3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\operatorname{coker}(f)=q\neq 0$.

$\endgroup$
  • $\begingroup$ Thank you very much! I checked that in fact all morphisms have kernels and cokernels. $\endgroup$ – Paul Frost Nov 18 '17 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 '18 at 13:39
2
$\begingroup$

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 $K/C \ne 0$.

$\endgroup$

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.