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 $ker(\beta)$.

  • $\begingroup$ Thanks for the visual improvement, I did not have the patience to do so. $\endgroup$ – MooS Feb 7 '15 at 8:50
  • $\begingroup$ This wasn't that different from the case in R-Mod, so I'm tempted to ask the corresponding question for $\alpha,\beta$ epi. $\endgroup$ – Lehs Feb 7 '15 at 9:02
  • $\begingroup$ 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. $\endgroup$ – MooS Feb 7 '15 at 9:07
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    $\begingroup$ Abstract nonsense is practical and useful then... $\endgroup$ – Lehs Feb 7 '15 at 9:11
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    $\begingroup$ The reason for this to seem harder might be the fact that quotients (which rely to the universal property of the cokernel) seem to be fancier than subobjects (which rely to the universal property of the kernel). But forgetting about that and just using arrows, it is literally the same. $\endgroup$ – MooS Feb 7 '15 at 9:13

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