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PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-29
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A map into a product is smooth iff its components are smooth

Statement

Let P, M, and N be smooth manifolds, let F:P→M×N be a map, and write

f:=πM∘F,g:=πN∘F,

where πM and πN are the product projections. Then F is smooth if and only if both component maps f:P→M and g:P→N are smooth.

Facts & Assumptions

Given: Smooth manifolds P, M, N; a map F:P→M×N; and its components f:=πM∘F, g:=πN∘F.

[F1]

The product M×N carries a canonical smooth structure whose smooth charts are represented by product charts (V×W, φ×ψ) built from smooth charts (V,φ) of M and (W,ψ) of N (Products of smooth manifolds have a canonical product smooth structure).

[F2]

A map between smooth manifolds is smooth exactly when it is continuous and, in smooth charts, one coordinate representative is smooth near each point (Cr and smooth maps between smooth manifolds).

[A1]

For open Euclidean sets A⊆Ra, B⊆Rb, and U⊆Rm, a map H:U→A×B is smooth if and only if its two components are smooth, because all coordinate partial derivatives of H are exactly the coordinate partial derivatives of the component maps.

Proof

technique · direct
1.1givenF1F2L1choose

Assume F is smooth, and fix p∈P with charts as below. By [F2] it is continuous, so [L1] makes both components f and g continuous. Fix p∈P and choose a smooth chart (U,α) of P at p, smooth charts (V,φ) of M at f(p) and (W,ψ) of N at g(p), and the product chart (V×W,φ×ψ) of M×N at F(p) from [F1].

2.1F1F2L1choose

Conversely assume that f and g are smooth, and fix p∈P with charts as below. Then [F2] makes them continuous, so [L1] makes F continuous. Fix p∈P and choose smooth charts (U,α) of P, (V,φ) of M, and (W,ψ) of N exactly as in step 1.1.

3.1F1F2A1step 1.1step 2.1

In the charts chosen in steps 1.1 and 2.1, the representative of F is the pair ((φ×ψ)∘F∘α−1)(x)=(φ∘f∘α−1(x),ψ∘g∘α−1(x)). Under the hypothesis of step 1.1, the left-hand side is smooth, so [A1] makes the two component representatives smooth; since p was arbitrary, f and g are smooth by [F2]. Under the hypothesis of step 2.1, the two component representatives are smooth by [F2], so [A1] makes the left-hand side smooth, and [F2] makes F smooth at p, hence everywhere.

4.1step 3.1∎

Step 3.1 proves both directions of the equivalence.

Depends on

Used by

Dependency tree · two levels

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Sources