By Dale Husemoller (auth.)
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Additional resources for Fibre Bundles
For ub(x 1 ,x 2 )EimubnW2 , we have wb(x 1 ,x 2 ,y)=0, and since wb is injective, we have x 1 = x 2 = y = 0. Then the quotient map (b, y) --+ (b, y mod(im u)) is a monomorphism U x W2 --+ E(coker ui U), and, for reasons of dimension, it is a U -isomorphism. Its inverse U -morphism is the factorization of the projection U x (W1 EB W2 ) -4 U x W2 through coker u. This proves the theorem. 3 Corollary. Let u: ~n--+ IJm be a B-morphism that is injective, or, equivalently, it is a monomorphism on each fibre of ~- Then im u and coker u are vector bundles.
Using this isomorphism and the functorial properties of EB, we have the next proposition. 3 Proposition. If (E 1 , p 1 , B) is a trivial bundle with fibre F 1 and if (E 2 , p2 , B) is a trivial bundle with fibre F2 , then (E 1 , p1 , B) EB (E 2 , p2 , B) is a trivial bundle with fibre F 1 x F2 • In the next proposition we compute the cross sections of a fibre product. 4 Proposition. The cross sections sofa fibre product (£ 1 EB £ 2 , q, B) are of the form s(b) = (s 1 (b), s2 (b)), where s 1 is a cross section of (E 1 , p, B) and s2 is a cross section of (E 2 , p 2 , B) uniquely defined by s.
8 Corollary. Let f: (Bt, Ad--+ (B, A) be a map of pairs, let g =fiAt: At --+ A, and let e be a bundle over B. Then g*(eiA) and f*(e)IAt are At-isomorphic. Proof Letj: A--+ B andjt: At--+ B 1 be the respective inclusion maps. 7), there is the following sequence of A 1 -isomorphisms: The next result is useful in discussing fibre bundles. 9 Proposition. Let e = (E, p, B) be a bundle, let f: B 1 --+ B be a map, and let f*(e) = (Et,Pt,Bd be the induced bundle of e under f If pis an open map, Pt is an open map.
Fibre Bundles by Dale Husemoller (auth.)