Even a yes or no would do. $\alpha^i,\beta^i\in\Omega^1M$ for $i=1,...,s$.
I was looking at a linear independent set ${\alpha^1(x),...,\alpha^s(x)}$ for all $x\in M$ and wedge product $\sum_i \alpha^i\land\beta^i=0$.
I'm thinking that since the wedge product is zero then $\beta$ would be the inverse mapping of $\alpha$. But linear independence necessitates (in one case) all non-zero (vectors). If $\beta^i=\sum_ja_j^i\alpha^j$ and $a^i_j:M\rightarrow \mathbb R$, $a^i_j=a^j_i$, would there be an everywhere non-zero flow if $a$ was a homotopy?