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.

Depends on

Used by

Dependency tree · two levels

22 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

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