3
$\begingroup$

$\DeclareMathOperator{\im}{im} \DeclareMathOperator{\coker}{coker} \require{AMScd}$ 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.
$\endgroup$
  • $\begingroup$ Are the rows are exact (and are we in a category where that makes sense)? $\endgroup$ – Arthur Nov 7 '17 at 6:55
  • $\begingroup$ Yeah, I'm assuming I'm in an abelian category, and rows are exact. Similar setup as the regular four lemma. $\endgroup$ – Henry Swanson Nov 7 '17 at 6:56
  • $\begingroup$ Cool. I just hadn't heard about the strong four lemma before, so I didn't know what the setup usually is. $\endgroup$ – Arthur Nov 7 '17 at 7:04
  • $\begingroup$ Tbh I don't know how well known it is, I saw it on nlab and it seemed pretty :p $\endgroup$ – Henry Swanson Nov 7 '17 at 7:05
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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.) $\endgroup$ – Henry Swanson Nov 7 '17 at 8:50
0
$\begingroup$

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)?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Henry Swanson Nov 9 '17 at 16:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.