then the two chain maps are equal.

Facts & Assumptions

Given: The two-term complex $0 \to Z \overset{1_{Z}}{\to} Z \to 0$ and the endomorphisms $f = 1_{C}$ and $g = 0_{C}$ of this complex.

[L1]

$A b$ is an abelian category (Abelian groups form an abelian category).

[L2]

Chain maps induce homology maps (A chain map induces a well-defined map on homology).

Refutation

technique · direct

1.1L1givenalgebra

The complex is acyclic: in degree $1$ the kernel of $1_{Z}$ is $0$ , and in degree $0$ the cokernel of $1_{Z}$ is also $0$ . Hence all its homology objects are zero. The maps $f$ and $g$ are distinct because $f_{1} = 1_{Z}$ while $g_{1} = 0$ .

2.1L2step 1.1∎

By [L2], both $f$ and $g$ induce the zero endomorphism on every homology object, since those homology objects are zero by step 1.1. Therefore equal homology maps do not force equality of chain maps.

Depends on

Used by

Two distinct chain maps inducing the same homology map Counterexample

Dependency tree · two levels

11 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, Chapter 1 of An Introduction to Homological Algebra (standard reference, not scraped)