Let $\omega $ be a $1$-form on $M$. For $f:M\to \mathbb R$, and $X\in \mathfrak{X}(M)$, Does the following true?
$f. i_X \omega = i_{f.x}\omega$
According to me proof is the following: Please point out the mistake in the following and if possible please make comment on above expression. Thanks.
For $p\in M$, we have $f.i_X\omega|_p= f(p). \omega_p(X_p)= \omega(f(p)X_p)= i_{fX}\omega|_p$