When is There a Solution to “Pullback Equation” of Differential Forms

All:

Let $f: M \to N$ be a smooth map between manifolds, and let $w$ be a $1$-form on $M$. Under what conditions is there a $1$-form $z$ defined on $N$ so that $w=f^*z$, i.e., so that $w$ is the pullback of the form $z$ by the map $f$?

All I can think of is considering the respective cohomologies of $M$, $N$, so that, e.g., if $H^1(N)=0$, then this would not be possible, or maybe we can use the (contravariant) map induced by $f$ in cohomology, but I cannot think of more general conditions.

Thanks for any suggestions.

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I assume you mean $w = f^*z$ (instead of $z = f^*w$). – Jesse Madnick Jul 26 '13 at 5:56
You always have pullbacks on differential forms. If $H^{1}(N) = 0$ that means there are no closed one forms which are not exact but there still are oodles of one-forms. – DBS Jul 26 '13 at 6:01
Thanks, Jesse ,just corrected it. – RickyBobby Jul 26 '13 at 6:02
DBS: true, but I don't think there is always an injection, i.e., I don't think the induced map on cohomology is necessarily a surjection (nor an injection, but that's a different issue) – RickyBobby Jul 26 '13 at 6:03
@DBS He's asking when one form is the pullback of another, not if forms can always be pulled back. – Potato Jul 26 '13 at 6:04