LemmaStatement: AI-generatedProof: AI-generated

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.

Maps to finite disjoint unions of the base

PDF p. 5: Maps to YA before equation (2.5)

Statement

Assume the Axiom of Choice. Let $k$ be a field, $A$ a finite set, and $X$ a Noetherian $k$ -scheme. Put $Y_{A} = ∐_{a \in A} Spec k$ . There are natural bijections between $k$ -morphisms $X \to Y_{A}$ , decompositions $X = ∐_{a \in A} X_{a}$ into pairwise disjoint open-and-closed subsets, and functions $π_{0} (X) \to A$ . Here empty $X_{a}$ are allowed and $π_{0} (X)$ is the set of connected components of the underlying topological space. If $U \subseteq X$ is dense, restriction from locally constant integer-valued functions on $X$ to such functions on $U$ is injective. In particular this holds when $U$ is dense and meets every component.

A subset is open-and-closed, or clopen, if it is both open and closed. A connected component is a maximal connected subset; connected means that the subspace cannot be partitioned into two nonempty open subsets. A function is locally constant if every point has an open neighbourhood on which it is constant. Noetherian means that descending chains of closed subsets stabilize. Density means that every nonempty open subset of $X$ meets $U$ .

Statement references

Schemes and morphisms over a base

A Noetherian space is a finite union of irreducible closed subsets

The Axiom of Choice

Facts & Assumptions

[A1]

The Axiom of Choice is assumed; it is used through the finite decomposition supplier below. The Axiom of Choice

[F1]

Under the Axiom of Choice, a Noetherian topological space is a finite union of irreducible closed subsets. An irreducible space is nonempty and cannot be expressed as a union of two proper closed subsets. A Noetherian space is a finite union of irreducible closed subsets

[F2]

A $k$ -scheme has a structure morphism to $Spec k$ , and a $k$ -morphism is a scheme morphism commuting with the structure morphisms. Scheme morphisms are morphisms of the underlying locally ringed spaces. Schemes and morphisms over a base Morphisms of schemes

[F3]

The scheme $Y_{A}$ is the disjoint union of copies of $Spec k$ : each copy is open and closed, and its structure morphism to $Spec k$ is the identity. Morphisms of locally ringed spaces glue uniquely over an open cover when their restrictions agree on overlaps; this follows by gluing their continuous maps and compatible sheaf maps, with locality checked on stalks. [F2]

Proof

Given: A field $k$ , a finite set $A$ , a Noetherian $k$ -scheme $X$ , and the Axiom of Choice.

Goal: Establish the stated bijections and injectivity of restriction.

1.1F1A1given

By [F1], write $X$ as a finite union of irreducible closed subsets $C_{1}, \dots, C_{n}$ , allowing $n = 0$ . Each $C_{i}$ is connected: a separation by two nonempty relatively open subsets would express it as a union of two proper relatively closed subsets. Form the finite graph with vertices $i$ and an edge whenever $C_{i} \cap C_{j} \neq \emptyset$ . For each graph component, the union of its $C_{i}$ is connected, since a finite chain of intersecting connected sets has connected union. These unions are pairwise disjoint, closed, and cover $X$ ; each is also open because its complement is a finite union of the other closed unions. Any connected subset meeting one such union lies entirely inside it, since otherwise the two clopen parts would separate that subset. Thus these unions are exactly the connected components of $X$ , and they are clopen. The empty-space case has no components.

1.2F2F3given

Let $f : X \to Y_{A}$ be a $k$ -morphism. The inverse images $X_{a} = f^{- 1} (Spec k in summand a)$ are pairwise disjoint clopen subsets covering $X$ . Conversely, given such a decomposition, the restriction of the structure morphism $X \to Spec k$ to $X_{a}$ , followed by the inclusion of summand $a$ , is a $k$ -morphism $X_{a} \to Y_{A}$ . These morphisms glue uniquely because the open sets are disjoint. On each $X_{a}$ any $k$ -morphism landing in summand $a$ must equal this composite: commuting with the structure map forces its map to that copy of $Spec k$ to be the structure morphism. The two constructions are therefore inverse, including when $A$ or $X$ is empty.

2.1step 1.1step 1.2

A clopen decomposition assigns each connected component to exactly one $X_{a}$ : the intersections of a connected component with the $X_{a}$ are disjoint open subsets covering it, so two nonempty intersections would separate it. Conversely, by [step 1.1], the union of components assigned to each $a$ is clopen: there are only finitely many components, each clopen. These constructions are inverse. Pullback of the decomposition shows the bijections are natural in $X$ ; relabelling summands shows naturality in $A$ .

3.1givenstep 1.1∎

Let $f, g : X \to Z$ be locally constant and agree on dense $U$ . The set $D = {x : f (x) \neq g (x)}$ is open: at each point, intersect neighbourhoods on which $f$ and $g$ are respectively constant. If $D$ were nonempty it would meet $U$ , contradicting agreement there. Thus $D$ is empty and $f = g$ ; restrictions remain locally constant in the subspace topology. In particular, a dense subset meets every nonempty clopen connected component by [step 1.1], so the stated special case follows.

Depends on

Used by · 2

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