I use the definition of abelian category by Stenstrom (Rings of quotients):

A category $C$ is abelian if:

1) $C$ is preadditive

2) Every finite family of objects has a product and coproduct

3) Every morphism has a kernel and a cokernel

4) $\bar f:\text{Coker}(\ker f)\to \ker(\text{coker} f)$ is an isomorphism for every morphism $f$

The fourth axiom can be replaced by the following axiom: $f=gh$ where $g$ is a kernel and $h$ a cokernel.

Prove that if $f:A\to B$ is a monomorphism, then $f=\ker(\text{coker} f)$

Now what I want to prove is just that $f$ is a kernel of some morphism because after that I have a result that guarantees the thesis.

The point is that I know that $\ker f=0$ and then $\text{coker}(\ker f)=A$ is isomorphic as object to $\ker(\text{coker} f)$ but I cannot see how it implies that the map $f$ satisfies the two axioms of kernel.

For me a kernel is a pair $(X,g)$ such that $g:X\to Y$ and in this case I want to prove that $(A,f)$ is a kernel.


As you said yourself, $f:A\to B$ can be written as $gh$ with $g:K\to B$ a kernel (of $p:B\to C$) and $h:A\to K$ a cokernel. Now if $f$ is a mono, so is $h$; and if $h$ is a mono and a cokernel, it has to be an isomorphism.

This implies that $f$ is the same as $g$ up to an isomorphism; since the definition of kernel is a universal property, this is enough to show that $f$ is a kernel of $p$. Indeed, consider $v:D\to B$ be such that $pv=0$. Since $g$ is the kernel of $p$, there must be a unique arrow $t:D\to K$ such that $v=gt$. Now if you define $t'=h^{-1}t$, then $ft'=ght'=ghh^{-1}t=gt=v$; and this is the only possible arrow with that property, since if $t''$ is such that $ft''=v$, then $ght''=ft''=v=gt$, hence $ht''=t$ (since $g$ is a kernel) and thus $t''=h^{-1}t=t'$.

The point here is that whenever you have a universal property for arrows to/from a certain object $X$ in a category, then any object $Y$ isomorphic to $X$ must satisfy the same universal property, since the isomorphism establishes a natural bijection between arrows to/from $X$ and arrows to/from $Y$.

  • $\begingroup$ Thank you for your help but the point is really the second part of your answer: it is not clear to me that $h$ isomorphism implies $f$ is a kernel. Could you show it with more details ? $\endgroup$ – Richard Apr 13 '17 at 11:26
  • $\begingroup$ Because say $g$ is a kernel of $p$. Of course we have again that $pf=pgh=0$ and so the first axiom of kernels is satisfied but I can't prove the second axiom actually. (By the second axiom I mean the universal property) $\endgroup$ – Richard Apr 13 '17 at 11:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.