Proving the (Strong) Four Lemma using the Snake Lemma

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

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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.

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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! $$ \ker p \to \ker \beta \to \ker \gamma \to \ker \delta \to \coker \beta \to \coker \gamma \to \coker \delta \to \coker i $$

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Unfortunately, this approach fails with any longer sequences, as far as I can tell.