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I don't know what to do to prove the following statement:

Let $U \subset \mathbb R^n$ be an open set and let $\alpha$ be a $k$-form on $U$ and $\beta$ be an $l$-form on $U$. Suppose both $\alpha, \beta$ are closed forms and that $\beta$ is exact. Then $\alpha \wedge \beta$ is exact.

Every assistance would be very much appreciated.

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  • $\begingroup$ Do you mean that $\alpha\wedge\beta$ is exact? $\endgroup$ – Olivier Bégassat Nov 9 '14 at 1:45
  • $\begingroup$ That's it, just fixed :) thank you $\endgroup$ – Vinicius Rodrigues Nov 9 '14 at 1:46
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    $\begingroup$ Hint: if $\beta=db$, then look at $\pm\alpha\wedge b$... $\endgroup$ – Olivier Bégassat Nov 9 '14 at 1:46
  • $\begingroup$ Oh, now I see exactly what I must do. If you wish, post this message as an answer so I can mark it as the answer :) $\endgroup$ – Vinicius Rodrigues Nov 9 '14 at 1:49
  • $\begingroup$ ok : ) ${}{}{}$ $\endgroup$ – Olivier Bégassat Nov 9 '14 at 1:55
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If $b$ satisfies $db=\beta$, and $\alpha$ is a closed $k$-form, then $$d\omega=\alpha\wedge\beta$$ where $\omega=(-1)^k\alpha\wedge b$. The proof works for forms on any manifold.

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