I want to proof the following statement:
The $\wedge$-product is "natural". Therefor $f^{\ast}\omega\wedge f^\ast\eta=f^\ast(\omega\wedge\eta)$ for every linear function $f:W\to V$ and every $\omega\in Alt^r V, \eta\in Alt^s V$.
With the definition $(f^\ast\omega)(w_1,\dotso, w_r):=\omega(f^\ast(w_1),\dotso,f^\ast(w_r))$ the proof is easy.
It is
$f^\ast(\omega\wedge\eta)(v_1,\dotso,v_{r+s})=(\omega\wedge\eta)(f(v_1),\dotso,f(v_{r+s})$
$=\frac{1}{r!s!}\sum_{\sigma\in S_{r+s}}sgn(\tau)\omega(f(v_{\sigma(1)},\dotso, f(v_{\sigma(r)})\eta(f(v_{\sigma(r+1)},\dotso,f(v_{\sigma(r+s)})$
$=\dotso=(f^\ast\omega\wedge f^\ast\eta)(v_1,\dotso, v_{r+s})$
Is this proof correct? Thanks in advance.