kernel pairKernelPair2008-09-02 00:42:562008-10-08 01:48:11Definitioncokernel pair\usepackage{amssymb,amscd}
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\newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}}Let $f:A\to B$ be a morphism in a category $\mathcal{C}$. The \emph{kernel pair} of $f$ is defined as the pair of morphisms $(k_1: K\to A, k_2:K\to A)$ such that
$$\xymatrix@+=4pc{
{K}\ar[r]^{k_1}\ar[d]_{k_2} &{A}\ar[d]^{f} \\
{A}\ar[r]_{f}&{B}
}
$$
is a pullback diagram.
Since
$$\xymatrix@+=4pc{
{A}\ar[r]^{1_A}\ar[d]_{1_A} &{A}\ar[d]^{f} \\
{A}\ar[r]_{f}&{B}
}
$$
is a commutative diagram, we have a unique morphism $g:A\to K$ such that
$$\xymatrix@+=4pc{
A\ar@/^1ex/[rrd]^{1_A} \ar@/_1ex/[rdd]_{1_A} \ar[rd]^g & & \\
& K \ar[d]^{k_2} \ar[r]_{k_1} & A\ar[d]^f \\
& A\ar[r]_f & B.
}
$$
is commutative. As a result, $k_1$ and $k_2$ are both monomorphisms: if $k_1\circ h_1 = k_1\circ h_2$, then $$h_1 = 1_A \circ h_1 = (g\circ k_1) \circ h_1 = g\circ (k_1 \circ h_1) =g\circ (k_1 \circ h_2) = (g\circ k_1) \circ h_2 = 1_A \circ h_2 = h_2.$$
For example, in \textbf{Set}, the category of sets, the kernel pair of a function $f:A\to B$ is the pair $p_1:K\to A$ and $p_2:K\to A$, given by $$K=\lbrace (a,b) \in A\times A \mid f(a)=f(b) \rbrace,$$ and $p_1$ and $p_2$ are given by $$p_1(a,b)=a \qquad \mbox{and} \qquad p_2(a,b)=b.$$
This is just the kernel of a function, in the sense of universal algebra. Please see \PMlinkname{this entry}{KernelOfAHomomorphismBetweenAlgebraicSystems} for more details.
The notion of \emph{cokernel pair} is dually defined.
\textbf{Remark}. $f:A\to B$ is a monomorphism iff the kernel pair of $f$ is $(1_A,1_A)$. Dually, $f$ is an epimorphism iff the cokernel pair of $f$ is $(1_A,1_A)$.
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\bibitem{fb} F. Borceux \emph{Basic Category Theory, Handbook of Categorical Algebra I}, Cambridge University Press, Cambridge (1994)
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