Q:

Tensor product of two quotient modules

I have a question from "Exercise 1.3, Chapter 1, Qing Liu, Algebraic Geometry and Arithmetic Curves": 

Let $A$ be a commutative ring with unit. Let $M,N$ be two $A-$modules,
  and $i:M' \to M, j:N' \to N$ submodules of $M$ and $N$, respectively.
  Then there exists a canonical isomorphism  $$\left( M/M' \right)
 \otimes_A \left( N/N' \right) \simeq \left( M \otimes_A N
 \right)/\left( \text{Im } i_N + \text{Im } j_M\right).$$

Notation: Let $f:N' \to N$ be a linear map of $A-$modules. For any $A-$module $M$, we denote the linear map $f \otimes_A \text{Id}_M : N' \otimes_A M \to N \otimes_A M$ by $f_M$. 
My attempt: 
Using the result 

Let $A$ be a commutative ring with unit. Let $M,N$ be two $A-$modules,
  and $i:M' \to M$ submodules of $M$.
  Then there exists a canonical isomorphism $$\left( M \otimes_A N \right)/\left( \text{Im } i_N \right) \simeq \left( M/M' \right)\otimes_A N,$$

we obtain that
\begin{align*}
\left( M/M' \right) \otimes_A \left( N/N' \right) &\simeq \left(N\otimes_A \left(M/M'\right)\right)/\left(\text{Im } j_{M/M'}\right)\\
&\simeq \left(\left(N\otimes_A M\right)/\text{Im } i_N\right)/\left(\text{Im } j_{M/M'}\right).
\end{align*}
However, I have a trouble when I try to prove 
$$\left(\left(N\otimes_A M\right)/\text{Im } i_N\right)/\left(\text{Im } j_{M/M'}\right) \simeq \left( M \otimes_A N
 \right)/\left( \text{Im } i_N + \text{Im } j_M\right).$$
Can you help me for this problem? Thank you very much.

A:

One way is to show that $M \otimes N$ modulo the the submodule $P$ generated by $m' \otimes n$ and $m \otimes n'$ (with $m' \in M', m \in M, n' \in N', n \in N$) directly satisfies the same universal property as $M/M' \otimes N/N'$.
You have a well defined $A$-bilinear map $j: M/M' \times N/N' \rightarrow (M \otimes N)/P$ given by 
$$(m+M',n+N') \mapsto m \otimes n + P$$
Suppose $Q$ is an $A$-module, and $f: M/M' \times N/N' \rightarrow Q$ is $A$-bilinear.  I claim that there exists a unique $A$-module homomorphism $\phi: (M\otimes N)/P \rightarrow Q$ such that $\phi \circ j = f$.  The uniqueness of the tensor product will then guarantee you an isomorphism $M/M' \otimes N/N' \rightarrow (M \otimes N)/P$ such that $(m +M') \otimes (n+N') \mapsto m \otimes n + P$.
The map $M \times N \rightarrow Q, (m,n) \mapsto f(m + M',n + N')$ is certainly $A$-bilinear, so there exists a unique $A$-linear map $\psi: M \otimes N \rightarrow Q$ given on generators by $\psi(m\otimes n) = f(m+M',n+N')$.  Also, for any $m' \in M', n' \in N$ you have 
$$\psi(m' \otimes n) = f(m' + M',n + N') = f(0 + M', n+ N') = 0$$
$$\psi(m \otimes n') = f(m + M', n' + N') = f(m+M',0+N') = 0$$
so $P$ is contained in the kernel of $\psi$.  Therefore the $A$-module homomorphism $\phi: (M \otimes N)/P \rightarrow Q$ defined by $\phi(x + P) = \psi(x)$ is well defined.
Now $\phi$ does what is required:
$$\phi \circ j(m+M',n+N') = \phi(m\otimes n + P) = \psi(m \otimes n) = j(m+M',n+N')$$
and the uniqueness of $\phi$ is easily seen from the uniqueness of $\psi$.

