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The pullback of a trivial smooth vector bundle is canonically trivial
Statement
Let be a smooth map and let be the trivial smooth real rank- bundle over . Then the pullback is canonically isomorphic to the trivial bundle : the constant frame of pulls back to the nowhere-zero global frame of , 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 and the trivial smooth rank- real bundle .
The pullback set is , with projection and fibrewise vector-space operations inherited from (Pullback vector bundles as fibre products, Smooth vector bundles, rank, fibres, and trivial bundles).
The pullback carries the smooth rank- vector-bundle structure constructed in The pullback fibre product is a smooth vector bundle: for a bundle chart of a bundle , the map with is a bundle chart over ; the trivial bundle has the single global bundle chart over .
A smooth rank- 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 determines the bundle isomorphism , .
Maps into a product of smooth manifolds are smooth exactly when their components are, and smooth maps are continuous ( and smooth maps between smooth manifolds).
Proof
Define by , using the description [F1]. Then is well defined, and its inverse is . On each fibre it is the linear isomorphism onto inverse to , and lies over . So is a fibrewise-linear bijection over the base.
is smooth with smooth inverse. Indeed, in the global pullback chart of [F2] coming from the global chart of , the chart map is exactly , i.e. the identity identification of ; and the inverse has components , and , which are smooth by [F4] since is smooth. Hence is a diffeomorphism, and consequently a smooth bundle isomorphism over .
Therefore as smooth vector bundles over , by the isomorphism , which was defined by an explicit formula using only and hence is canonical. Equivalently, the constant sections of pull back to the sections of , none of whose values is the zero vector; these pullbacks are smooth because in the global pullback chart of step 2.1 the section reads as the constant map , and they form a global frame of that converts into the standard frame of . 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 gives the zero bundle over , and the empty or disconnected base is covered verbatim.
Depends on
- Smooth vector bundles, rank, fibres, and trivial bundles
- Pullback vector bundles as fibre products
- The pullback fibre product is a smooth vector bundle
- Local and global frames of a vector bundle
- A vector bundle is trivial if and only if it has a global frame
- $C^r$ and smooth maps between smooth manifolds
Used by
- The pullback of the Euclidean tangent bundle is canonically trivial Corollary
- An embedding into Euclidean space gives a rank-(n-m) stable normal inverse Lemma
- An immersion into Rⁿ gives a rank-(n-m) representative of the stable normal bundle Lemma
- Smale-Hirsch makes rank reduction sufficient for Euclidean immersion in positive codimension Proposition
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
- Ralph L. Cohen, Bundles, Manifolds, and Homotopy (author draft, complete 568-page text) (standard reference, not scraped)
- Ralph L. Cohen, Immersions of Manifolds and Homotopy Theory (lecture notes, 30 June 2022; complete 46-page text) (standard reference, not scraped)