PropositionStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-29

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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

1.1

Claim 1: for a smooth chart $(U, φ)$ , the representative of [given, F1, F2] ${i d}_{M}$ with respect to $(U, φ)$ and $(U, φ)$ is $φ \circ {i d}_{M} \circ φ^{- 1} = {i d}_{φ (U)}$ , which is smooth, each coordinate partial being a constant function; [F2] then declares ${i d}_{M}$ smooth at every point, and continuity holds by Smooth maps are continuous.

givenF1F2

1.2

Claim 2, continuity: both maps are continuous (smooth maps are continuous), [given, choose] so $G \circ F$ is continuous; for $p \in M$ choose a chart $(W, χ)$ of $P$ at $G (F (p))$ and then charts $(V, ψ)$ of $N$ at $F (p)$ with $G (V) \subseteq W$ and $(U, φ)$ of $M$ at $p$ with $F (U) \subseteq V$ , which is possible because $F$ and $G$ are continuous and charts exist.

givenchoose

1.3

The representative of the composite with respect to $(U, φ)$ and [given, F2, L1] $(W, χ)$ is $χ \circ (G \circ F) \circ φ^{- 1} = (χ \circ G \circ ψ^{- 1}) \circ (ψ \circ F \circ φ^{- 1})$ on $φ (U \cap F^{- 1} (V) \cap (G \circ F)^{- 1} (W))$ . The two factors are smooth by the smoothness of $F$ and $G$ through [F2], so [L1] makes their composite smooth. Hence $G \circ F$ is smooth at $p$ by [F2].

givenF2L1

2.1

Applying step 1.3 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.

step 1.1step 1.3∎

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)