CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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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.

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Every short exact sequence of smooth vector bundles splits

Statement

Every short exact sequence of smooth vector bundles over one base,

$0 \to E \overset{i}{\to} G \overset{q}{\to} F \to 0,$

admits a smooth splitting $s : F \to G$ with $q \circ s = {id}_{F}$ .

Facts & Assumptions

Given: A short exact sequence of smooth vector bundles over one base $M$ .

[L1]

Constant-rank kernels and images of bundle maps over one base are smooth subbundles (Constant-rank kernels and images of bundle maps over one base are subbundles).

[L2]

Every vector subbundle has a smooth complement (Every vector subbundle has a smooth complement).

[L3]

A fibrewise bijective smooth bundle map over the identity is a bundle isomorphism (A fibrewise bijective smooth bundle map over a diffeomorphism is a bundle isomorphism).

Proof

technique · direct

1.1

Exactness gives $i (E) = \ker q \subseteq G$ . By [L1], this image is a smooth subbundle of $G$ . Choose a smooth complement $H \subseteq G$ to $i (E)$ by [L2], so $G_{p} = i (E_{p}) \oplus H_{p}$ for every $p \in M$ .

L1L2givenchoose

2.1

Because $\ker (q ∣_{H_{p}}) = H_{p} \cap i (E_{p}) = 0$ and $q_{p}$ is surjective, $q ∣_{H} : H \to F$ is fibrewise bijective. By [L3] it is a smooth bundle isomorphism, so its inverse $s : F \to H \subseteq G$ is smooth and satisfies $q \circ s = {id}_{F}$ . Thus the sequence splits.

L3step 1.1algebra∎

Depends on

Used by

A short exact sequence of vector bundles has a canonical splitting False statement

Dependency tree · two levels

15 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)