A theorem about the tensor product of modules is that
If $\varphi: B \to C$ and $\psi: B' \to C'$ are surjective, then $$\llap{\text{the kernel of} \quad} \varphi \otimes \psi: B \otimes B' \to C \otimes C'$$ $$\llap{\text{is} \qquad}i(\operatorname{ker} \varphi \otimes B') + j(B \otimes \operatorname{ker}\psi),$$ where $i$ and $j$ are the tensorings of the inclusions with the other module.
There are a couple of proofs for this. One, given here, is to write down the inverse of the map $$\frac{B \otimes B'}{i(\operatorname{ker} \varphi \otimes B') + j(B \otimes \operatorname{ker}\psi)} \longrightarrow C \otimes C'.$$
Another way, mentioned here by Martin Brandenberg, is to note that given two exact sequences $$A \to B \xrightarrow{\varphi} C \to 0 \rlap{\qquad \text{and}}$$ $$A' \to B' \xrightarrow{\psi} C' \to 0,$$ we can see that the sequence $$(A \otimes B') \oplus (B \otimes A') \to B \otimes B' \xrightarrow{\varphi \otimes \psi} C \otimes C' \to 0$$ is exact, and we can do so in a purely categorical way.
What troubles me about both these proofs is that, if $\sum b_i \otimes b'_i \overset{\varphi \otimes \psi}{\mapsto} 0 \in C \otimes C'$, I don't see any particular way of writing $\sum b_i \otimes b'_i$ like $i(\operatorname{ker} \varphi \otimes B') + j(B \otimes \operatorname{ker}\psi)$. I feel like there should be a way.
Does anyone have some input on this?
