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

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.

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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

1.1F1L1given

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].

1.2F1L1given

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

Cotangent pullback is contravariantly functorial

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources