snake lemma, proof ofProofOfSnakeLemma2004-02-14 12:03:472004-02-14 12:03:47Proofsnake lemma0exact sequencekernel-cokernel sequence% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\DeclareMathOperator{\coker}{coker}
\DeclareMathOperator{\im}{im}\PMlinkescapeword{commutative}
\PMlinkescapeword{commutativity}
\PMlinkescapeword{sequence}
\PMlinkescapeword{sequences}
\PMlinkescapeword{exact}
\PMlinkescapeword{row}
\PMlinkescapeword{square}
\PMlinkescapeword{presentation}
\PMlinkescapeword{lemma}
%\begin{proof}
Suppose we are given a commutative diagram
\[\xymatrix{
0\ar[r] &
A_1\ar[d]_{\alpha}\ar[r] &
B_1\ar[d]_{\beta}\ar[r] &
C_1\ar[d]_{\gamma}\ar[r] &
0 \\
0\ar[r] &
A_2\ar[r] &
B_2\ar[r] &
C_2\ar[r] &
0 \\
}\]
with \PMlinkname{exact rows}{ExactSequence2}.
We wish to prove that the sequence
\[0\to\ker\alpha\to\ker\beta\to\ker\gamma\to
\coker\alpha\to\coker\beta\to\coker\gamma\to 0\]
is exact.\footnote{This proof was reconstructed without any notes, but the style of the proof is influenced by a presentation by Edgar Enochs of the zig-zag lemma.}
First we claim that if any square
\[\xymatrix{
X_1\ar[d]_{\varphi}\ar[r] & Y_1\ar[d]_{\psi} \\
X_2\ar[r] & Y_2
}\]
is commutative, then there are well-defined morphisms
$\ker\varphi\to\ker\psi$ and $\coker\varphi\to\coker\psi$.
For example, if $x\in\ker\varphi$, then the square
\[\xymatrix{
x\ar@{|->}[d]_{\varphi}\ar@{|->}[r] & y\ar@{|->}[d]_{\psi} \\
0\ar@{|->}[r] & 0
}\]
must commute, and so the image of $x$ in the top row must be
in $\ker\psi$. The proof of the claim for cokernels is similar.
Thus we have two sequences,
\[0\to\ker\alpha\to\ker\beta\to\ker\gamma\text{\ and\ }
\coker\alpha\to\coker\beta\to\coker\gamma\to 0,\]
each of which inherits being a complex from the original diagram.
Suppose $x\in\ker\beta$ is sent to $0\in\ker\gamma$. By exactness,
$x$ has a preimage $x'\in A_1$. Because the diagram
\[\xymatrix{
x'\ar@{|->}[d]_{\alpha}\ar@{|->}[r] & x\ar@{|->}[d]_{\beta} \\
y'\ar@{|->}[r] & 0
}\]
is commutative and the bottom morphism is injective, $y'=0$ and so
$x'\in\ker\alpha$. So the sequence
\[0\to\ker\alpha\to\ker\beta\to\ker\gamma\]
is exact. The proof of the claim for the cokernel sequence is
similar.
So now all we need to do is find a connecting morphism
$\ker\gamma\to\coker\alpha$ such that the resulting sequence
is exact at both of those points.
Suppose $x''\in\ker\gamma$. Then $x''$ has at least one preimage in $B_1$.
So let $x$ and $\widehat{x}$ be preimages of $x''$.
Thus $\widehat{x}-x\mapsto 0$ and so by exactness has a
preimage $x'\in A_1$.
By commutativity of the diagram, $\beta(x)$ has a preimage $y'$, which is
unique by injectivity of the morphism $A_2\to B_2$. But we know
that the square
\[\xymatrix{
x'\ar@{|->}[d]_{\alpha}\ar@{|->}[r] & \widehat{x}-x\ar@{|->}[d]_{\beta} \\
\alpha(x')\ar@{|->}[r] & \beta(\widehat{x}-x)
}\]
is commutative.
We wish to define $\ker\gamma\to\coker\alpha$ by
$x''\mapsto y'+\im\alpha$.
Observe that
\[y'+\alpha(x')\mapsto\beta(x)+\beta(\widehat{x}-x)=\beta(\widehat{x}),\]
and so the choice of preimage of $x''$ does not affect which cokernel
element we ultimately select. So now we have our connecting
morphism. By applying this definition we see that
\[\ker\beta\to\ker\gamma\to\coker\alpha\to\coker\beta\]
is a complex.
Suppose $x''\in\ker\gamma$ is sent to $0$ by the connecting morphism.
Thus we have a diagram
\[\xymatrix{
x'\ar@{|->}[d]_{\alpha} & x\ar@{|->}[d]_{\beta}\ar@{|->}[r] & x''\ar@{|->}[d]_{\gamma} \\
y'\ar@{|->}[r] & \beta(x)\ar@{|->}[r] & 0
}\]
which is commutative. Let $\widehat{x}$ be the image of $x'$ under the
morphism $A_1\to B_1$. Exactness of the diagram implies that $x-\widehat{x}$
is a preimage of $x''$. But $\beta(x-\widehat{x})=0$. So the kernel-cokernel
sequence is exact at $\ker\gamma$. The proof that it is exact at
$\coker\alpha$ is similar.\qedhere
%\end{proof}