Alphabeta Math
TheoremStatement: Literature-sourcedProof: Literature-sourcedprecheck passaudited 2026-08-29
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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.

Member equivalence is transitive

Statement

For members of one object in an abelian category, the relation ≡ is transitive.

Facts & Assumptions

Given: An abelian category and members x:X→A, y:Y→A, and z:Z→A with x≡y and y≡z.

[L1]

The relation x≡y is witnessed by epimorphisms from one common domain as in Equivalence of members.

[L2]
[L3]

The pullback of an epimorphism is an epimorphism (The pullback of an epimorphism is an epimorphism).

Proof

technique · direct
1.1L1choose

Choose epimorphisms u:W1→X, v:W1→Y, w:W2→Y, and r:W2→Z with xu=yv and yw=zr, using [L1].

1.2L2L3construct

Form the pullback of v and w, with projections v′:P→W1 and w′:P→W2. By [L3], both v′ and w′ are epic.

2.1L1step 1.1step 1.2algebra

The pullback equation gives vv′=ww′, so xuv′=yvv′=yww′=zrw′. Hence the epimorphisms uv′:P→X and rw′:P→Z witness x≡z.

3.1step 2.1∎

Therefore member equivalence is transitive.

Depends on

Used by

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