How to prove Matsumura, Appendix A, Tensors, Formula 7? Let all tensors be tensor products of $A$-modules or the tensoring of two functions $f\otimes g : M\otimes N \to M' \otimes N'$ where $f: M \to M', \ g : N \to N'$ are two $A$-module homs.

Formula 7. Suppose given exact sequences
$$0 \to K \xrightarrow{i} M \xrightarrow{f}M' \to 0 \text{ and } 0 \to L \xrightarrow{j} N \xrightarrow{g} N' \to 0$$
then $M' \otimes N' \simeq (M \otimes N)/T$, where $T = (i \otimes 1)(K \otimes N) + (1 \otimes j) (M \otimes L)$.

I know that the map $f\otimes g$ is surjective by the exactness at $M', N'$ of the two sequences.  So I merely need to show that $T = \ker f \otimes g$, where $\ker f \otimes g = $ the submodule of $M\otimes N$ generated by all $x\otimes y$ such that $f(x) = 0$ or $g(y) = 0$.
It's easy for me to see that $T$ must be a submodule of $\ker f \otimes g$, and since $f\otimes g$ is generated by simple tensors, that in order to show the reverse inclusion, I need to show that $x \otimes y \in T$ for any simple tensor $x \otimes y \in M \otimes N$ such that $f(x) = 0$ or $g(y) = 0$.

Can we find two simple tensors $k \otimes n \in K \otimes N$ and $m \otimes l \in M \otimes L$ such that $i(k)\otimes n + l \otimes j(m) = x \otimes y$?
I thought that would be the easiest to solve for if it indeed had a solution.  Although, it's still not obvious to me what we should try to put for $k,n,l,m$.
 A: The most straightforward way to do this is probably to use the universal property of $M' \otimes N'$ as a tensor product.  Since the kernel $K$ of the $A$-linear map $f \otimes g: M \otimes N \rightarrow M' \otimes N'$ contains $T$, there is an obvious induced $A$-linear map
$$\alpha: (M \otimes N)/T \rightarrow M' \otimes N'.$$
What is less straightforward is to show that there exists an $A$-linear map going in the opposite direction
$$\beta: M' \otimes N' \rightarrow (M \otimes N)/T$$
such that the composition in each direction is the identity.  This will imply that $\alpha$ is an isomorphism and that the kernel of $f \otimes g$ is exactly $T$.
To define $\beta$, define an $A$-bilinear map
$$\mathcal B: M' \times N' \rightarrow (M \otimes N)/T$$
as follows: given $(m',n') \in M' \times N'$, choose $m \in M$ and $n \in N$ such that $f(m) = m'$ and $g(n) = n'$.  Then set
$$\mathcal B(m',n') = m \otimes n + T.$$
It is not difficult to see that this is well defined, and therefore induces an $A$-linear map $\beta$ of the type suggested above.  It is defined on elementary tensors by
$$\beta(f(m) \otimes g(n) ) = m \otimes n + T.$$
