TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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A coproduct of projectives is projective and a product of injectives is injective

Statement

Assume an abelian category satisfies AB3 and AB3*.

Every finite coproduct of projective objects is projective, and every finite product of injective objects is injective.
For an arbitrary small family, the same conclusion holds provided one may choose one lift or one extension for each index in each lifting problem; in particular it holds under the Axiom of Choice.

Facts & Assumptions

Given: An abelian category satisfying AB3 and AB3*, and small families $(P_{i})$ of projective objects and $(I_{i})$ of injective objects.

[L1]

AB3 and AB3* supply the required coproducts and products (The axioms AB3 and AB3*).

[L2]

Projective objects are characterized by the lifting property against epimorphisms, and injective objects dually by the extension property against monomorphisms (Projective object characterisations, Injective object characterisations).

Proof

technique · direct

1.1L1L2chooseconstruct

By [L1], let $P = ∐_{i} P_{i}$ . Given an epimorphism $q : E ↠ M$ and a morphism $f : P \to M$ , write $f_{i} = f ι_{i}$ on the coproduct summands. Because each $P_{i}$ is projective, [L2] gives a lift ${\tilde{f}}_{i} : P_{i} \to E$ of $f_{i}$ . For a finite family these lifts are chosen explicitly; for an arbitrary small family they are exactly the stated choice-dependent data. The coproduct universal property then assembles the ${\tilde{f}}_{i}$ into a lift $\tilde{f} : P \to E$ , so $P$ is projective.

1.2L1L2chooseconstruct

The injective claim is dual. By [L1], let $I = \prod_{i} I_{i}$ . Given a monomorphism $m : M ↣ E$ and a morphism $f : M \to I$ , write $f_{i} = π_{i} f$ . Each injective object $I_{i}$ admits an extension ${\tilde{f}}_{i} : E \to I_{i}$ by [L2]. The product universal property assembles them into $\tilde{f} : E \to I$ extending $f$ . So $I$ is injective.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the finite case without extra choice and the arbitrary small-family case with the stated choice boundary.

Depends on

Used by

Dependency tree · two levels

9 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

Pavel Etingof, Shlomo Gelaki, Dmitri Nikshych, and Victor Ostrik, Tensor Categories, Section 1.6 (standard reference, not scraped)