TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-28

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Assuming the Axiom of Choice, abelian groups satisfy AB4*

Statement

Assume the Axiom of Choice. Then the abelian category $A b$ satisfies AB4*.

Facts & Assumptions

Given: The Axiom of Choice and a small family of epimorphisms of abelian groups $e_{i} : A_{i} ↠ B_{i}$ .

[L1]

AB4* means that small products of epimorphisms remain epimorphic (The axioms AB4 and AB4*).

[L2]

The Axiom of Choice gives a choice function for every family of nonempty sets (The Axiom of Choice).

[F1]

In $A b$ , epimorphisms are exactly surjective homomorphisms.

Proof

technique · direct

1.1L2F1construct

Let $b = (b_{i}) \in \prod_{i} B_{i}$ . By [F1], each $e_{i}$ is surjective, so every fibre $e_{i}^{- 1} (b_{i})$ is nonempty. By [L2], choose $a_{i} \in A_{i}$ with $e_{i} (a_{i}) = b_{i}$ for every index $i$ . Then $a = (a_{i}) \in \prod_{i} A_{i}$ satisfies $(\prod_{i} e_{i}) (a) = b .$ So $\prod_{i} e_{i}$ is surjective, hence epic in $A b$ .

2.1L1step 1.1∎

This is exactly the AB4* condition of [L1]. Therefore, assuming the Axiom of Choice, $A b$ satisfies AB4*.

Depends on

Used by

FALSE under the Axiom of Choice: AB4 implies AB5 False statement

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

Charles A. Weibel, An Introduction to Homological Algebra, Appendix A.4 (standard reference, not scraped)