The pull back of interior multiplication

Question

Let $M$ be a smooth manifold, and $X$ a vector field, $w$ a $k$ form on $X$. $f: M \to M$ is a smooth map. Is it true that $f^{\ast}(i_Xw) = i_Xf^{\ast}w$?

I think it is true because:

$f^{\ast}(i_Xw)(X_1, \ldots, X_{k-1}) = i_Xw(DfX_1, \ldots, DfX_k) = w(X, DfX_1, \ldots, DfX_{k-1})$, where as $i_Xf^{\ast}w(X_1, \ldots, X_{k-1}) = w(X, DfX_1, \ldots, DfX_{k-1})$

Answer 1

No, it is not true in general, as you made a mistake in your last step. What you have is

$$\left(\iota_Xf^*w\right)\left({X_1},\ldots,{X_{k-1}}\right)=(f^*w)\left(X,X_1,\ldots,X_{k-1}\right)=w\left(Df(X),Df(X_1),\ldots,Df(X_{k-1})\right)$$

Answer 2

Consider $M=\mathbb{R}$, $\omega=dx, f(x)=x^2$ and $X(x)=1$.

$i_X\omega=\omega(X(x))=1$; $f^*(i_X\omega)=1$.

$(f^*\omega)_x(u)=\omega_{f(x)}((df_x.u))=dx_{x^2}(2xu)=2xu$ this implies that $f^*\omega=2xdx$.

$i_Xf^*\omega=i_X(2xdx)=2x$.