Kernel of quotient-like homomorphism induced by tensor product Let $k$ be a field and $V,W$ vector spaces over $k$. Suppose $V'\subset V$ and $W'\subset W$ are vector subspaces. We have the morphisms $q_V: V\to V/V'$ and $q_W: W\to W/W'$ by taking quotients in the usual way. It seems to be the case that this induces a morphism $q_{V\otimes W}: V\otimes W\to V/V' \otimes W/W'$ in the obvious way. I have checked that this map is indeed well-defined, since it comes from a bilinear and $k$-balanced map on $V\times W$.
Is there a reasonable description of the kernel of this map? My instinct was to guess $\ker  q_V \otimes \ker q_W$, but it is in fact clear that it must be strictly larger, since elements $v\otimes w$ where $v\in \ker q_V$ and $w$ is any element of $W$ are killed.
 A: It is $(\ker q_V \otimes W) + (V \otimes \ker W)$. You just argued that $\ker q_V \otimes W$ and $V \otimes \ker q_W$ are in $\ker q_{V \otimes W}$, so it follows that the sum of these two subspaces is also in the kernel (why?). For the other inclusion, take a pure tensor $v \otimes w \in \ker q_{V \otimes W}$. Then
\begin{equation}
0 = q_{V\otimes W}(v\otimes w) = q_V(v) \otimes q_W(w) 
\end{equation}
so that at least one of the two factors has to vanish, i.e. either $v \in \ker q_V(V)$ or $w \in \ker q_W(W)$. Any other element of $\ker q_{V\otimes W}$ is a linear combination of pure tensors of this form, which shows that $\ker q_{V \otimes W} \subseteq (\ker q_V \otimes W) + (V \otimes \ker W)$. This is not the case as Eric pointed out. The result is still true nonetheless, see Theorems 2.19 and 5.5 in these notes by Keith Conrad.
A: This has been answered here already: What is the kernel of the tensor product of two maps?
More generally, if $A \to B \to C \to 0$ and $A' \to B' \to C' \to 0$ are exact sequence in a monoidal category which has finite colimits which are preserved by $\otimes$ in both variables (this is automatic when the monoidal structure is closed), then we get an exact seqeunce $(A \otimes B') \oplus (A' \otimes B) \to B \otimes B' \to C \otimes C' \to 0$. In fact, you can verify directly that $B \otimes B' \to C \otimes C'$ is a cokernel of $(A \otimes B') \oplus (A' \otimes B) \to B \otimes B'$. See Lemma 3.1.20 here.
