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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.

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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 $Y$.

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  • $\begingroup$ 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

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