Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedprecheck pass
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.

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

The pullback of a trivial smooth vector bundle is canonically trivial

Statement

Let f:N→M be a smooth map and let εMr=M×Rr be the trivial smooth real rank-r bundle over M. Then the pullback f∗εMr is canonically isomorphic to the trivial bundle εNr=N×Rr: the constant frame e1,…,er of εMr pulls back to the nowhere-zero global frame x↦(x,ej) of f∗εMr, and the isomorphism is the one determined by that frame (Local and global frames of a vector bundle, A vector bundle is trivial if and only if it has a global frame). This product-bundle assertion is choice-free.

Facts & Assumptions

Given: A smooth map f:N→M and the trivial smooth rank-r real bundle εMr=M×Rr→M.

[F1]

The pullback set is f∗εMr={(q,e)∈N×(M×Rr):f(q)=pr⁡1(e)}={(q,(f(q),v)):q∈N, v∈Rr}, with projection (q,(f(q),v))↦q and fibrewise vector-space operations inherited from εMr (Pullback vector bundles as fibre products, Smooth vector bundles, rank, fibres, and trivial bundles).

[F2]

The pullback carries the smooth rank-r vector-bundle structure constructed in The pullback fibre product is a smooth vector bundle: for a bundle chart Φα:E∣Uα→Uα×Rr of a bundle E, the map (q,e)↦(q,v) with Φα(e)=(f(q),v) is a bundle chart over f−1(Uα); the trivial bundle εMr has the single global bundle chart Φ(p,v)=(p,v) over M.

[F3]

A smooth rank-r vector bundle is trivial if and only if it has a global frame (Local and global frames of a vector bundle, A vector bundle is trivial if and only if it has a global frame); a global frame (s1,…,sr) determines the bundle isomorphism N×Rr→E, (q,(λ1,…,λr))↦∑jλjsj(q).

[F4]

Maps into a product of smooth manifolds are smooth exactly when their components are, and smooth maps are continuous (Cr and smooth maps between smooth manifolds).

Proof

1.1F1F4

Define Φ:f∗εMr→N×Rr by Φ(q,(f(q),v))=(q,v), using the description [F1]. Then Φ is well defined, and its inverse is (q,v)↦(q,(f(q),v)). On each fibre it is the linear isomorphism onto {q}×Rr inverse to v↦(f(q),v), and Φ lies over id⁡N. So Φ is a fibrewise-linear bijection over the base.

2.1F2F4step 1.1

Φ is smooth with smooth inverse. Indeed, in the global pullback chart of [F2] coming from the global chart Φ(p,v)=(p,v) of εMr, the chart map is exactly Φ, i.e. the identity identification of N×Rr; and the inverse (q,v)↦(q,(f(q),v)) has components q↦q, q↦f(q) and v↦v, which are smooth by [F4] since f is smooth. Hence Φ is a diffeomorphism, and consequently a smooth bundle isomorphism over N.

3.1F2F3step 1.1step 2.1∎

Therefore f∗εMr≅εNr as smooth vector bundles over N, by the isomorphism Φ, which was defined by an explicit formula using only f and hence is canonical. Equivalently, the constant sections sj(p)=(p,ej) of εMr pull back to the sections f∗sj(q)=(q,(f(q),ej)) of f∗εMr, none of whose values is the zero vector; these pullbacks are smooth because in the global pullback chart of step 2.1 the section f∗sj reads as the constant map q↦ej, and they form a global frame of f∗εMr that Φ converts into the standard frame of N×Rr. No selection of local trivializations, complements or representatives has been made: the chart of [F2] used above is the single global chart of the product bundle, so the argument uses no choice principle. The case r=0 gives the zero bundle over N, and the empty or disconnected base is covered verbatim.

Depends on

Used by

Dependency tree · two levels

21 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