# $f^*(w_1 + w_2) = f^*w_1 + f^* w_2$

A few pullback identities are required to prove on Guillemin and Pollack's Differential Topology on Page 164. I am not sure if I get it since it is the first time I play with them.

$w$ is an alternating $p$-tensor. Prove $f^*(w_1 + w_2) = f^*w_1 + f^* w_2$

\begin{eqnarray} f^*(w_1 + w_2) &= & (df_x)^*(w_1 + w_2)(f)\\ & = &(df_x)^* w_1 (f) + (df_x)^* w_2 (f)\\ & = &f^*w_1 + f^* w_2, \end{eqnarray} since $p$-tensors are multilinear by definition.

I hope I got this right?

One of the other two problems is $f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$

Assuming $f:X \longrightarrow Y$ is a smooth map of manifolds, let $V_1, \dots, V_p$ be a collection of $p$ vector fields on $X$. Then we have that for any $x \in X$, \begin{align} f^\ast(\omega_1 + \omega_2)_x(V_1, \dots, V_p) & = (\omega_1 + \omega_2)_{f(x)}(df_x V_1, \dots, df_x V_p) \\ & = (\omega_1)_{f(x)}(df_x V_1, \dots, df_x V_p) + (\omega_2)_{f(x)}(df_x V_1, \dots, df_x V_p) \\ & = (f^\ast \omega_1)_x(V_1, \dots, V_p) + (f^\ast \omega_2)_x(V_1, \dots, V_p) \\ & = (f^\ast \omega_1 + f^\ast \omega_2)_x(V_1, \dots, V_p). \end{align} Hence $$f^\ast (\omega_1 + \omega_2) = f^\ast \omega_1 + f^\ast \omega_2.$$ From the computation one sees that this identity basically follows from the definition of the the sum $\omega_1 + \omega_2$.