How to complete an exact sequence

Question

Suppose that $\require{AMScd}$ \begin{CD} @. @. @. K\\ @. @. @. @V\alpha VV \\ 0@>>>L @>{\phi}>>M @>{\psi}>>N @>>>0 \end{CD} is a diagram of $R$-modules and homomorphisms whose row is exact. I want to show that one can turn this diagram into a commutative diagram with exact rows and exact columns of the following form $\require{AMScd}$ \begin{CD} @[email protected] @.0\\ @. @. @VVV @VVV\\ @.@. Ker \beta @>{\cong}>>Ker \alpha \\ @.@.@VVV @VVV\\ 0@>>>L@>>>P@>>>K@>>>0\\ @.@V{1_L}VV @V\beta VV @V\alpha VV \\ 0@>>>L @>{\phi}>>M @>{\psi}>>N @>>>0\\ @. @. @VVV @VVV \\ @. @. \frac{M}{Im \beta} @>{\cong}>> \frac{N}{Im \alpha} \\ @. @. @VVV @VVV \\ @. @. 0 @. 0 \\ \end{CD} First, I want to know how to find such a $P$. I think it must be formed using both $Im \phi$ and $K$ but then I will face a problem on checking the commutativity. I would appreciate any hint or solution.

Answer 1

Hint : Take $P$ to be the pullback $M\times_N K$ of $\psi$ and $\alpha$. In other words $$P=\{(m,k)\in M\times K \mid \psi(m)=\alpha(k)\},$$ with maps $P\to M$ and $P\to K$ given by the restrictions of the two projections of $M\times K$. From there, it's easy to prove that the induced maps on the kernels must be isomorphisms, and for the cokernels you can use the Nine Lemma.