2023-09-17

368: Restriction of C^\infty Vectors Bundle on Regular Submanifold Base Space Is C^\infty Vectors Bundle

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A description/proof of that restriction of C vectors bundle on regular submanifold base space is C vectors bundle

Topics


About: C manifold

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that for any C vectors bundle, the restriction of the vectors bundle on any regular submanifold base space is a C vectors bundle.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.


Main Body


1: Description


For any C vectors bundle, π:M1M2, and any regular submanifold, M3M2, the restriction of π on M3, πM3:π1(M3)M3, is a C vectors bundle.


2: Proof


π is a submersion, π is a transversal map to M3, and π1(M3) is a regular submanifold of M1 as a preimage of a regular submanifold under a transversal map, by the proposition that for any transversal map from any C manifold to any regular submanifold, the preimage of the regular submanifold under the transversal map is a regular submanifold of the domain.

πM3 is C as a restriction of a C map on a regular submanifold domain and a regular submanifold codomain, by the proposition that for any C map between C manifolds, the restriction of the map on any regular submanifold domain and any regular submanifold codomain is C. πM3 is obviously surjective.

For any pM3, πM31(p) is a real vectors space, because πM31(p)=π1(p).

For any pM3, there is an open neighborhood, UpM2, such that there is a fiber-preserving diffeomorphism, f:π1(Up)Up×Rr. There is the fiber-preserving bijection, f=f|π1(UpM3):π1(UpM3)(UpM3)×Rr, because f is fiber-preserving.

Is f diffeomorphic? π1(Up) is an open subset of M1, because π is continuous, and is a C manifold, by the proposition that for any C manifold and its any regular submanifold, any open subset of the super manifold is canonically a C manifold, and the intersection of the open subset and the regular submanifold is a regular submanifold of the open subset manifold. Up×Rr is an open subset of M2×Rr, and is a C manifold likewise.

π1(UpM3)=π1(Up)π1(M3), by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets. So, π1(UpM3) is a regular submanifold of π1(Up), by the proposition that for any C manifold and its any regular submanifold, any open subset of the super manifold is canonically a C manifold, and the intersection of the open subset and the regular submanifold is a regular submanifold of the open subset manifold. (UpM3)×Rr=(Up×Rr)(M3×Rr), by the proposition that the intersection of the same-indices-set products of possibly uncountable number of sets is the product of the intersections of the subsets, and M3×Rr is a regular submanifold of M2×Rr, by the proposition that for the product of any 2 C manifolds, the product for which one of the constituents is replaced with any regular submanifold is a regular submanifold of the original product, and (UpM3)×Rr is a regular submanifold of Up×Rr likewise.

So, f=f|π1(UpM3):π1(UpM3)(UpM3)×Rr is a restriction of the C f on a regular submanifold domain and a regular submanifold codomain, and is C, by the proposition that for any C map between C manifolds, the restriction of the map on any regular submanifold domain and any regular submanifold codomain is C. As f is a bijection, there is the inverse, f1, which is a restriction of the C f1 on a regular submanifold domain and a regular submanifold codomain, and is C, likewise.

For any point, pUpM3, f|π1(p):π1(p){p}×Rr is a 'vectors spaces - linear morphisms' isomorphism, because it is the same with f|π1(p).


References


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