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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1L1L2

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

2.1L1step 1.1algebra∎

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

Depends on

Dependency tree · 2 levels

Sources

The Stacks Project, Section 12.7, Additive functors