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I am trying to understand more geometrically the relation between triviality

of bundles and existence of global sections. This is what I have for now. Please

comment/critique:

Let p:E-->B be a fiber bundle :

consider E embedded in B as the 0 section. Then , if the bundle is trivial, E

is a (global) product space; E=BxF )so that p(b,f)=b).

Then every continuous map from B to E is a global section:

I am trying to understand why we can define a global section on S^1xI as

a bundle over S^1 , but not in M, the Mobius band , as a bundle over S^1 .

It seems that the "torsion" of M (I think it is measured in Chern Classes, or

Characteristic classes. ) prevents this from happening. Anyone know, at

least intuitively how the twisting prevents a global section?.

My friend told me that we take the Mobius band as an identification/quotient space

( a square, with sides given orientations , and sides identified), say

S=IxI , and we consider the strip {1/2xI}, as the base, embedded in S .

Then, if we have a continuous map f from {1/2xI} into I , which is nowhere-zero, the

twisting (when identifying the sides of S with opposite orientation) will force f to

take on both negative and positive values . This seems intuitively correct, but too

fuzzy. How do we rigorize this?.

Then, by continuity , f must be zero at some point.

Thanks.

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# Bundles and global sections, triviality.

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