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}
Again I have no idea how to do, it seems so complicated, please helps.
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 $i^\prime = i \circ u$.
By standard category theoretic considerations, the universal property implies that any two kernels are uniquely isomorphic. For more on this, see the Wikipedia article on kernels.