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I know that in a regular category, every morphism $f$ can be factored as $f=ip$, with $i$ a monomorphism and $p$ an epimorphism. In abelian category, the codomain of $p$=domain of $i$ is the cokernel of the kernel of $f$. Assuming the existence of kernels and cokernels, is this true also in regular categories? If yes, I would also be interested to a sketch of the proof.

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    $\begingroup$ I guess you mean "regular epimorphism" instead of just "epimorphism"? Otherwise, your question would not really make sense, since there can be more than one epi-mono in general. $\endgroup$ – Arnaud D. Oct 4 '17 at 15:45
  • $\begingroup$ yes , I meant regular. But if I have many, are they isomorphic, aren't they? $\endgroup$ – bateman Oct 4 '17 at 15:53
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    $\begingroup$ Those where $p$ is regular (or even strong), yes. But for example the injection $\Bbb Z\to \Bbb Q$ is both an epi and a mono in the category of rings, so it has at least two non-isomorphic factorization as an epi followed by a mono. $\endgroup$ – Arnaud D. Oct 4 '17 at 15:58
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It's not necessary true in any pointed regular category : for example, in the category of monoids, the addition map $$p:\mathbb{N}\times \mathbb{N}\to \mathbb{N}:(x,y)\mapsto x+y $$ is a regular epimorphism, since it has a section; but it is not the cokernel of its kernel, because said kernel is simply $0\to \mathbb{N}\times \mathbb{N}$, whose cokernel is the identity.

What is true, however, is that if $f=ip$ with $i:I\to Y$ mono and $p:X\to I$ a regular epi, then $p$ is the coequalizer of the kernel pair of $f$. Indeed, any epimorphism is the coequalizer of its kernel pair, and since for any two parallel arrows $u,v:W\to X$ we have $fu=fv$ if and only if $pu=pv$, $p$ and $f$ have the same kernel pair.

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  • $\begingroup$ I was convinced that in ANY category (with kernels and cokernels), every cokernel is the cokernel of its kernel, as proved here: math.stackexchange.com/questions/2185870/… in the dual case of kernels. What am I missing? $\endgroup$ – bateman Oct 4 '17 at 16:06
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    $\begingroup$ It's certainly true that any cokernel is the cokernel of its kernel. The map $p$ is just not a cokernel. $\endgroup$ – Arnaud D. Oct 4 '17 at 16:08

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