When is a fibration a fiber bundle?

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.

• An obvious necessary condition is that the fibers over every point in the base must be homeomorphic, and not just homotopy equivalent. – Dan 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
• @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
• @JasonDeVito could you please give an sketch about how to prove that a smooth fibration is a submersion, please? – D1811994 May 21 '17 at 14:48

You are right. If the fibration is smooth and the spaces involved are compact, then it is a fiber bundle. Just to make clear what I mean by smooth fibration:

Definition. A smooth map $p\colon E \to B$ is said to satisfy the homotopy lifting property in the smooth category if given the following commutative diagram where all maps are smooth: there exists an smooth map $\widetilde{F}$ making the following diagram smooth: Definition. A smooth map is said to be a smooth (Hurewicz) fibration if it satisfies the homotopy lifting property in the smooth category for all manifolds $Y$.

Definition. A smooth map is said to be a smooth Serre fibration if it satisfies the homotopy lifting property in the smooth category for all discs $I^n$, $n\ge 0$.

Now the idea of the proof:

1) A (Serre) fibration $p\colon E \to B$ where $B$ is path-connected and $E\neq \emptyset$ is surjective.

2) A smooth (Serre) fibration is a submersion. I asked the question and it was answered here.

3) Compactness combined with the above guarantees we can apply Ereshmann, as you said, and we are done.

Remark. This works for weak or Serre fibrations. You don't need to assume you are working with Hurewicz fibrations. So the answer is a bit more general than the question you asked.

I hope this helps! Regarding the topological case (not assuming smoothness) I am really interested in your question. So, hopefully, someone will shed some light on this point.

• Is any locally trivial smooth fibre bundle a smooth Serre/ Hurewicz fibration(as defined in your answer)? Can you suggest any literature reference regarding it ? Or else a proof is also very much appreciated . I asked this question here math.stackexchange.com/questions/3363514/… – Adittya Chaudhuri. Sep 21 at 11:52