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kernel of a morphism
Let $\mathcal{C}$ be a category with zero object $O$. Given objects $A,B$, we can define a zero morphism, or null morphism to be the morphism $o_{{A,B}}$ of the composition of the two unique morphisms $A\to O$ and $O\to B$ in $\operatorname{Hom}(A,B)$:
$\xymatrix@+=4pc{A\ar[r]^{{o_{{A,B}}}}&B}=\xymatrix@1{A\ar[r]&O\ar[r]&B}.$ 
This morphism is unique with respect to $A$ and $B$, because $A\to O$ and $O\to B$ are both unique, and any two zero objects are isomorphic. Instead of writing $o_{{A,B}}$, we may drop the subscript and write $o$ if there is no confusion. It is easy to see that a composition of a morphism with a zero morphism is a zero morphism.
With the zero morphism, we define the kernel of a morphism $f:A\to B$ to be the equalizer of $f$ and the corresponding zero morphism $o:A\to B$. This means that, if $k:K\to A$ is the kernel of $f$, then $f\circ k=o$, and if $f\circ g=o$ for some morphism $g$, there is a unique morphism $h$ such that $g=k\circ h$. Diagrammatically, this means that
$\xymatrix@+=3pc{K\ar[r]^{k}&A\ar[r]^{f}&B}=0,$ 
and if
$\xymatrix@+=3pc{C\ar[r]^{g}&A\ar[r]^{f}&B}=0,$ 
then there is a unique $h:C\to K$ such that
$\xymatrix@R=1pc{C\ar[dr]^{g}\ar[dd]_{h}\\ &A\\ K\ar[ur]_{k}}$
is a commutative diagram.
By the universality of equalizers, $(K,k)$ is unique up to isomorphism, if it exists. We usually call $K$ the kernel of $f$, and denote it by $\ker(f)$, since $k$ is determined by $K$, up to isomorphism. A kernel is the kernel of some morphism. A kernel is always a monomorphism:
Proof.
Suppose $k:K\to A$ is the kernel of $f:A\to B$, and $g,h:D\to K$ are morphisms such that $r:=k\circ g=k\circ h:D\to A$. Then $f\circ r=f\circ(k\circ g)=(f\circ k)\circ h=0$, so that there is a unique morphism $s:D\to K$ such that $k\circ s=r$. But then $k\circ s=k\circ g=k\circ h$ also. Since $s$ is unique, $g=h$. ∎
Dually, we can define the cokernel of a morphism $g$, $\operatorname{coker}(g)$, to be the coequalizer of $g$ and $o$. A cokernel is the cokernel of a morphism. A cokernel is always an epimorphism.
Remark. A category with zero object is said to have kernels if every morphism has a kernel. Dually, it is said to have cokernels if every morphism has a cokernel.
Examples.

The category Grp of groups has kernels. For any group homomorphism $\phi:G\to H$, let $K=\ker(\phi)$, the kernel of $\phi$ (in the sense of group theory), and $k:K\to G$ be the canonical injection. We shall see presently that $K$ is the kernel of $f$, in the sense of category theory. First, $\phi\circ k(a)=\phi(a)=e$ for every $a\in K=\ker(\phi)$. Let $\sigma:S\to G$ be another group homomorphism with $\phi\circ\sigma=1$, the trivial map from $S$ to $H$. Define $\psi:S\to K$ by $\psi(s)=\sigma(s)$, which is a welldefined, because $\phi(\sigma(s))=e$, or $\sigma(s)\in\ker(\phi)=K$. Furthermore, $\sigma(s)=\psi(s)=k(\psi(s))$. It is easy to see that $\psi$ is unique. Thus, $\ker(\phi)$ is the kernel of $\phi:G\to H$ in Grp.

In addition, Grp has cokernels. Let $\phi:G\to H$ be as above, and $C:=\phi(G)^{H}$ the normal closure of $\phi(G)$ in $H$. Form $Q=H/C$ and let $p:H\to Q$ be the canonical projection. For any $a\in G$, we have $p\circ\phi(a)=\phi(a)C=C$, since $\phi(a)\in\phi(G)\subseteq C$. Let $q:H\to T$ be a group homomorphism with $q\circ\phi=1$. Define $\mu:Q\to T$ by $\mu(bC)=q(b)$. This is welldefined, for if $cC=bC$, then $cb^{{1}}\in C$, which means that $cb^{{1}}$ is a finite product of elements of the form $d\phi(a)d^{{1}}$ where $a\in G$ and $d\in H$. Since $q(d\phi(a)d^{{1}})=q(d)q(d^{{1}})=e\in T$, we have that $q(c)q(b)^{{1}}=e$. Again, $\mu$ is easily seen to be unique. This shows that $H/C$ is the cokernel of $\phi:G\to H$ in Grp.
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