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In a pointed category $\mathbf{C}$ with kernels and cokernels each morphism $f : A \to B$ factors as $f = ker(coker(f)) \circ f'$ with some morphism $f' : A \to K$. In abelian categories $f'$ is an epimorphism. This is also true in some nonabelian categories, for example in the category of pointed sets. The category of groups is an example where $f'$ is in general no epimorphism, but it is a homomorphism with $coker(f') = 0$.

In abelian categories and in the category of pointeds sets epimorphims coincide with morphisms having zero cokernel. The question is:

In the above factorization, is it always true that $coker(f') = 0$?

A counterexample would be welcome.

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I believe the following counterexample works.

Define $\mathcal{C}$ to be the category with $4$ objects $0,A,B,C$, where $0$ is the zero object, and with, in addition to the identities and zero arrows :

  • one arrow $f:A\to B$
  • one arrow $q:B\to C$
  • a collection of arrows $\beta_n:B\to B$ for $n\geq 1$.

The compositions between these arrows are given by $q\circ f=0=q\circ \beta_n$, $\beta_n\circ f=f$ and $\beta_n\circ\beta_m = \beta_{n+m}$ for all $n,m\geq 1$.

Then $q$ is the only non-zero arrow $g$ such that $g\circ f$ exists and is zero, and thus it must be its cokernel. Moreover, any arrow $h$ with codomain $B$ other than the identity $1_B$ is such that $q\circ h=0$, and also factors uniquely through $\beta_1$ (because $\beta_1$ is a monomorphism, and $\beta_n=\beta_1\circ\beta_{n-1}$ for all $n> 1$ and $f=\beta_1\circ f$). Thus $\beta_1$ is the kernel of $q$.

Thus in this case the factorisation of $f$ as $\ker(\operatorname{coker}(f))\circ f'$ is simply $f=\beta_1\circ f$. But $\operatorname{coker}(f)=q\neq 0$.

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  • $\begingroup$ Thank you very much! I checked that in fact all morphisms have kernels and cokernels. $\endgroup$
    – Paul Frost
    Nov 18, 2017 at 15:55
  • $\begingroup$ I have to correct my above claim about the category of groups. One does not always have $coker(f') = 0$ . $\endgroup$
    – Paul Frost
    Jan 5, 2018 at 13:39
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My claim concerning the category of groups was wrong. Here is a counterexample. Let $D_8$ be the dihedral group of order $8$ which has the presentation $\langle x, a \mid a^4 = x^2 = xax^{-1}a = e \rangle $. Let $K = \lbrace e, x, a^2, a^2x \rbrace$ which is one of the Klein groups of order $4$ sitting in $D_8$ and let $C = \lbrace e, x \rbrace$ which is a non-normal subgroup of $D_8$, but a normal subgroup of $K$. If $i : C \to K$, $j : C \to D_8$ and $k : K \to D_8$ denote inclusions, we get $ker(coker(j)) = k$. Therefore $j = ker(coker(j)) \circ i$, but $coker(i) \ne 0$ since $K/C \ne 0$.

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  • $\begingroup$ Please strive not to vote to reopen dupes of FAQs, cf. recent site policy announcement here. $\endgroup$ Jul 7, 2021 at 15:48
  • $\begingroup$ And you did so once again, why? $\endgroup$ Mar 15, 2023 at 21:42
  • $\begingroup$ @BillDubuque The OP did not ask for a proof, but wanted to have feedback concerning his understanding of a certain proof. I think this is a legitimate reason to reopen it. I absolutely understand if you do not share this point if view, but you will rarely find an issue where all people have the same opinion. $\endgroup$
    – Paul Frost
    Mar 15, 2023 at 23:18
  • $\begingroup$ Hmm, looking closer, the OP has significantly edited the question since it was closed (probably that's what you saw in review), but it will still end up being a dupe (we have many tens if not hundreds of answers on this basic topic). $\endgroup$ Mar 15, 2023 at 23:37
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    $\begingroup$ I appreciate your explanation of your thought processes. The reason I care so much is that the more we allow rampant duplication of FAQs like this, the more cluttered the search results are for students searching for answers, and the less hope they have of locating our "proofs from the book" (which hopefully our answers will evolve to after years of feedback). If worst comes to worst the site may devolve into an ephemeral stream of low-quality quick answers, instead of a library of polished answers (in which case all our prior work has been mostly for nought). $\endgroup$ Mar 15, 2023 at 23:55

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