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I will be infinitely grateful to the one who could give a thorough introduction with examples on fiber bundles or a link to a document that deals with it. I was desperately looking on the web for something interesting but I either find 200 pages notes or a poor paragraphe.

Again thank you for being always kind and helpful :)

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I'd start with trying to understand the tangent bundle and the Hopf fibration thoroughly. These are two good examples that show many of the inherent intricacies. Then I think good place to look at is the book by Bott and Tu on differential forms in algebraic topology. One of the standard references on fibre bundles is Husemöller's book. – t.b. May 18 '11 at 12:11
The section on fibre bundles in Hatcher's Algebraic Topology is a very good introduction (p375-384), and his Vector Bundles book is fab too (the first chapter, which goes up to classification, is about 35 pages long, but it's a very easy read). You can find both books available free online on Hatcher's website. For more depth, Husemöller (mentioned above by Theo) is excellent. – Alex May 18 '11 at 13:04
Many thanks Alex and Theo – El Moro May 18 '11 at 15:02
Steenrod's Topology of Fiber Bundles is a classic that will get your hands dirty. – Manuel Nov 23 '11 at 2:14

Are you looking for an introduction to fiber bundles themselves or to methods of computing the cohomology of the various pieces?

The comments already have some very good recommendations on sources for learning about what a fiber bundle is. As far as their topology goes, I suppose I'll mention two facts:

  • The first is about the Euler characteristic and actually follows in some ways from the next point. In fact, one only needs a fibration $p : E \rightarrow B$ here, although some technical conditions are needed: $B$ should be path-connected and the fibration should be orientable over a field. Let $F$ be the fiber of the fibration. Then $\chi(E) = \chi(B)\chi(F)$.

Here's an example of this in practice. Identify $S^3$ with $SU(2)$ and implement the Hopf fibration $S^1 \rightarrow S^3 \rightarrow S^2$. The subgroup $U(1)$ is realized as $S^1$ and the quotient $SU(2)/U(1)$ is realized as $S^2$. This gives rise to a fibration $SU(2) \rightarrow SU(2)/U(1)$ with fiber $U(1)$; hence $\chi(SU(2)) = \chi(S^2)\chi(S^1) = 0$.

(Note that this can be seen also by observing that $SU(2)$ is a compact Lie group of positive dimension; hence it admits a nowhere vanishing vector field and must have Euler characteristic zero.)

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