False statementConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-08-27

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FALSE: preserving zero morphisms is enough for additivity

Statement

False claim: a functor between additive categories is additive whenever it preserves zero morphisms.

Facts & Assumptions

Given: The functor $T : A b \to A b$ sending an abelian group $A$ to the reduced free abelian group $Z [U (A)] / Z [0]$ on its underlying pointed set.

[L1]

Every additive functor preserves zero morphisms (An additive functor preserves zero morphisms).

Refutation

technique · direct

1.1

If $0 : A \to B$ is the zero homomorphism, then the induced set map on underlying pointed sets sends every element to the basepoint $0 \in B$ . Therefore $T (0)$ sends every generator to $[0] = 0$ , so $T$ preserves zero morphisms.

givenL1

1.2

Let $u = v = 1_{Z}$ . In $T (Z)$ write $[n]$ for the class of the generator $n \in Z$ . Then $T (u + v) ([1]) = T (2) ([1]) = [2]$ , while $(T (u) + T (v)) ([1]) = [1] + [1] = 2 [1]$ . These are distinct elements of the free abelian group on the nonzero integers, so $T (u + v) \neq T (u) + T (v)$ .

given

2.1

Thus $T$ preserves zero morphisms without being additive, refuting the claim.

step 1.1step 1.2∎

Depends on

An additive functor preserves zero morphisms

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Peter Freyd, Abelian Categories, Exercise 2A (standard reference, not scraped)