Let $F:U \rightarrow V$ be a diffeomorphism between open sets in $\mathbb{R}^n$. Let $\mathbb{Y}$ be a vector field on $V$ and $\omega$ a $k$-form on $V$. Show that
$F^*(L_{\mathbb{Y}}\omega)=L_{\mathbb{X}}(F^*\omega)$
where $\mathbb{X}=F_*^{-1}\mathbb{Y}$
I have no idea how this could be done.
Notice that, for any form $\omega$, $i_X(F^*\omega)=F^*(\iota_Y\omega)$. Hence, using the fact that $d$ commutes to $F^*$, and the Cartan formula, \begin{eqnarray} L_X(F^*\omega)&=&\iota_X(d(F^*\omega))+d(\iota_X(F^*\omega))\\ &=&\iota_X(F^*d\omega)+d(F^*\iota_Y\omega)\\ &=&F^*\iota_Y(d\omega)+F^*d(\iota_Y\omega)\\ &=&F^*(L_Y\omega)\,. \end{eqnarray} EDIT. To prove $i_X(F^*\omega)=F^*(\iota_Y\omega)$, you simply evaluate both sides on vector fields. Let $X_2,\dots,X_k$ be vector fields on $U$, then by definition $$\Big(i_X(F^*\omega)\Big)(X_2,\dots,X_k)=(F^*\omega)(X,X_2,\dots,X_k)=\omega(Y,F_*X_2,\dots,F_*X_k)$$ and $$\Big(F^*(\iota_Y\omega)\Big)(X_2,\dots,X_k)=(\iota_Y\omega)(F_*X_2,\dots,F_*X_k)=\omega(Y,F_*X_2,\dots,F_*X_k)$$ This proves the assertion.