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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.

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You want the first square to be a pushout square. So take $Z=(X\oplus Y)/N$ where $N=\{(f(m),-g(m)):m\in M\}$.

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