# Another basic short exact sequence problem

In the following commutative diagram of R-modules, all of the rows and columns are exact. Prove that $K$ is isomorphic to $L$.

\begin{array}{ccccccccccc} &&&&&&&&0 &&\\ &&&&&&&&\downarrow &&\\ &&&&&& 0 & & L &&\\ &&&&&& \downarrow && \downarrow &&\\ &&&& M^{\prime\prime} & \rightarrow& N^{\prime\prime} & \rightarrow & P^{\prime\prime} & \rightarrow& 0\\ &&&& \downarrow & & \downarrow && \downarrow &&\\ && 0 & \rightarrow & M & \rightarrow & N & \rightarrow & P & \rightarrow & 0\\ &&&& \downarrow & & \downarrow && \downarrow &&\\ 0 & \rightarrow & K & \rightarrow & M^\prime & \rightarrow & N^\prime& \rightarrow & P^\prime & \rightarrow & 0\\ &&&& \downarrow & & \downarrow && \downarrow &&\\ &&&& 0 & & 0 && 0 && \end{array}

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Hah, I wish I could tell you why right away :) Rather than diagram chasing, I bet it's an application of a diagram lemma you learned recently. Do you have any such lemmas on hand? –  rschwieb May 17 '13 at 13:10
snake lemma!! Let me think about it:) –  Ishigami May 17 '13 at 13:11

HINT The snake lemma tells us that there is an exact sequence $$0 \to L \to M^\prime.$$
HINT 2 Kernels of $R$-modules have a universal property.
Added: The universal property of a kernel is this. A kernel of a morphism $f:M \to N$ is a morphism $i:K \to M$ such that $f \circ i = 0$ and such that is universal with respect to this property, meaning that if $i^\prime:K^\prime \to M$ is any other morphism with $f \circ i^\prime = 0$, then there exists a unique morphism $u:K^\prime \to K$ such that $i^\prime = i \circ u$.
If you want to apply the universal property of the kernel, don't you need to say that $0 \rightarrow L \rightarrow M' \rightarrow N'$ is exact (which it is)? –  tkr May 17 '13 at 14:47