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Let $M$ be an $m$-dimensional oriented manifold with boundary $\partial M$, that is connected and oriented. I now have to show that if $M$ is compact then if $\alpha$ is a volume form on $\partial M$, there does not exist a $\beta \in \Omega^{m-1}(M)$ such that $d\beta = 0$ and $\iota^*\beta = \alpha$ where $\iota$ is the inclusion map.

I think this proof is correct, but I would like to get some confirmation:

Assume that there is a $\beta \in \Omega^{m-1}(M)$ that has these properties. Because $M$ is compact, we now that $\beta$ has compact support and hence, we may apply Stokes's Theorem, which says that \begin{equation*} 0 = \int_M d\beta = \int_{\partial M} \iota^*\beta = \int_{\partial M } \alpha \end{equation*} But $\alpha$ is a volume form on so $\int_{\partial M } \alpha \neq 0$, which concludes the proof.

I also have to check whether this holds when $M$ is not compact but I don't really have an idea on how to start with this statement.

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Your proof is fine.

For the noncompact case, what's one of the simplest non-compact manifolds $M$ with boundary you can think of? (You can give easy examples with $\partial M$ non-compact or compact.)

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  • $\begingroup$ Can we take the closed half plane, with $\alpha= \beta = dx$? $\endgroup$
    – Mee98
    Dec 28, 2019 at 17:33
  • $\begingroup$ Good. And, analogously, on the cylinder $x^2+y^2=1$, $z\ge 0$? $\endgroup$
    – Ted Shifrin
    Dec 28, 2019 at 19:07
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    $\begingroup$ You can reduce to the connected case by working with each component. But you're exactly right that if the oriented manifold is disconnected, a top-degree form could be nowhere-zero and yet have zero integral. $\endgroup$
    – Ted Shifrin
    Jan 2, 2020 at 22:06

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