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.

The graph is a pullback of the diagonal

Statement

For an $S$ -morphism $u : X \to Y$ , put $H = (u {pr}_{X}, {pr}_{Y}) : X \times_{S} Y \to Y \times_{S} Y$ . The square with top arrow $Γ_{u} : X \to X \times_{S} Y$ , bottom arrow $Δ_{Y / S} : Y \to Y \times_{S} Y$ , left arrow $u$ , and right arrow $H$ is Cartesian.

Facts & Assumptions

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

[F1]

For an $S$ -morphism $u : X \to Y$ (as in def-scheme-over-base), the graph morphism is $Γ_{u} = ({id}_{X}, u) : X \to X \times_{S} Y$ , supplied by thm-fibre-products-of-schemes-exist. Its first projection is the identity and its second projection is $u$ . The definition alone does not assert that its image is closed. (The graph morphism over a base)

[F2]

For $X \to S$ , the diagonal morphism is the unique $Δ_{X / S} : X \to X \times_{S} X$ satisfying ${pr}_{1} Δ_{X / S} = {id}_{X} = {pr}_{2} Δ_{X / S}$ . It exists by thm-fibre-products-of-schemes-exist. For any test scheme $T$ , it takes an $S$ -morphism $a : T \to X$ to the compatible pair $(a, a)$ . (The diagonal morphism)

[F3]

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

By F1 and F2 both composites around the square are $(u, u)$ . A compatible test pair consists of $a : T \to X \times_{S} Y$ and $c : T \to Y$ satisfying $H a = Δ c$ . Write $a = (a_{X}, a_{Y})$ . Equality means exactly $u a_{X} = c$ and $a_{Y} = c$ .

givenF1F2

2.1

Thus $a_{X}$ is the unique map $T \to X$ whose graph composite is $a$ and whose $u$ -composite is $c$ . Conversely any map $d : T \to X$ supplies the pair $((d, u d), u d)$ . These operations are inverse, so the square has the pullback universal property, with its canonical uniqueness as in F3. The argument includes empty schemes and $u = id$ , and imposes no reducedness or separation hypothesis.

F1F3step 1.1∎

Depends on

Used by

The ideal of a polynomial graph Example

Dependency tree · two levels

6 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 11.1.18, diagram (11.1.18.1), p.232 (standard reference, not scraped)