PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-29

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.

Restrictions, corestrictions, and products of smooth maps are smooth

Statement

Let $F : M \to N$ and $G : P \to Q$ be smooth maps of smooth manifolds.

If $U \subseteq M$ is open, with its restricted smooth structure, then the restriction $F ∣_{U} : U \to N$ is smooth.
If $W \subseteq N$ is open and $F [M] \subseteq W$ , then the corestriction $F^{W} : M \to W$ is smooth for the restricted smooth structure on $W$ .
The product map $F \times G : M \times P \to N \times Q, (m, p) \mapsto (F (m), G (p)),$ is smooth for the canonical product smooth structures.

Facts & Assumptions

Given: Smooth maps $F : M \to N$ and $G : P \to Q$ of smooth manifolds.

[F1]

An open subset of a smooth manifold carries a canonical restricted smooth structure (An open subset of a smooth manifold has a canonical restricted smooth structure).

[F2]

Identity maps and composites of smooth maps are smooth (Identity maps and composites of smooth maps are smooth).

[F3]

A map into a product smooth manifold is smooth exactly when its component maps are smooth (A map into a product is smooth iff its components are smooth).

[F4]

Products of smooth manifolds carry canonical product smooth structures (Products of smooth manifolds have a canonical product smooth structure).

Proof

technique · direct

1.1F1F2

Let $U \subseteq M$ be open. [F1, F2] By [F1] the subset $U$ is a smooth manifold, and the inclusion $j : U \to M$ has identity coordinate representative in restricted charts, hence is smooth. Since $F ∣_{U} = F \circ j$ , [F2] makes $F ∣_{U}$ smooth.

1.2F2F3F4

By [F4] the source and target products are smooth manifolds. [F2, F3, F4] The first component of $F \times G$ is $F \circ π_{M}$ , and the second is $G \circ π_{P}$ , where the projections are smooth in product charts because they are ordinary Euclidean coordinate projections. Hence [F2] makes both components smooth, and then [F3] makes $F \times G$ smooth.

2.1F1F2step 1.1

Let $W \subseteq N$ be open and suppose $F [M] \subseteq W$ . [F1, F2, step 1.1] By [F1] the subset $W$ is a smooth manifold, and the inclusion $i : W \to N$ is smooth by the same restricted-chart identity argument as in step 1.1. In charts of $W$ inherited from $N$ , the corestriction $F^{W} : M \to W$ has exactly the same Euclidean representative as $F$ , so it is smooth.

3.1step 1.1step 1.2step 2.1∎

Step 1.1 proves the restriction claim, step 2.1 proves the corestriction claim, and step 1.2 proves the product claim.

Depends on

Used by

Transverse fibre products are embedded submanifolds Theorem

Dependency tree · two levels

26 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

Rob van der Vorst, Introduction to differentiable manifolds, §2 (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds, §2.4 (standard reference, not scraped)