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I'm trying to prove that the wedge of a closed form $\xi$ with an exact form $\omega$ is exact. We already have that half of it is exact. Maybe we can use the equation of $\xi$ being closed to rewrite the wedge. Other than that I am sort of stuck.

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    $\begingroup$ If $d \alpha = 0$, then $\alpha \wedge d\beta = \pm d\left(\alpha \wedge \beta\right)$, which is clearly exact. $\endgroup$ – darij grinberg Sep 23 '18 at 23:33

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