Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

Iterated base change

Statement

For S′′→kS′→hS and an S-scheme X, there is a canonical isomorphism (X×SS′)×S′S′′≅X×SS′′. It is functorial in X and compatible with the induced maps of S-schemes.

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

Let h:S′→S. For an S-scheme f:X→S, its base change is XS′=X×SS′, with structure map the second projection. For an S-morphism u:X→Y, define uS′:XS′→YS′ by its projections u∘pr⁡X and pr⁡S′. Existence and uniqueness follow from thm-fibre-products-of-schemes-exist; the meaning of S-morphism is def-scheme-over-base. These formulas preserve identities and composition because their projections do, so they define a functor. A property of morphisms is stable under arbitrary base change when every pullback of a morphism with that property again has it. No restriction such as flatness is implicit in “arbitrary”. (Base change of objects, morphisms and properties)

[F2]

For S-schemes X,Y,Z there are natural projection-compatible isomorphisms X×SY≅Y×SX,(X×SY)×SZ≅X×S(Y×SZ),X×SS≅X≅S×SX. Any coherence identity between these identifications holds whenever both sides induce the same ordered projections to the original factors. (Symmetry, associativity and units)

Proof

1.1givenF1

Using F1, a map T→(X×SS′)×S′S′′ is a triple (a,b,c) into X,S′,S′′ satisfying fa=hb and b=kc. Eliminating b gives exactly a pair (a,c) satisfying fa=hkc.

2.1F2step 1.1

The inverse operation is (a,c)↦(a,kc,c). They yield inverse morphisms by the projection-compatible identifications of F2. Empty schemes and identity base maps obey these same formulas.

3.1F1step 1.1step 2.1∎

For u:X→Y the corresponding pair becomes (ua,c) on both sides. Thus identities and compositions commute with the isomorphism, proving functoriality without a choice of points. More generally, if a property P is preserved by both base change and composition, the product of two S-morphisms u:X→Y and v:X′→Y′ with P also has P: factor it as X×SX′→Y×SX′→Y×SY′. The first arrow is the pullback of u along Y×SX′→Y, and the second the pullback of v along Y×SY′→Y′. Their test pairs verify these pullback identifications.

Depends on

Used by

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