# Show that any monomorphism and a homomorphism can be embedded into a commutative diagram with exact rows.

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.

## 1 Answer

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