PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30

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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.
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Cotangent pullback is contravariantly functorial

Statement

For smooth maps $F : M \to N$ and $G : N \to P$ , one has $(G \circ F)_{p}^{*} = F_{p}^{*} \circ G_{F (p)}^{*}$ for every $p \in M$ , and $({id}_{M})_{p}^{*} = {id}_{T_{p}^{*} M}$ .

Facts & Assumptions

Given: Smooth maps $F : M \to N$ and $G : N \to P$ and a point $p \in M$ .

[F1]

Pullback is defined by composition with the differential (Pullback of a cotangent vector).

[L1]

Differentials satisfy the chain rule (The chain rule for differentials of smooth maps).

Proof

technique · direct

1.1

For $ω \in T_{G (F (p))}^{*} P$ , one has $(G \circ F)_{p}^{*} (ω) = ω \circ d (G \circ F)_{p} = ω \circ d G_{F (p)} \circ d F_{p} = F_{p}^{*} (G_{F (p)}^{*} (ω))$ by [F1] and [L1].

F1L1given

1.2

Applying the same formula to the identity map gives $({id}_{M})_{p}^{*} = {id}_{T_{p}^{*} M}$ .

F1L1given

2.1

Therefore cotangent pullback is contravariantly functorial.

step 1.1step 1.2∎

Depends on

Used by

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)