CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Two morphisms agreeing on every member need not be equal

Statement refuted

If two morphisms $f, g : A \to B$ satisfy $f x \equiv g x$ for every member $x$ of $A$ , then $f = g$ .

Facts & Assumptions

Given: The abelian category $A b$ and the two endomorphisms $1_{Z}, - 1_{Z} : Z \to Z$ .

[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 (Each object has a zero member and each member has a negative).

Counterexample

technique · direct

1.1

Let $x : X \to Z$ be any member. Then $(- 1_{Z}) x = x (- 1_{X})$ , and the automorphism $- 1_{X}$ is epic. Therefore $1_{Z} x \equiv (- 1_{Z}) x$ .

L1L2algebra

1.2

Nevertheless $1_{Z} \neq - 1_{Z}$ , since they send $1 \in Z$ to different integers. So memberwise equivalence of composites does not force equality of the morphisms themselves.

L1algebra

2.1

This refutes the statement.

step 1.1step 1.2∎

Depends on

Used by

FALSE: two morphisms that agree on every member are equal False statement

Dependency tree · two levels

12 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)