Smooth representative $f: S^{2n - 1} \to S^n$, do we have $f^*\omega = d\alpha$?

Question

Let $[f] \in \pi_{2n - 1}(S^n)$. Choose a smooth representative $f: S^{2n - 1} \to S^n$. Let $\omega$ be a smooth $n$-form on $S^n$ with$$\int_{S^n} \omega = 1.$$Do we have that$$f^*\omega = d\alpha$$for some $(n -1 )$-form $\alpha$ on $S^{2n - 1}$?

Answer 1

Yes, you have $d(f^*\omega)=f^*d\omega=0$. So $f^*\omega$ is closed thus exact since $H^n(S^{2n-1},R)=0$.