# 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.
• Are the rows are exact (and are we in a category where that makes sense)? – Arthur Nov 7 '17 at 6:55
• Yeah, I'm assuming I'm in an abelian category, and rows are exact. Similar setup as the regular four lemma. – Henry Swanson Nov 7 '17 at 6:56
• Cool. I just hadn't heard about the strong four lemma before, so I didn't know what the setup usually is. – Arthur Nov 7 '17 at 7:04
• Tbh I don't know how well known it is, I saw it on nlab and it seemed pretty :p – Henry Swanson Nov 7 '17 at 7:05

$\DeclareMathOperator{\im}{im} \DeclareMathOperator{\coker}{coker}$ 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.

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 $c\in \im g$, say $c=g(b)$. Now $g'\beta(b)=\gamma g(b)=\gamma(c)=0$, so $\beta(b)\in \ker g'=\im \alpha$. 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.

• Yeah, one can do the standard diagram chase. But one of the reasons I was motivated to look for a snake lemma proof was because I was trying to prove this in an arbitrary abelian category. (I don't want to just invoke Freyd-Mitchell either, because I don't know how to prove that.) – Henry Swanson Nov 7 '17 at 8:50

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)?

• Yeah, it's a restriction of $\gamma$, or equivalently, a projection of $\beta$ (the map $B \to B' \to B'/\ker g'$ factors through $B/\ker g$). The cokernel stuff is: a class in $\mathop{coker} \beta$ maps to zero iff its representative in $B'$ maps to something in $\mathop{im} \gamma$. So if it's not injective, there's some element outside $\mathop{im} \beta$ that maps to \mathop{im} \gamma\$, and vice versa. – Henry Swanson Nov 9 '17 at 16:28