Pull back of wedge of differential forms Suppose that $\phi : M \rightarrow N$ is a smooth map between differential manifolds. Let $\omega, \eta$ be forms on $N$. Is there an easy proof for the fact that
$$\phi^*(\omega \wedge \eta) = \phi^* \omega \wedge \phi^* \eta?$$
In other words, $\phi^* : \Omega(N) \rightarrow \Omega(M)$ is a graded algebra homomorphism. (By easy, I mean does not require expanding both sides using the definition.)
 A: Let $p$ be a point $M$, $v,w \in T_pM$ and $\omega_1, \omega_2, \omega$ and $\eta  $ be 1-forms of $M$ . We have to use that:
$$(\omega_1 \wedge \omega_2)_p(v,w) =
\begin{vmatrix} 
{\omega_1}_p(v) & {\omega_1}_p(w) \\ 
{\omega_2}_p(v) & {\omega_2}_p(w)
\end{vmatrix}$$
$\phi^*(\omega \wedge\eta)_p(v,w)
  = (\omega \wedge \eta)_{\phi (p)}(d \phi_p(v),d \phi_p(w)) = \begin{vmatrix} 
{(\omega)}_{\phi(p)}(d \phi_p(v)) & {(\omega)}_{\phi(p)}(d \phi_p(w)) \\ 
{(\eta)}_{\phi(p)}(d \phi_p(v) & {(\eta)}_{\phi(p)}(d \phi_p(w))
\end{vmatrix} = \begin{vmatrix} 
{(\phi^*\omega)}_p(v) & {(\phi^*\omega)}_p(w) \\ 
{(\phi^*\eta)}_p(v) & {(\phi^*\eta)}_p(w)
\end{vmatrix} = (\phi^*\omega) \wedge (\phi^*\eta)_p(v,w) $. 
So we have $\phi^*( \omega \wedge \eta) = \phi^*(\omega) \wedge \phi^*(\eta).$
A: I've read your question wrong, here is a more general answer you want, I guess.
We will use this following proprieties: 
$$\phi^*(\omega_1+\omega_2)= \phi^*(\omega_1)+\phi^*(\omega_2)$$
$$\omega \wedge \eta = \frac {(m+l)!}{m!l!}Alt(\omega \otimes \eta) $$
$$Alt(\omega \otimes \eta)=\frac {1}{(m+l)!} \sum_{\sigma \in S_ {m+l }} (\omega \otimes \eta)\circ\sigma $$
By that,  $$\phi^*(\omega \wedge \eta)= \frac {(m+l)!}{m!l!}\phi^*(Alt(\omega \otimes \eta)) = \frac {(m+l)!}{m!l!}\phi^*(\frac {1}{(m+l)!} \sum_{\sigma \in S_ {m+l }} (\omega \otimes \eta)\circ\sigma) =\\ =\frac {1}{m!l!} \sum_{\sigma \in S_ {m+l }}\phi^* (\omega \otimes \eta)\circ\sigma $$
Now if we prove that $\phi^*(\omega \otimes \eta)\circ\sigma = (\phi^*\omega \otimes \phi^*\eta)\circ\sigma $, we have that
$$\frac {1}{m!l!} \sum_{\sigma \in S_ {m+l }}\phi^* (\omega \otimes \eta)\circ\sigma = \frac {1}{m!l!} \sum_{\sigma \in S_ {m+l }} (\phi^*\omega \otimes\phi^*\eta)\circ\sigma = \phi^*(\omega)\wedge \phi^*(\eta)$$
Proving that $\phi^*(\omega \otimes \eta)=(\phi^*\omega \otimes \phi^*\eta)$ is easy.
