\[ \vcenter { \xymatrix{ A & A' \ar[l] & I \ar[l] \\ B \ar[u] & B' \ar[l] \ar[u] & J \ar[l] \ar[u] } } \quad \text{and}\quad \vcenter { \xymatrix{ M & M' \ar[l] & K \ar[l] \\ N \ar[u]_\varphi & N' \ar[l] \ar[u] & L \ar[l] \ar[u] } } \]

where $I, J, K, L$ are the kernels of the horizontal maps of the original diagrams. We present the proof as a sequence of observations:

$K = IM'$ (see statement lemma),

$B' \to B$ is surjective with kernel $J$ and $J \to I$ is bijective,

$N' \to N$ is surjective with kernel $L$ and $L \to K$ is bijective,

$JN' \subset L$,

$\mathop{\mathrm{Im}}(N \to M)$ generates $M$ as an $A$-module (because $N \otimes _ B A = M$),

$\mathop{\mathrm{Im}}(N' \to M')$ generates $M'$ as an $A'$-module (because it holds modulo $K$ and $L$ maps isomorphically to $K$),

$JN' = L$ (because $L \cong K = I M'$ is generated by images of elements $x n'$ with $x \in I$ and $n' \in N'$ by the previous statement),

$N' \otimes _{B'} B = N$ (because $N = N'/L$, $B = B'/J$, and the previous statement),

there is a map $\gamma : N' \otimes _{B'} A' \to M'$,

$\gamma $ is surjective (see above),

the kernel of the composition $N' \otimes _{B'} A' \to M' \to M$ is generated by elements $l \otimes 1$ and $n' \otimes x$ with $l \in K$, $n' \in N'$, $x \in I$ (because $M = N \otimes _ B A$ by assumption and because $N' \to N$ and $A' \to A$ are surjective with kernels $L$ and $I$),

any element of $N' \otimes _{B'} A'$ in the submodule generated by the elements $l \otimes 1$ and $n' \otimes x$ with $l \in L$, $n' \in N'$, $x \in I$ can be written as $l \otimes 1$ for some $l \in L$ (because $J$ maps isomorphically to $I$ we see that $n' \otimes x = n'x \otimes 1$ in $N' \otimes _{B'} A'$; similarly $x n' \otimes a' = n' \otimes xa' = n'(xa') \otimes 1$ in $N' \otimes _{B'} A'$ when $n' \in N'$, $x \in J$ and $a' \in A'$; since we have seen that $JN' = L$ this proves the assertion),

the kernel of $\gamma $ is zero (because by (10) and (11) any element of the kernel is of the form $l \otimes 1$ with $l \in L$ which is mapped to $l \in K \subset M'$ by $\gamma $).

This finishes the proof.
$\square$

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