The tensor quotient by a one-dimensional subspace and its kernel

By [F2] the four tensors $e_i\otimes f_j$ form a basis of $V\otimes W$, and the classes $[e_2]\in V\mathrm{/}U$ and $[f_1]\in W\mathrm{/}Z$ form bases of the one-dimensional quotients, so $[e_2]\otimes [f_1]$ is the product basis of $(V\mathrm{/}U)\otimes (W\mathrm{/}Z)\cong k\otimes k\cong k$ by [F2]. Since $p$ kills $e_1$ and fixes the class of $e_2$, while $q$ kills $f_2$ and fixes the class of $f_1$, the map $p\otimes q$ kills exactly those basis tensors with $e_1$ in the first factor or $f_2$ in the second, namely $e_1\otimes f_1$, $e_1\otimes f_2$ and $e_2\otimes f_2$, and sends $e_2\otimes f_1$ to the basis element $[e_2]\otimes [f_1]$; in particular $p\otimes q$ is surjective by [F1] and [F2].

Since $(e_i\otimes f_j)$ is a basis of $V\otimes W$ by [F2], a tensor ${\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ lies in the kernel of the linear map $p\otimes q$ exactly when ${\sum }_{i,j}{c}_{ij}(p\otimes q)(e_i\otimes f_j)=0$; step 1.1 shows that $(p\otimes q)(e_i\otimes f_j)=0$ for the three basis tensors $e_1\otimes f_1$, $e_1\otimes f_2$, $e_2\otimes f_2$, while $(p\otimes q)(e_2\otimes f_1)=[e_2]\otimes [f_1]\mathrm{\ne }0$ is a nonzero product-basis element, so in the basis expansion the kernel condition is the vanishing of the single coefficient of $e_2\otimes f_1$ and the kernel is exactly $\mathrm{span}\{e_1\otimes f_1,e_1\otimes f_2,e_2\otimes f_2\}$. That span equals $U\otimes W+V\otimes Z$ inside $V\otimes W$ because $U\otimes W=\mathrm{span}\{e_1\otimes f_1,e_1\otimes f_2\}$ and $V\otimes Z=\mathrm{span}\{e_1\otimes f_2,e_2\otimes f_2\}$ as the product bases of the respective tensor products, and by the basis expansion of [F2] its dimension is $3$. The dimension formula reads $\dim U\dim W+\dim V\dim Z-\dim U\dim Z=1\cdot 2+2\cdot 1-1\cdot 1=3$, in agreement with the kernel just computed and with the prediction of [F3], whose complement argument is not needed for this explicit computation.