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\title
{The Snake Diagram}

\author[Starr]{Jason Michael Starr}
\address{Department of Mathematics \\
  Stony Brook University \\ Stony Brook, NY 11794}
\email{jstarr@math.sunysb.edu} 



\date{\today}

\begin{document}

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\begin{abstract}
This is a test.  This is only a test.  Do not attempt to adjust your
set.  Do not contact emergency personnel.
\end{abstract}

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\section{Statement of the lemma} \label{sec-intro}
\marpar{sec-intro}

Let $\mc{C}$ be an Abelian category.  In particular, every image in
$\mc{C}$ equals the coimage.  Thus we make no distinction between
images and coimages in what follows.  One of the fundamental notions
of homological algebra is the following.

\begin{defn} \label{defn-ses}
\marpar{defn-ses}
A \emph{short exact sequence} 
$$
\begin{CD}
\Sigma_A: \ \ @.  0 @>>> A' @> q_A >> A @> p_A >> A'' @>>> 0
\end{CD}
$$
is a pair of morphisms in $\mc{C}$
$$
\Sigma_A = (q_A:A'\to A,p_A:A \to A)
$$
such that all of the following hold: 
\begin{enumerate}
\item[(i)]
$q_A$ is a monomorphism,
\item[(ii)]
$p_A$ is an epimorphism, and
\item[(iii)]
the image of $q_A$ equals the kernel of $p_A$.
\end{enumerate}
\end{defn}

There is a category whose objects are short exact sequences in
$\mc{C}$.  Here is the notion of morphism in this category.

\begin{defn} \label{defn-sesmor}
\marpar{defn-sesmor}
Let $\Sigma_A=(q_A,p_A)$ and $\Sigma_B=(q_B,p_B)$ be short exact
sequences in $\mc{C}$.  A 
\emph{morphism} $\Sigma_f$ from $\Sigma_A$ to $\Sigma_B$,
$$
\begin{CD}
\Sigma_A: \ \ @.  0 @>>> A' @> q_A >> A @> p_A >> A'' @>>> 0 \\
@V \Sigma_f VV @. @V f' VV @VV f V @VV f'' V \\
\Sigma_B: \ \ @.  0 @>>> B' @>> q_B > B @>> p_B > B'' @>>> 0
\end{CD},
$$
is a triple of morphisms in $\mc{C}$ 
$$
\Sigma_f = (f':A'\to B', f:A\to B, f'':A''\to B'')
$$
such that every square commutes, i.e., both of the following hold:
\begin{enumerate}
\item[(i)]
$q_B\circ f'$ equals $f\circ q_A$, and
\item[(ii)]
$p_B\circ f$ equals $f''\circ p_A$.
\end{enumerate}
\end{defn}

In the category of short exact sequences the identity morphisms and
the compositions are the obvious notions.  The category of short exact
sequences is an additive category.  

Let $\Sigma_f$ be a morphism of short exact sequences as above.
Denote the kernels of $f'$, respectively $f$, $f''$ by,
$$
i':K_{\Sigma_f}' \to A', \ \text{resp.} \ \ i:K_{\Sigma_f} \to A, \
i'':K_{\Sigma_f}'' \to A''.
$$
Similarly, denote the cokernels of $f'$, respectively $f'$, $f''$ by,
$$
s':B' \to C_{\Sigma_f}', \ \text{resp.} \ \ s:B\to C_{\Sigma_f},
\ s'':B'' \to C_{\Sigma_f}''.
$$
Because $q_B\circ f'$ equals $f\circ q_A$, also $f\circ (q_A\circ i')$ equals
$q_B\circ (f'\circ i')$, which equals $q_B\circ 0 = 0$.  Thus, by the
universal property of the kernel, there is a unique morphism 
$$
q_K:K_{\Sigma_f}' \to K_{\Sigma_f}
$$
such that $i\circ q_K$ equals $q_A\circ i'$.  For a similar reason,
there is a unique morphism
$$
p_K:K_{\Sigma_f} \to K_{\Sigma_f}''
$$
such that $i''\circ p_K$ equals $p_A\circ i$.  And by analogous
arguments there are unique morphisms
$$
q_C:C_{\Sigma_f}' \to C_{\Sigma_f}, \ \ p_C:C_{\Sigma_f} \to C_{\Sigma_f}''
$$
such that $q_C\circ s'$ equals $s\circ q_B$, and $p_C\circ s$ equals
$s''\circ p_B$.  To summarize, we have that the following diagram is
commutative.
$$
\begin{CD}
@. @. K_{\Sigma_f}' @> q_K >> K_{\Sigma_f} @> p_K >> K_{\Sigma_f}'' \\
@. @. @V i' VV @VV i V @VV i'' V \\
\Sigma_A: \ \ @.  0 @>>> A' @> q_A >> A @> p_A >> A'' @>>> 0 \\
@V \Sigma_f VV @. @V f' VV @VV f V @VV f'' V \\
\Sigma_B: \ \ @.  0 @>>> B' @>> q_B > B @>> p_B > B'' @>>> 0 \\
@. @. @V s' VV @VV s V @VV s'' V \\
@. @. C_{\Sigma_f}' @>> q_C > C_{\Sigma_f} @>> p_C > C_{\Sigma_f}'' \\
\end{CD}
$$

By hypothesis, both $f''\circ p_A$ and $p_B\circ f$ are equal.  Denote
by $t$ this common morphism
$$
t:A \to B''.
$$
Denote the kernel of $t$ by 
$$
j:K_t \to A.
$$
Now $f''\circ(p_A\circ j)$ equals $t\circ j$, which is $0$.  By the
universal property of the kernel of $f''$, there is a unique morphism
$$
\widetilde{p_A}:K_t \to K_{\Sigma_f}''
$$
such that $i''\circ \widetilde{p_A}$ equals $p_A\circ j$.  Similarly,
$p_B\circ (f\circ j)$ equals $t\circ j$, which is $0$.  By the
universal property of the kernel of $p_B$, there is q unique morphism
$$
\widetilde{f}:K_t \to B'
$$
such that $q_B\circ \widetilde{f}$ equals $f\circ j$.  

\begin{lem}[The Snake Lemma] \label{lem-snake}
\marpar{lem-snake}
For a morphism $\Sigma_f$ of commutative diagrams as above, all of the
following hold.
\begin{enumerate}
\item[(i)]
The morphism $q_K$ is a monomorphism, and the morphism $p_C$ is an epimorphism.
\item[(ii)]
The image of $q_K$ equals the kernel of $p_K$, and the kernel of $p_C$
equals the image of $q_C$.
\item[(iii)]
There is a unique morphism $\delta_{\Sigma_f}:K_{\Sigma_f}'' \to
C_{\Sigma_f}'$ such that $\delta_{\Sigma_f} \circ \widetilde{p_A}$
equals $s'\circ\widetilde{f}$ as morphisms $K_t\to C_{\Sigma_f}'$.  
\item[(iv)]
The image of $p_K$ equals the kernel of $\delta_{\Sigma_f}$, and the
kernel of $q_C$ equals the image of $\delta_{\Sigma_f}$.
\end{enumerate}
In summary, the following long sequence is exact,
$$
\begin{CD}
0 @>>> K_{\Sigma_f}' @> q_K >> K_{\Sigma_f} @> p_K >> K_{\Sigma_f}''
@> \delta_{\Sigma_f} >> \dots \\
\\
\dots @>> \delta_{\Sigma_f} > C_{\Sigma_f}' @>> q_C > C_{\Sigma_f} @>>
p_C > C_{\Sigma_f}'' @>>> 0.
\end{CD}
$$
\end{lem}

This entire situation is often summarized with the following large
diagram.
$$
\begin{CD}
@. @. 0 @. 0 @. 0 \\
@. @. @VVV @VVV @VVV \\
@. 0 @>>> K_{\Sigma_f}' @> q_K >> K_{\Sigma_f} @> p_K >>
K_{\Sigma_f}'' @> \delta_{\Sigma_f} >> \dots \\
@. @. @V i' VV @VV i V @VV i'' V \\
\Sigma_A: \ \ @.  0 @>>> A' @> q_A >> A @> p_A >> A'' @>>> 0 \\
@V \Sigma_f VV @. @V f' VV @VV f V @VV f'' V \\
\Sigma_B: \ \ @.  0 @>>> B' @>> q_B > B @>> p_B > B'' @>>> 0 \\
@. @. @V s' VV @VV s V @VV s'' V \\
@. \dots @>> \delta_{\Sigma_f} >  
C_{\Sigma_f}' @>> q_C > C_{\Sigma_f} @>> p_C > C_{\Sigma_f}''
@>>> 0 \\
@. @. @VVV @VVV @VVV \\
@. @. 0 @. 0 @. 0
\end{CD}
$$


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