TITLE: Show that any monomorphism and a homomorphism can be embedded into a commutative diagram with exact rows.
QUESTION [3 upvotes]: Let $R$ be a commutative ring with identity. Show that the diagram of $R$-module homomorphisms with the row exact
\begin{matrix}
0&\to&M&\mathop{\to}\limits^{f}&X\\
&&\downarrow^g\\
&&Y&\\
\end{matrix}
can be embedded into the following commutative diagram with exact rows.
\begin{matrix}
0&\to&M&\mathop{\to}\limits^{f}&X&\to&{\rm coker}(f)&\to&0\\
&&\downarrow^g&&\downarrow^\beta&&\downarrow\\
0&\to&Y&\mathop{\to}\limits^\alpha&Z&\to&{\rm coker}(f)&\to&0\\
\end{matrix}
What I have tried is to set $Z={\rm coker}(f)\oplus Y$ and let $\alpha:y\mapsto(\bar 0,y)$ and $\beta:x\mapsto(\bar x,g(a))$, where $a=f^{-1}(x)$. But this definition fails for the second component of $\beta$ when $x$ is not in ${\rm im}f$.
Any suggestions would be appreciated.

REPLY [3 votes]: You want the first square to be a pushout square. So take $Z=(X\oplus Y)/N$
where $N=\{(f(m),-g(m)):m\in M\}$.
{"set_name": "stack_exchange", "score": 3, "question_id": 3255970}