LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Iterated base change

Statement

For $S^{''} \overset{k}{\to} S^{'} \overset{h}{\to} S$ and an $S$ -scheme $X$ , there is a canonical isomorphism $(X \times_{S} S^{'}) \times_{S^{'}} S^{''} ≅ X \times_{S} S^{''} .$ 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 $h : S^{'} \to S$ . For an $S$ -scheme $f : X \to S$ , its base change is $X_{S^{'}} = X \times_{S} S^{'}$ , with structure map the second projection. For an $S$ -morphism $u : X \to Y$ , define $u_{S^{'}} : X_{S^{'}} \to Y_{S^{'}}$ by its projections $u \circ {pr}_{X}$ 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 $X \times_{S} Y ≅ Y \times_{S} X, (X \times_{S} Y) \times_{S} Z ≅ X \times_{S} (Y \times_{S} Z), X \times_{S} S ≅ X ≅ S \times_{S} X .$ Any coherence identity between these identifications holds whenever both sides induce the same ordered projections to the original factors. (Symmetry, associativity and units)

Proof

1.1

Using F1, a map $T \to (X \times_{S} S^{'}) \times_{S^{'}} S^{'^{'}}$ is a triple $(a, b, c)$ into $X, S^{'}, S^{'^{'}}$ satisfying $f a = h b$ and $b = k c$ . Eliminating $b$ gives exactly a pair $(a, c)$ satisfying $f a = h k c$ .

givenF1

2.1

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

F2step 1.1

3.1

For $u : X \to Y$ the corresponding pair becomes $(u a, c)$ 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 $u : X \to Y$ and $v : X^{'} \to Y^{'}$ with $P$ also has $P$ : factor it as $X \times_{S} X^{'} \to Y \times_{S} X^{'} \to Y \times_{S} Y^{'}$ . The first arrow is the pullback of $u$ along $Y \times_{S} X^{'} \to Y$ , and the second the pullback of $v$ along $Y \times_{S} Y^{'} \to Y^{'}$ . Their test pairs verify these pullback identifications.

F1step 1.1step 2.1∎

Depends on

Used by

Fibres after base change Lemma
Independence of the chosen algebraic closure Lemma
The diagonal commutes with base change Lemma

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

Vakil proof 10.1.1 Step 1; 10.3.C (standard reference, not scraped)