CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-29

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The members of an object do not form a group

Statement refuted

For every object in an abelian category, its members modulo equivalence form an group under the ambient addition of arrows.

Facts & Assumptions

Given: The abelian category $A b$ and the identity member $1_{Z} : Z \to Z$ .

[L1]

The category $A b$ is abelian (Abelian groups form an abelian category).

[L2]

Every member has a negative, and equivalence to zero is literal equality to the zero morphism (Each object has a zero member and each member has a negative).

Counterexample

technique · direct

1.1L1L2algebra

In $A b$ , the identity member satisfies $1_{Z} \equiv - 1_{Z}$ because $1_{Z} (- 1_{Z}) = (- 1_{Z}) 1_{Z}$ and the automorphism $- 1_{Z}$ is epic.

1.2L2assume-hypalgebra

If member classes carried a group law induced from arrow addition, then $[1_{Z}] + [1_{Z}] = [1_{Z}] + [- 1_{Z}] = [0]$ . But the left-hand class is represented by $2 \cdot 1_{Z} : Z \to Z$ , and [L2] says $2 \cdot 1_{Z} \equiv 0$ would force $2 \cdot 1_{Z} = 0$ , which is false in $A b$ .

2.1step 1.1step 1.2∎

Therefore the member classes of an object need not carry an induced group law.

Depends on

Used by

FALSE: the members of an object form an abelian group False statement

Dependency tree · two levels

11 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

Saunders Mac Lane, Categories for the Working Mathematician, VIII.4 (standard reference, not scraped)