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| Title of object: |
kernel pair |
| Canonical Name: |
KernelPair |
| Type: |
Definition |
| Created on: |
2008-09-02 00:42:56 |
| Modified on: |
2008-09-02 00:42:56 |
| Classification: |
msc:18A30 |
| Defines: |
cokernel pair |
Preamble:
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Content:
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 $$h_1 = 1_A \circ h_1 = (g\circ k_1) \circ h_1 = g\circ (k_1 \circ h_1) =g\circ (k_1 \circ h_1) = (g\circ k_1) \circ h_2 = 1_A \circ h_2 = h_2.$$
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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