Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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.

A square commutes if and only if its transposed square commutes

Statement

Let F⊣G be an adjunction between locally small categories. Suppose

u:Fc→d,u′:Fc′→d′,a:c→c′,b:d→d′.

Then

b∘u=u′∘F(a)⟺G(b)∘u♭=(u′)♭∘a.

The analogous equivalence holds after applying inverse transposition to a square between morphisms c→Gd.

Facts & Assumptions

Given: The adjunction and the four typed morphisms in the Statement.

[L1]

Transposition is a bijection natural in both variables: for a:c→c′, b:d→d′ and v:Fc′→d, one has Φc,d′(b∘v∘F(a))=G(b)∘Φc′,d(v)∘a (Under local smallness, transposition gives the natural hom-set bijection, and conversely).

Proof

technique · direct
1.1L1

If b∘u=u′∘F(a), apply transposition. By [L1], the transpose of the left side is G(b)∘u♭, while the transpose of the right side is (u′)♭∘a, so the transposed square commutes.

1.2L1

Conversely, if the transposed square commutes, apply the inverse bijection to its two sides. The two inverse images are b∘u and u′∘F(a) by [L1], so the original square commutes.

2.1step 1.1step 1.2L1∎

Repeating steps 1.1 and 1.2 with the inverse bijections proves the analogous assertion for inverse transposition.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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