LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

Finite biproducts of injective objects are injective

Statement

In an abelian category, a finite biproduct of injective objects is injective, including the empty biproduct. No choice axiom is required.

Facts & Assumptions

Given: Injective objects $I_{1}, \dots, I_{m}$ with $m \geq 0$ .

[F1]

An abelian category is additive, so it has finite biproducts (Abelian category).

[F2]

Injectivity means extension of a map across any monomorphism (Injective object).

Proof

1.1

Let $u : X ↣ Y$ and $f : X \to I = ⨁_{j = 1}^{m} I_{j}$ . By the finite product property, $f$ is determined by the components $f_{j} = π_{j} f$ . For each $j$ , injectivity gives $g_{j} : Y \to I_{j}$ with $g_{j} u = f_{j}$ . The finite conjunction of these existence assertions follows by induction on $m$ in ordinary first-order logic; it uses no infinite choice.

givenF1F2

2.1

The product property supplies $g : Y \to I$ with $π_{j} g = g_{j}$ , and hence $π_{j} g u = π_{j} f$ for every $j$ . Uniqueness in the product property gives $g u = f$ , proving injectivity of $I$ . For $m = 0$ , $I = 0$ and both maps to it are unique, so the same extension property holds; for $m = 1$ the construction is precisely the extension property of $I_{1}$ . Zero summands and zero $X$ or $Y$ obey the same equations.

F1F2step 1.1∎

Depends on

Used by

A Cartan-Eilenberg resolution totalizes to an injective replacement Lemma

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

Weibel, 5.7.2 (finite-diagonal Cartan-Eilenberg totalization) (standard reference, not scraped)
Stacks Project, Tag 015G (Cartan-Eilenberg resolutions) (standard reference, not scraped)