Proving a diagram chase result from standard lemmas Consider three exact sequences of modules over a commutative ring
$0\to A\to M\to N\to B\to 0$,
$0\to E\to M\to K\to F\to 0$,
$0\to C\to N\to K\to D\to 0$,
where the maps $M\to N$ and $N\to K$ and $M\to K$ commute. I believe that diagram chasing gives that there is an exact sequence
$0\to A\to E\to C\to B\to F\to D\to 0$
that commutes with the maps in the other sequences. Is it possible to prove the exactness of this sequence from the standard lemmas in homological algebra?
Also, this appears to be a fairly simple result in that it states that for any commuting triangle, there is an exact sequence involving the kernels and cokernels of the three maps. Does this result have a name or is there a reference for this result?
 A: Apply the snake lemma to the diagram
$$\require{AMScd}
\begin{CD}
0@>>>M@>\binom{id}{0}>> M\oplus N@>(0\;\; id)>> N@>>>0\\
@.@VfVV@VV\begin{pmatrix}f & id\\0 & -g \end{pmatrix}V@VV-gV\\
0@>>>N@>>\binom{id}{0}> N\oplus K@>>(0\;\;id)> K@>>>0
\end{CD}
$$
Note that the kernel of the middle map is the set of pairs $(m,n)$ such that $(f(m)+n,-g(n))=(0,0)$ which is naturally isomorphic to the kernel of $gf$ (so $E$). The same holds for the cokernel of the middle map which is isomorphic to the cokernel of $gf$ (that is $F$).

A word about this construction : the map of left is the mapping cone of $f$ (with $M$ in degree $-1$). The map of the right is the mapping cone of $-g$, or rather this is $Cone(g[-1])[1])$. There is a natural map
$\varphi:Cone(g[-1])\rightarrow Cone(f)$ given by the morphism of complexes :
$$ 
\begin{CD}
{}@. M\\
@.@VVfV\\
N@=N\\
@VgVV@.\\
K
\end{CD}
$$
The mapping cone of $\varphi$ the complex $M\oplus N\overset{\begin{pmatrix}f & id\\0 & -g \end{pmatrix}}\longrightarrow N\oplus K$. So the diagramm above is exactly the short exact sequence of complexes :
$$ 0\rightarrow Cone(f)\rightarrow Cone(\varphi)\rightarrow Cone(g[-1])[1]\rightarrow 0$$
and the six terms exact sequence you wrote is exactly the long exact sequence of cohomology of this short exact sequence.
A: I just found an easy proof in James Milne's class field theory notes. Apply the snake lemma to the diagram
$$\require{AMScd}
\begin{CD}
@.M@>>>N@>>>B@>>>0\\
@.@VVV@VVV@VVV\\
0@>>>K@>>>K@>>>0
\end{CD}
$$
and you get the long exact sequence
$$0\to\text{ker}(M\to N)\to\text{ker}(M\to K)\to\text{ker}(N\to K)\to\text{ker}(B\to 0)\to\text{coker}(M\to K)\to\text{coker}(N\to K)\to\text{coker}(B\to 0)$$
which can be rewritten as
$$0\to A\to E\to C\to B\to F\to D\to0.$$
