If we have a set of 1-forms $w_1, ... w_n$ on a smooth manifold $X$ I can show that $w_i$ are linearly dependent if and only if $w_1 \wedge ... \wedge w_n = 0$.
I wondered if this is also true in the cohomology $H(X)$ using the product induced from the wedge product. Namely is it true that cohomology classes $[w_i]$ are linearly dependent if and only if:
$$[w_1][w_2]...[w_n] = [0]$$
I tried to prove this using the same method as for differential 1-forms. If the classes are linearly dependent then we can write $[w_1] = \sum c_i [w_i]$ and then terms like $[w_2][w_2][w_3]...[w_n]$ are the zero class because $w'_2 \wedge w''_2$ will be exact ($w'$ some representatives for $[w_2]$) and then the whole wedge product will be exact because the cohomology classes are closed.
I'm struggling to show the other direction, namely if $[w_1][w_2]...[w_n] = [0]$ are the classes linearly dependent in $H(X)$? I'm not even sure if it is true.
Thanks for any help