PropositionStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

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.

Identity maps and composites of smooth maps are smooth

Statement

Let $M$ , $N$ , $P$ be smooth manifolds.

The identity map ${i d}_{M} : M \to M$ is smooth.
If $F : M \to N$ and $G : N \to P$ are smooth, then $G \circ F : M \to P$ is smooth.

Facts & Assumptions

Given: Smooth manifolds $M, N, P$ and smooth (respectively $C^{r}$ ) maps $F : M \to N$ , $G : N \to P$ .

[F1]

Smooth charts are members of the maximal atlas of the smooth structure (Smooth manifolds and their smooth charts), and a chart is a homeomorphism onto an open Euclidean set (Manifold charts, coordinate domains, and coordinate functions).

[F2]

A map is $C^{r}$ at $p$ when its representative with respect to one — hence, by chart independence, every — suitable chart pair is $C^{r}$ ( $C^{r}$ and smooth maps between smooth manifolds, Chart independence of $C^{r}$ smoothness).

[L1]

If $u : W \to W^{'}$ is smooth and $g : W^{'} \to W^{''}$ is $C^{r}$ , then $g \circ u$ is $C^{r}$ ; and if $h : W \to W^{'}$ is $C^{r}$ and $v : W^{'} \to W^{''}$ is smooth, then $v \circ h$ is $C^{r}$ (Compatibility of smooth atlases is an equivalence relation, and smooth Euclidean maps compose).

Proof

technique · direct

1.1givenF1F2

Claim 1: ${i d}_{M}$ is continuous. For a smooth chart $(U, φ)$ , its representative with respect to $(U, φ)$ on both sides is $φ \circ {i d}_{M} \circ φ^{- 1} = {i d}_{φ (U)}$ , which is smooth: its first coordinate partials are constant, and all higher partials vanish. Thus [F2] makes ${i d}_{M}$ smooth at every point.

1.2givenF1

Claim 2, continuity: $F$ and $G$ are continuous by Smooth maps are continuous, so $G \circ F$ is continuous. For $p \in M$ choose a chart $(W, χ)$ of $P$ at $G (F (p))$ , then a chart $(V, ψ)$ of $N$ at $F (p)$ with $G (V) \subseteq W$ , and finally a chart $(U, φ)$ of $M$ at $p$ with $F (U) \subseteq V$ ; continuity permits these restrictions.

2.1F2L1step 1.2

On $φ (U)$ , the representative of the composite with respect to $(U, φ)$ and $(W, χ)$ is $χ \circ (G \circ F) \circ φ^{- 1} = (χ \circ G \circ ψ^{- 1}) \circ (ψ \circ F \circ φ^{- 1})$ . Both factors are smooth near the relevant points by [F2], so their composite is smooth by [L1]. Since step 1.2 already established continuity, [F2] makes $G \circ F$ smooth at $p$ .

3.1step 1.1step 2.1∎

Applying step 2.1 at every point shows $G \circ F$ is smooth on all of $M$ , and step 1.1 gives smoothness of the identity. Hence both claims are proved.

Identity maps and composites of smooth maps are smooth

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources