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.

Uniqueness of the fibre product

Statement

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 $q^{'} u = q$ .

Facts & Assumptions

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

[F1]

Let $f : X \to S$ and $g : Y \to S$ be morphisms of schemes. A fibre product is a scheme $P$ , with projections $p : P \to X$ and $q : P \to Y$ , such that $f p = g q$ and, for every scheme $T$ and morphisms $a : T \to X$ , $b : T \to Y$ with $f a = g b$ , there is exactly one $h : T \to P$ satisfying $p h = a$ and $q h = b$ . Thus, naturally in every test scheme $T$ , $Hom (T, P) ≅ Hom (T, X) \times_{Hom (T, S)} Hom (T, Y) .$ Write $P = X \times_{S} Y$ . The commutative square with edges $p, q, f, g$ is Cartesian when it has this universal property. Morphisms here are morphisms of locally ringed spaces, as in def-morphism-of-schemes. No existence assertion is part of the definition. (Fibre product of schemes)

Proof

1.1givenF1

Apply the universal property of $P^{'}$ to the compatible maps $p, q$ . It supplies a unique map $u : P \to P^{'}$ with the required projections. This works also for $P = \emptyset$ .

2.1F1step 1.1

Apply the universal property of $P$ to $p^{'}, q^{'}$ to obtain $v : P^{'} \to P$ . Both $v u$ and ${id}_{P}$ have projections $p, q$ , so uniqueness gives $v u = {id}_{P}$ ; likewise $u v = {id}_{P^{'}}$ .

3.1step 1.1step 2.1∎

Thus $u$ is an isomorphism, and any projection-compatible isomorphism must equal the map already uniquely obtained. No condition on the number of points or on local nilpotents was used.

Depends on

Fibre product of schemes

Used by

Group schemes over a base scheme Definition
Gluing fibre products along open covers Lemma
Restricting fibre products to open subschemes Lemma
Symmetry, associativity and units Lemma
The graph is a pullback of the diagonal Lemma

Dependency tree · two levels

2 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 and 10.1.5 (standard reference, not scraped)