Proving the (Strong) Four Lemma using the Snake Lemma $\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\coker}{coker} \require{AMScd}$
The usual formulation of the Strong Four Lemma is: given the diagram below, if the rows are exact, $\alpha$ is epic, and $\delta$ is monic, then $g(\ker \beta) = \ker \gamma$ and $\im \beta = g'^{-1}(\im \gamma)$:
\begin{CD}
A @>{f}>> B @>{g}>> C @>{h}>> D \\
@VV{\alpha}V @VV{\beta}V @VV{\gamma}V @VV{\delta}V \\
A' @>{f'}>> B' @>{g'}>> C' @>{h'}>> D' \\
\end{CD}
This is equivalent to saying that the induced maps $\ker \beta \to \ker \gamma$ and $\coker \beta \to \coker \gamma$ are epic and monic, respectively.
Upon drawing the kernels and cokernels of the sequences, it seems like this could be interpreted as a consequence of the Snake Lemma.
\begin{CD}
@. \ker \beta @>>> \ker \gamma @>>> 0 @. \textrm{(complex)} \\
@.@VVV @VVV @VVV \\
A @>{f}>> B @>{g}>> C @>{h}>> D @. \textrm{(exact)} \\
@VV{\alpha}V @VV{\beta}V @VV{\gamma}V @VV{\delta}V \\
A' @>{f'}>> B' @>{g'}>> C' @>{h'}>> D' @. \textrm{(exact)} \\
@VVV @VVV @VVV @. \\
0 @>>> \coker \beta @>>> \coker \gamma @. @. \textrm{(complex)} \\
\end{CD}
Is this intuition misguided, or have I simply not found the right arrows?
Some thoughts:


*

*The element-chasing proof dances through all eight objects, so I suspect the diagrammatic proof must use them all as well. So it's not just a result of some "smaller" lemma, like the left-exactness of kernels.

*I could replace $A'$ with $A$, and $D$ with $D'$, without changing the exactness of the rows. Maybe this makes things clearer, maybe it doesn't.

*This lemma is all about the central square $(g, g', \beta, \gamma)$. The functions $f$ and $h$ really don't seem to matter very much; it's not hard to cook up $f$ and $h$ making the sequence exact. But this result isn't true for an arbitrary commutative square, so they must play some important role, though it's not obvious what it is.

*EDIT: This theorem is equivalent to the special case where $A = \ker g$, $A' = \ker g'$, $D = \coker g$, and $D' = \coker g'$. This explains why I had trouble figuring out the significance of $f$ and $h$ earlier.

 A: $\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\coker}{coker}$
Two long bus rides later, I've got it! Posting it here so others may read.

First, we make the diagram more snake-ready: replace the lower-left corner by $\ker g'$, and the upper-left by $\coker g$:
\begin{CD}
@. A @>f>> B @>g>> C @>>> \coker g @>>> 0 \\
@. @VVpV @VV{\beta}V @VV{\gamma}V @VViV @. \\
0 @>>>\ker g' @>>> B' @>g'>> C' @>h'>> D' @. \\
\end{CD}
Note that $p$ is epi, because it's the composite of two epi maps, $A \to A'$ and $A' \to \im f' = \ker g'$. Likewise, $i$ is mono.
We can't apply the snake lemma to a four-term sequence. But where one snake fails, two may do. We can break the exact sequences in half, forming two smaller diagrams:
\begin{CD}
@. A @>>> B @>>> \im g @>>> 0 \\
@. @VVpV @VV{\beta}V @VV{\varphi}V @. \\
0 @>>>\ker g' @>>> B' @>>> \im g' @>>> 0 \\
\end{CD}
and
\begin{CD}
0 @>>> \im g @>>> C @>>> \coker g @>>> 0 \\
@. @VV{\varphi}V @VV{\gamma}V @VViV @. \\
0 @>>> \im g' @>>> C' @>>> D' @. \\
\end{CD}
The snake lemma gives us two exact sequences:
$$ \ker p \to \ker \beta \to \ker \varphi \to 0 \to \coker \beta \to \coker \varphi \to 0 $$
$$ 0 \to \ker \varphi \to \ker \gamma \to 0 \to \coker \varphi \to \coker \gamma \to \coker i $$
Thus, $\ker \beta \twoheadrightarrow \ker \gamma$ and $\coker \beta \hookrightarrow \coker \gamma$, as desired.

I tried extending this approach to longer sequences, and it's got at least one interesting consequence.
Consider the following diagram, with $\alpha$ epi and $\epsilon$ mono:
\begin{CD}
A @>f>> B @>g>> C @>h>> D @>k>> E \\
@VV{\alpha}V @VV{\beta}V @VV{\gamma}V @VV{\delta}V @VV{\epsilon}V \\
A' @>f'>> B' @>g'>> C' @>h'>> D' @>k'>> E' \\
\end{CD}
Just as before, we break the sequence into smaller chunks, getting three diagrams this time. (Also, we can do the same trick with the (co)kernels in the (co?)corners.)
\begin{CD}
@. A @>>> B @>>> \im g @>>> 0 \\
@. @VVpV @VV{\beta}V @VV{\varphi}V @. \\
0 @>>>\ker g' @>>> B' @>>> \im g' @>>> 0 \\
\end{CD}
and
\begin{CD}
0 @>>> \im g @>>> C @>>> \im h @>>> 0 \\
@. @VV{\varphi}V @VV{\gamma}V @VV{\psi}V @. \\
0 @>>> \im g' @>>> C' @>>> \im h @>>> 0 \\
\end{CD}
and
\begin{CD}
0 @>>> \im h @>>> D @>>> \coker h @>>> 0 \\
@. @VV{\psi}V @VV{\delta}V @VViV @. \\
0 @>>> \im h' @>>> D' @>>> E' @. \\
\end{CD}
Three diagrams means three exact sequences:
$$ \ker p \to \ker \beta \to \ker \varphi \to 0 \to \coker \beta \to \coker \varphi \to 0 $$
$$ 0 \to \ker \varphi \to \ker \gamma \to \ker \psi \to \coker \varphi \to \coker \gamma \to \coker \psi \to 0 $$
$$ 0 \to \ker \psi \to \ker \delta \to 0 \to \coker \psi \to \coker \delta \to \coker i $$
All together, this gives us one long snake-lemma-like exact sequence!
$$ \ker p \to \ker \beta \to \ker \gamma \to \ker \delta \to \coker \beta \to \coker \gamma \to \coker \delta \to \coker i $$

Unfortunately, this approach fails with any longer sequences, as far as I can tell.
A: For convenience, I suppose that we are working in the category of some $R$-modules so that we are free to talk about the "elements" in $A,B,$ etc. The proof is nothing more than a standard diagram-chasing,  
For example, let's show the map $\ker \beta\to \ker \gamma$ is epic. Assume that $c$ is an element in $\ker \gamma$. Now $\delta h(c)=h'\gamma(c)=0$, so $h(c)\in \ker \delta$.But $\delta$ is monic, so $h(c)=0$ and hence $c\in \operatorname{im} g$, say $c=g(b)$. Now $g'\beta(b)=\gamma g(b)=\gamma(c)=0$, so $\beta(b)\in \ker g'=\operatorname{im} \alpha$. Moreover, since $\alpha$ is epic, one can find $a\in A$ such that $f'\alpha(a)=\beta(b)$, so $\beta(f(a)-b)=f'\alpha(a)-\beta(b)=0$, which means $f(a)-b\in\ker \beta$. However, we do also have $g(b-f(a))=g(b)=c$, hence the claim follows.
For the dual statement, I guess the argument is quite similar, or you can just apply the duality principle to save time.
A: the map between im(g) and im(g') is a restriction of gamma to im(g) or am i mistaken?
Also, how did you conclude that coker(beta) can be embedded into coker(gamma)?
