TITLE: Help with diagram chasing
QUESTION [7 upvotes]: Given the diagram
$\require{AMScd}$
\begin{CD} 
0 @>>> A @>f>> B @>g>> C @>>> 0 \\
@. @V\alpha VV \#@V\beta V V\# @VV\gamma V @. \\
0  @>>> {A'} @>>{f'}> {B'} @>>{g'}> {C'} @>>> 0\\
\end{CD}
where $\alpha,\gamma$ are monic and the rows are exact. Prove that also $\beta$ is a monomorphism. 
In R-Mod this is rather straightforward: given $\beta(b)=0$, show that $b=0$; 
$g(b)=0\Rightarrow \exists a\in A: f(a)=b\Rightarrow f'\alpha(a)=0\Rightarrow a=0$.
But how to do this using only the universal properties (of zero-elements, kernels etc)?

REPLY [5 votes]: Take a morphism $T \to B$ such that $T \to B \to B'$ is zero. We have to show that $T \to B$ is zero.
$\require{AMScd}$
\begin{CD}
@.T@=T\\ 
@.@VVV\#@VVV\\
0 @>>> A @>f>> B @>g>> C @>>> 0 \\
@. @V\alpha VV \#@V\beta V V\# @VV\gamma V @. \\
0  @>>> {A'} @>>{f'}> {B'} @>>{g'}> {C'} @>>> 0\\
\end{CD}
By commutativity of the diagram the morphism $T \to B \to C \to C'$ is zero, hence the morphism $T \to B \to C$ is zero, since $C \to C'$ is monic.
But this shows that $T \to B$ factors over $T \to A \to B$ and by commutativity we get that $T \to A \to A' \to B'$ is zero. Since $A \to A' \to B'$ is monic, we deduce that $T \to A$ is zero, hence the result.
What I somehow did here by hand: I showed exactness of the snake lemma exact sequence at $ker(\beta)$.