Let $M$ be connected, and let $\pi: M\times N \rightarrow N$ be the natural projection. Prove that a $p$-form $w$ on $M\times N$ is $\delta \pi(\alpha)$ for some $p$-form $\alpha$ on $N$ if and only if $i(X)w = 0$ and $L_X w=0$ for every vector field $X$ on $M\times N$ for which $d\pi(X(m,n)) = 0$ at each point $(m,n) \in M\times N$.
$i(X)$ is the interior multiplication my the vector field $X$, and $L_X$ is the Lie-derivative w.r.t. the vector field $X$. $d$ is exterior differentiation, and $\delta$ is its transpose or adjoint.
Could you give me some hints for this problem? I tried to do the forward direction just by applying the formulas in the book, and I do not know how to approach the reverse direction.
What I have done so far, first I want to prove the forward direction, given $w=\delta\pi(\alpha)$, then for any vector fields $Y_1, \cdots Y_{p-1}$ on $M\times N$ we have $$ \begin{align} i(X)(\delta\pi(\alpha))(Y_1, \cdots Y_{p-1}) (m,n) &=(\delta\pi(\alpha))(X,Y_1, \cdots Y_{p-1}) (m,n) \\ &= \alpha_{\pi((m,n))}\big(d\pi(X(m,n)), \cdots, d\pi(Y_{p-1}(m,n))\big)\\ &= \alpha_{\pi((m,n))}\big(0, \cdots, d\pi(Y_{p-1}(m,n))\big)\\ &=0 \end{align}$$
Next to show $L_X(w) = 0$, we use the formula that $L_X = i(X)\circ d + d\circ i(X)$. The second part is zero by above calculation, and for the first part, note that $$i(X) d(\delta\pi(\alpha)) = i(X) \delta \pi(d\alpha)$$ then this would be a similar calculation as above, just $d\alpha$ instead of $\alpha$.
Are my calculations correct? and could you give me some hints for the reverse direction.