TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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The differential sends curve velocities to composite curve velocities

Statement

If $γ$ is a smooth curve with $γ (0) = p$ and $F : M \to N$ is smooth, then $d F_{p} (\dot{γ} (0)) = \frac{d}{d t} (F \circ γ) (0)$ as derivations at $F (p)$ .

Facts & Assumptions

Given: A smooth curve $γ$ through $p$ and a smooth map $F : M \to N$ .

[F1]

The differential acts by pullback of target germs (The differential of a smooth map).

[F2]

The velocity derivation of a curve sends $[g]$ to $(g \circ γ)^{'} (0)$ (The velocity derivation of a smooth curve).

Proof

technique · direct

1.1

Let $[g]$ be a smooth germ at $F (p)$ . By [F1], $d F_{p} (\dot{γ} (0)) ([g]) = \dot{γ} (0) ([g \circ F])$ .

F1F2given

2.1

By [F2], the right-hand side is $((g \circ F) \circ γ)^{'} (0) = (g \circ (F \circ γ))^{'} (0)$ , which is exactly the value of the velocity derivation of $F \circ γ$ on $[g]$ .

F2step 1.1

3.1

Since the two derivations agree on every germ $[g]$ , they are equal.

step 2.1∎

Depends on

Used by

A smooth function with zero differential is constant on each connected component Proposition

Dependency tree · two levels

4 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

John M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)