Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In this question I am using Wiki's definitions for fibration and fiber bundle. I want to be general in asking my question, but I am mostly interested in smooth compact manifolds and smooth fibrations and bundle projection between them. Under some mild topological assumption on the base space (of course verified in the case of manifolds) a fiber bundle always gives rise to a fibration; so in this context I consider fiber bundles as particular examples of fibrations.

My question: in which cases a general fibration turns out to be a fiber bundle?


EDIT: in this question on MO, they suggest that it is probably true that the projection of a smooth fibration is a submersion. Does anyone have a reference/counterexample for this?

If this happens to be true, then we can apply Ehresmann and obtain a fiber bundle (I am restricting to the case of compact smooth manifolds here). This would solve the problem at least in the case I was interested in.

share|improve this question
    
An obvious necessary condition is that the fibers over every point in the base must be homeomorphic, and not just homotopy equivalent. –  Daniel Rust Jun 9 '13 at 11:26
    
Sure this has to hold: I am actually trying to understand when a fibration is a bundle, so that I can unambiguously speak of a "submanifold fiber". –  Lor Jun 9 '13 at 11:44
    
@DanielRust : I have found something on MO which may be related and edited the question accordingly, please have a look. –  Lor Jun 10 '13 at 12:27
1  
@Lor: According to wikipedia en.wikipedia.org/wiki/Submersion_%28mathematics%29, a smooth fibration is a submersion. If "smooth fibration" just means $\pi$ is smooth, I think I have a counterexample. If "smooth fibration" means "$\pi$ is a smooth and lifts are smooth", then I agree with wikipedia. My counterexample is the map $\pi:\mathbb{R}\rightarrow\mathbb{R}$ given by $\pi(x) = x^3$. This is a smooth homeomorphism (whose inverse is not differentiable at $0$). This map is not a submersion because $0$ isn't a regular value. –  Jason DeVito Jun 11 '13 at 2:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.