Exact one form on non-simply-connected manifold

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Let $M$ be a smooth $n-$dimensional manifold and $\alpha\in\Lambda^1(M)$ a smooth $1-$form. Let $N\subset M$ be an embedded submanifold where $\alpha|_N=0$. If $d\alpha=0$ on the whole $M$, can I say that $\alpha$ is exact? $M$ is not simply-connected so this is not a trivial fact in principle since by Poincaré theorem I can just say it $\alpha$ is locally exact.

My idea is to work with the immersion map $I:N\rightarrow M$ so that the pull-back of $\alpha$ vanishes: $I^*\alpha =0$. Then can I conclude by linearity of the pull-back or something similar?