$f^*(w \wedge \theta) = (f^*w) \wedge (f^* \theta)$

Solution 1:

I read your approach but I am giving another approach which might be helpful. Lets start with the definition. If $f:X\rightarrow Y$ is a map and $\omega$ is a $p$ form in $Y$ then its pull back is defined as $f^*(\omega)v=\omega (f_*v).,$ where $f_*v=v(f),$ for any $v\in T_pX.$ Therefore Expand (assuming $\omega$ is a p form and $\theta$ is a q form) the term (where $v_i$and $w_i$ are in $T_pX$ ) $f^*(\omega\wedge \theta)(v_1,\cdots ,v_p,w_1,\cdots ,w_q)=(\omega\wedge \theta)(f_*(v_1),\cdots ,f_*(v_p),f_*(w_1),\cdots ,f_*(w_q)).$ using the summation formula for wedge product. Which will be same as $\omega(f_∗(v_1),⋯,f_∗(v_p))\wedge \theta(f_*(w_1),⋯,f_∗(w_q)).$ Which is (by definition) same as $f^*(\omega)\wedge f^*(\theta).$