Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-29
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.

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Two morphisms agreeing on every member need not be equal

Statement refuted

If two morphisms f,g:A→B satisfy fx≡gx for every member x of A, then f=g.

Facts & Assumptions

Given: The abelian category Ab and the two endomorphisms 1Z,−1Z:Z→Z.

[L1]

The category Ab 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
1.1L1L2algebra

Let x:X→Z be any member. Then (−1Z)x=x(−1X), and the automorphism −1X is epic. Therefore 1Zx≡(−1Z)x.

1.2L1algebra

Nevertheless 1Z≠−1Z, since they send 1∈Z to different integers. So memberwise equivalence of composites does not force equality of the morphisms themselves.

2.1step 1.1step 1.2∎

This refutes the statement.

Depends on

Used by

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