PlanetMath: kernel of a morphism

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kernel of a morphism

(Definition)

Let $\mathcal{C}$ be a category with zero object . Given objects , 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 $\operatorname{Hom}(A,B)$ :

$\displaystyle \xymatrix@+=4pc{A\ar[r]^{o_{A,B}}&B}=\xymatrix@1{A\ar[r]&O\ar[r]&B}.$

This morphism is unique with respect to

and

, 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

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 and the corresponding zero morphism $o:A\to B$ . This means that, if $k:K\to A$ is the kernel of , then $f\circ k=o$ , and if $f\circ g=o$ for some morphism , there is a unique morphism such that $g=k\circ h$ . Diagrammatically, this means that

$\displaystyle \xymatrix@+=3pc{K\ar[r]^k & A\ar[r]^f &B}=0,$

and if

$\displaystyle \xymatrix@+=3pc{C\ar[r]^g & A\ar[r]^f &B}=0,$

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, is unique up to isomorphism, if it exists. We usually call the kernel of , and denote it by $\ker(f)$ , since is determined by , 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

is unique,

. $\qedsymbol$

Dually, we can define the cokernel of a morphism , $\operatorname{coker}(g)$ , to be the coequalizer of and . 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 is the kernel of , 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 to . 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 $% latex2html id marker 575 $ \textbf{Grp}$$ .
In addition, $% latex2html id marker 577 $ \textbf{Grp}$$ has cokernels. Let $\phi: G\to H$ be as above, and $C:=\phi(G)^H$ the normal closure of $\phi(G)$ in . Form 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 , 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 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.

"kernel of a morphism" is owned by CWoo.

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See Also: kernel is an inverse limit

Other names:

null morphism, has kernels, has cokernels, Ab1 axiom

Also defines:

kernel, cokernel, zero morphism, have kernels, have cokernels, Ab1

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Cross-references: product, finite, canonical projection, addition, well-defined, map, category theory, canonical injection, theory, range, identity, domain, mapping, group homomorphism, groups, epimorphism, coequalizer, monomorphism, isomorphism, universality, commutative diagram, equalizer, easy to see, subscript, isomorphic, composition, morphism, objects, zero object, category
There are 25 references to this entry.

This is version 19 of kernel of a morphism, born on 2008-08-29, modified 2008-09-05.
Object id is 10968, canonical name is KernelOfAMorphism.
Accessed 992 times total.

Classification:

AMS MSC:	18A20 (Category theory; homological algebra :: General theory of categories and functors :: Epimorphisms, monomorphisms, special classes of morphisms, null morphisms)
	18A30 (Category theory; homological algebra :: General theory of categories and functors :: Limits and colimits )

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