Tensoring injections and the kernel of a tensor product of quotient maps over a field

Injectivity of $f\otimes {\mathrm{i}\mathrm{d}}_W$: by [A1] write $V^{\mathrm{\prime }}=f(V)\oplus C$ and define $g:V^{\mathrm{\prime }}\to V$ by $g(f(v)+c):=v$; this is well defined and linear because every element of $V^{\mathrm{\prime }}$ has a unique decomposition, and $g\circ f={\mathrm{i}\mathrm{d}}_V$. By [F2], $(g\otimes {\mathrm{i}\mathrm{d}}_W)\circ (f\otimes {\mathrm{i}\mathrm{d}}_W)=(g\circ f)\otimes {\mathrm{i}\mathrm{d}}_W={\mathrm{i}\mathrm{d}}_V\otimes {\mathrm{i}\mathrm{d}}_W={\mathrm{i}\mathrm{d}}_{V\otimes W}$, so $f\otimes {\mathrm{i}\mathrm{d}}_W$ has a left inverse and is injective.

Containment: for $u\in U$ and $w\in W$ one has $(p\otimes q)(u\otimes w)=p(u)\otimes q(w)=0\otimes q(w)=0$ by [F1], because $p$ kills $U$; likewise $p\otimes q$ kills every $v\otimes z\in V\otimes Z$ because $q$ kills $Z$. Hence $\mathrm{im}(U\otimes W)$ and $\mathrm{im}(V\otimes Z)$ lie in the kernel $\ker (p\otimes q)$, which is a subspace by [F3], so their sum $U\otimes W+V\otimes Z$ lies in the kernel as well.

Complements and an isomorphism: by [A1] write $V=U\oplus V_1$ and $W=Z\oplus W_1$. The restrictions $p_1:=p{\mathrm{\mid }}_{V_1}$ and $q_1:=q{\mathrm{\mid }}_{W_1}$ are isomorphisms onto $V\mathrm{/}U$ and $W\mathrm{/}Z$: a coset $v+U$ with $v=u+v_1$ equals $v_1+U$, so $p_1$ is surjective, and if $p(v_1)=0$ then $v_1\in U\cap V_1=0$ by the direct-sum condition, so $p_1$ is injective (and likewise for $q_1$); by [F2] the map $p_1\otimes q_1:V_1\otimes W_1\to (V\mathrm{/}U)\otimes (W\mathrm{/}Z)$ is then an isomorphism with inverse $s\otimes r$, where $s=p_1^{-1}$ and $r=q_1^{-1}$. Every element of $V$ is uniquely $u+x$ with $u\in U$, $x\in V_1$, so the component maps $v\mapsto u$ and $v\mapsto x$ are well defined and linear, and likewise for $W=Z\oplus W_1$; consequently each of the three inclusion-induced maps ${\iota }_U\otimes {\mathrm{i}\mathrm{d}}_W:U\otimes W\to V\otimes W$, ${\mathrm{i}\mathrm{d}}_V\otimes {\iota }_Z:V\otimes Z\to V\otimes W$ and ${\iota }_{V_1}\otimes {\iota }_{W_1}:V_1\otimes W_1\to V\otimes W$ has a left inverse induced by the corresponding component projection (for the first, $({\pi }_U\otimes {\mathrm{i}\mathrm{d}}_W)\circ ({\iota }_U\otimes {\mathrm{i}\mathrm{d}}_W)=({\pi }_U\circ {\iota }_U)\otimes {\mathrm{i}\mathrm{d}}_W={\mathrm{i}\mathrm{d}}_{U\otimes W}$ by [F2], and similarly for the others), hence is injective and exhibits its domain inside $V\otimes W$ through the injections of the statement; under these identifications the composite $V_1\otimes W_1\to V\otimes W\to (V\mathrm{/}U)\otimes (W\mathrm{/}Z)$ is exactly $p_1\otimes q_1$, since both send $x\otimes y$ to $p_1(x)\otimes q_1(y)$.

Kernel: let $t\in \ker (p\otimes q)$ and write $t={\sum }_i{v}_i\otimes w_i$ as a finite sum of elementary tensors by [F1]. Decompose $v_i=u_i+x_i$ with $u_i\in U$, $x_i\in V_1$ and $w_i=z_i+y_i$ with $z_i\in Z$, $y_i\in W_1$ by the direct sums of step 1.3 and expand bilinearly by [F1]: $t={\sum }_i{u}_i\otimes w_i+{\sum }_i{x}_i\otimes z_i+{\sum }_i{x}_i\otimes y_i$, where the first summand lies in $\mathrm{im}(U\otimes W)$, the second in $\mathrm{im}(V_1\otimes Z)\subseteq \mathrm{im}(V\otimes Z)$, and the third is the image of ${\sum }_i{x}_i\otimes y_i\in V_1\otimes W_1$. Applying $p\otimes q$ kills the first two summands by step 1.2, so $0=(p\otimes q)(t)=(p_1\otimes q_1)({\sum }_i{x}_i\otimes y_i)$ by the identification of step 1.3; injectivity of $p_1\otimes q_1$ gives ${\sum }_i{x}_i\otimes y_i=0$ in $V_1\otimes W_1$, hence the third summand is $0$ in $V\otimes W$, and $t\in U\otimes W+V\otimes Z$.

Steps 1.2 and 2.1 give the two inclusions, so $\ker (p\otimes q)=U\otimes W+V\otimes Z$ for the subspaces exhibited through the injections of step 1.1, and step 1.1 itself is the first assertion of the statement.