# Help with diagram chasing

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)?

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 $ker(\beta)$.
• This wasn't that different from the case in R-Mod, so I'm tempted to ask the corresponding question for $\alpha,\beta$ epi. – Lehs Feb 7 '15 at 9:02
• This is just the dual statement, isn't it? So you take a morphism $B' \to T$, such that $B \to B' \to T$ is zero. Then you do literally the same thing. So you show that $B' \to T$ factors over $C'$ by the universal property of the cokernel. And then by surjectivity of $B \to C \to C'$, you can deduce that $C' \to T$ is zero. – MooS Feb 7 '15 at 9:07