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

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 the trivial group as the zero object, and any trivial group homomorphism, mapping every element in the domain to the identity in the range, as a zero morphism.
• 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 well-defined, 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 well-defined, 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.

Remark. A category is said to be Ab1, a la Grothendieck, if it satisifes the Ab1 Axiom: it has kernels and cokernels. From the examples above, Grp is Ab1.