Edit: Exact Question. my question is b part
$\phi:E\times F\to G$ be bilinear
$\psi:E\times F\to H$ be bilinear
Given $N_1(\phi)\subset N_1(\psi)$ and $N_2(\phi)\subset N_2(\psi)$ Show that there exist a linear function $f:G\to H$ such that $f\cdot\phi=\psi$ where $N_1(\phi) = \{x|\phi(x,y)=0\forall y\in F\}$ and similarly $N_2$ for the second coordinate
My "counterexample"
$F=\mathbb R^3$
$H=E=\mathbb R^2$
$G=\mathbb R^5$
$\phi(e,f)=(e_1f_1,e_1f_2,e_1f_3,e_2f_1,e_2f_2)$
$\psi(e,f)=(e_2f_2,e_2f_3)$
$N_1(\phi)=\{(0,0)\}\subset\{(x,0)\}=N_1(\psi)$
$N_2(\phi)=\{(0,0,0)\}\subset\{(x,0,0)\}=N_2(\psi)$
Please point out the mistake in the "counterexample". Also please provide hints for the problem
Your counterexample is correct, it shows the claim made before cannot be correct (there is no way to get $e_2f_3$ as a linear combination of the components of $\phi(e,f)$). Are you sure you copied the problem correctly, especially the definition of $N_1, N_2$?