Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Pullback is functorial up to canonical bundle isomorphism

Statement

For a smooth vector bundle π:E→M, there are canonical bundle isomorphisms

id⁡M∗E≅E,(f∘g)∗E≅g∗(f∗E).

Facts & Assumptions

Given: A smooth vector bundle E→M and smooth maps g:P→N and f:N→M.

[L1]

The pullback bundle is the fibre-product set with its smooth bundle structure (Pullback vector bundles as fibre products, The pullback fibre product is a smooth vector bundle).

Proof

technique · direct
1.1L1givenconstruct

For the identity map, define I:id⁡M∗E→E by I(p,e)=e. This is well defined because p=π(e) in the identity pullback, and its inverse is e↦(π(e),e).

2.1L1step 1.1algebra∎

An element of g∗(f∗E) is a pair (x,(g(x),e)) with f(g(x))=π(e). Send it to (x,e)∈(f∘g)∗E. The inverse is (x,e)↦(x,(g(x),e)). In the pulled-back bundle charts of [L1], both maps are the identity on the fibre coordinate, so they are smooth vector bundle isomorphisms.

Depends on

Used by

Dependency tree · two levels

12 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