PropositionStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-27

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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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An additive functor preserves zero morphisms

Statement

If $F : C \to D$ is an additive functor between preadditive categories, then for every pair of objects $A, B$ ,

$F (0_{A, B}) = 0_{F A, F B} .$

Facts & Assumptions

Given: An additive functor $F : C \to D$ and objects $A, B \in C$ .

[L1]

In a preadditive category each hom-set is an abelian group, so each $D (F A, F B)$ has a zero element $0_{F A, F B}$ (Preadditive category).

[L2]

An additive functor induces a homomorphism on each hom-group (Additive functor).

Proof

technique · direct

1.1

In the abelian group $C (A, B)$ one has $0_{A, B} = 0_{A, B} + 0_{A, B}$ . Applying $F$ and using [L2] gives $F (0_{A, B}) = F (0_{A, B}) + F (0_{A, B})$ .

L1L2

2.1

Add the inverse of $F (0_{A, B})$ in the hom-group $D (F A, F B)$ to both sides of the equality from step 1.1. The result is $0_{F A, F B} = F (0_{A, B})$ .

L1step 1.1algebra∎

Depends on

Used by

FALSE: preserving zero morphisms is enough for additivity False statement
An additive functor preserves finite biproducts Theorem

Dependency tree · two levels

4 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

The Stacks Project, Section 12.7, Additive functors (standard reference, not scraped)