Let $f:M\to N$ be a differentiable map between two manifolds. $e_i$ is a basis vector of $N$ with respect to some chart and $e_i'$ its dual (i.e. $e_i'(e_j)=\delta_{ij}$). How do I prove the following? $$d(f^*e_i')=0$$ By definition of $f^*$ we have $$d(f^*e_i')=d(e_i'\circ df).$$ But how to proceed?
Edit: My definition of the differential $d$: Let $\omega$ be a differentiable $k$-form on $M$ (i.e. $\omega\in \Lambda^kM$) and $(b_i)_{1\le i \le n}$ the standard basis with respect to some chart $x$ and $(b_i')_{1\le i \le n}$ its dual basis. Then we have $$d\omega = \sum_{|I|=k} \sum_{r=1}^n \frac{\partial\omega_I}{\partial x_r}b_r'\wedge b_I'$$ in the domain of $x$. In our special case above $f$ is a 0-form and $f^*e_i'$ is a 1-form.