For a topological space M, It is known from homotopy theory that the elements of the first cohomology $H^1(M;\mathbb{Z})$ are in 1-1 correspondence with homotopy classes of maps $[M,S^1]$


In my case of interest M is a smooth manifold. Take $\alpha$ and take smooth $f\colon M \to S^1$ representing $\alpha$ under the above identification.


My question is, under what (sufficient / necessary / equiv) conditions $f$ gives a locally trivial fibration (Fiber bundle) of M over $S^1$?


I would prefer conditions in terms of homology / cohomology of M.


What comes to my mind is that, according to Ehrshmann theorem, since $S^1$ is compact,this map is a fibration iff surjective + submersion.
Surjectivity can be assured by picking nontrivial cohomology class.
As for submersion. I have no idea how to characterize.


Thanks 



When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$
Let me assume throughout this answer that $M$ is closed, oriented, and connected. Here are some necessary conditions. 


If you ask for a smooth fiber bundle, then a necessary condition is that the tangent bundle of $M$ has a trivial quotient of rank $1$, or equivalently a trivial subbundle of rank $1$. This is possible iff the Euler class $e(M)$ vanishes. This gives



> 
> **Condition #1:** $\chi(M) = 0$ (automatic when $\dim M$ is odd). 
> 
> 
> 


Next, if $F$ denotes the fiber of $f$, then the long exact sequence in homotopy for the fibration $F \to M \to S^1$ takes the form


$$1 \to \pi\_1(F) \to \pi\_1(M) \to \mathbb{Z} \to \pi\_0(F) \to 1$$


(and for $n \ge 2$ the maps $\pi\_n(F) \to \pi\_n(M)$ are isomorphisms). This gives $\pi\_1(F) = \text{ker} \left( \pi\_1(M) \to \mathbb{Z} \right)$. Since $M$ is compact, so is $F$, so $\pi\_0(F)$ is finite. It follows that $\text{ker}(\mathbb{Z} \to \pi\_0(F))$ is nonzero, so we get



> 
> **Condition #2:** $\pi\_1(M) \to \mathbb{Z}$ is nonzero.
> 
> 
> 


This is equivalent to the condition that the corresponding cohomology class in $H^1(M)$ is nonzero. If we furthermore assume that $F$ is connected, then $\pi\_1(M) \to \mathbb{Z}$ must be surjective, which is equivalent to the condition that the corresponding cohomology class is indivisible.


Next, if $M$ is a closed smooth manifold, then so is $F$. This gives



> 
> **Condition #3:** $\text{ker} \left( \pi\_1(M) \to \mathbb{Z} \right)$ is finitely presented.
> 
> 
> 


Of course this kernel must in fact be the fundamental group of a closed manifold of dimension $\dim M$, so if $\dim M \le 4$ (so that $\dim F \le 3$) then that puts some extra restrictions on it. Beyond this I don't know if there's anything easy to say. 


