Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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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.

A square commutes if and only if its transposed square commutes

Statement

Let FG be an adjunction between locally small categories. Suppose

u:Fcd,u:Fcd,a:cc,b:dd.

Then

bu=uF(a)G(b)u=(u)a.

The analogous equivalence holds after applying inverse transposition to a square between morphisms cGd.

Facts & Assumptions

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

[L1]

Transposition is a bijection natural in both variables: for a:cc, b:dd and v:Fcd, one has Φc,d(bvF(a))=G(b)Φc,d(v)a (Under local smallness, transposition gives the natural hom-set bijection, and conversely).

Proof

technique · direct
1.1

If bu=uF(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.

L1
1.2

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

L1
2.1

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

step 1.1step 1.2L1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources