Let $F:V \rightarrow W$ be a linear map. Show that $F^{\ast}(\omega \wedge \eta)=(F^{\ast}\omega) \wedge (F^{\ast}\eta)$ for all $\omega \in \Lambda^{p}(W) , \eta \in \Lambda^{q}(W)$.
Where $F^{\ast}\omega$ denotes the pullback of $\omega$, $\wedge$ is the wedge product and $\Lambda^{p}(W)$ is set of all alternating $p$-tensors on W.
So far I have $F^{\ast}(\omega \wedge \eta)(v_{1},\dotsc,v_{p+q})=\omega \wedge \eta(F(v_{1}),\dotsc, F(v_{p+q}))$
$=\frac{(p+q)!}{p!q!}A(\omega \otimes \eta)(F(v_{1}),\dotsc, F(v_{p+q}))$
$=\frac{(p+q)!}{p!q!}\frac{1}{(p+q)!}\displaystyle\sum_{\sigma \in S_{p+q}}\operatorname{sgn}(\sigma)(\omega \otimes \eta)(w_{\sigma(1)},\dotsc,w_{\sigma(p+q)})$
Where $w_{i}$ denotes $F(v_{i})$. I am unsure on how to proceed from here, so any help would be greatly appreciated!
Best Answer
Try starting from the other direction and use
$(F^*\omega) \otimes (F^*\eta)(v_{\sigma(1)},\ldots, v_{\sigma(p+q)}) = (F^*\omega)(v_{\sigma(1)},\ldots,v_{\sigma(p)})\cdot (F^*\eta)(v_{\sigma(p+1)},\ldots, v_{\sigma(p+q)}) = \omega(F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p)}))\cdot\eta(F(v_{\sigma(p+1)}),\ldots,F(v_{\sigma(p+q)})) = \omega \otimes \eta (F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p+q)}))$
and then compare to what you already have.