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.

Symmetry, associativity and units

Statement

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.

Facts & Assumptions

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

[F1]

Every diagram $X \to S \leftarrow Y$ of schemes has a fibre product. Given an affine cover $S = ⋃_{i} Spec A_{i}$ and affine covers $f^{- 1} (Spec A_{i}) = ⋃_{j} Spec B_{i j}$ and $g^{- 1} (Spec A_{i}) = ⋃_{k} Spec C_{i k}$ , the product has open affine cover $Spec (B_{i j} \otimes_{A_{i}} C_{i k}) .$ (Existence of all scheme fibre products)

[F2]

If $(P, p, q)$ and $(P^{'}, p^{'}, q^{'})$ are fibre products of the same pair $X \to S \leftarrow Y$ , there is a unique isomorphism $u : P \to P^{'}$ with $p^{'} u = p$ and $q^{'} u = q$ . (Uniqueness of the fibre product)

Proof

1.1

All products exist by F1. For symmetry swap the two projections; applying this operation twice restores them. For either triple product a map from $T$ is exactly three maps to $X, Y, Z$ with the same composite to $S$ . Thus the projections construct mutually inverse maps between the two bracketings.

givenF1

2.1

The pair $({id}_{X}, f)$ supplies $X \to X \times_{S} S$ and the first projection supplies the inverse. A pair into $X, S$ over $S$ has its second coordinate forced by the first. Every inverse assertion follows by uniqueness, as in F2. This includes empty factors and identity structure maps.

F2step 1.1

3.1

Naturality and all claimed coherence equations are checked after each original projection. Both sides then give exactly the same coordinate maps. Repeated uniqueness in the binary universal property makes the maps equal. This holds for arbitrary test schemes with nilpotents as well as one-point tests.

F1F2step 1.1step 2.1∎

Depends on

Used by

Intersections of subschemes Lemma
Iterated base change 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 10.1.3; proof 10.1.1 Step 1 (standard reference, not scraped)