When can a class in $H^1(M;\mathbb{Z})$ be represented by a fiber bundle over $S^1$ 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 
 A: 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. 
A: Here is another way to approach this problem. Say $M$ is closed and orientable. The correspondance $H^1(M,\mathbb{Z})$ with $[M,S^1]$ works by pulling back the generator of $H^1(S^1,\mathbb{Z})$ along $f$ to get $\alpha$. 
Using Poincare duality and universal coefficient theorem (and forgetting about torsion) we can view $H^1(M,\mathbb{Z})$ as a lattice inside $H^1(M,\mathbb{R})$. By de Rham's theorem, $\alpha$ is represented by a closed $1$-form so that if we pick $f$ to be a smooth representative in the homotopy class, then we have $\alpha = f^*dx$ where $dx$ is the $1$-form on $S^1$ generating $H^1_{dR}(S^1)$. Then $f$ is a submersion if and only if $f^*dx$ is a nonvanishing $1$-form on $M$. So $\alpha$ corresponds to a fiber bundle if and only if, after tensoring with $\mathbb{R}$, it can be represented by a nonvanishing closed $1$-form.
In fact one can prove something slightly stronger as is done in this paper. A closed orientable manifold $M$ is fibered over $S^1$ if and only if it has a nonvanishing closed $1$-form and in this case the fibers are the leaves of the foliation generated by this $1$-form. Furthermore, if we have a nonvanishing closed $1$-form $\omega$ that is a nonero real multiple of $f^*dx$, then the fiber bundle map induced by $\omega$ is isotopic to $f$. 
A: If $M$ is a compact and irreducible 3-manifold, one answer is provided by a theorem of Stallings, in his 1962 paper "On fibering certain 3-manifolds": $\alpha$ is represented by a fibration $f : M \to S^1$ if and only if the kernel of the associated homomorphism $\pi_1(M) \to \mathbb{Z}$ is finitely generated and that homomorphism is nontrivial.
