1
$\begingroup$

Let's assume we have a category which has all kernels and cokernels. If needed we can also assume the category to abelian.

I am currently struggling to show the "first homomorphism theorem"/"fundamental theorem on homomorphisms" for such categories, i.e. let $\phi: A \to B$ be a morphism, then naturally $\phi$ factors through the $C=coimage(\phi)=coker(ker(\phi))$ and the corresponding morphism $\varphi: C\to B$ is epi if $\phi$ is epi.

The problem is now to show, that $\varphi$ is also mono.

Does anyone know if this holds (and how to show this)/what assumptions on the category we need e.g. preadditive. (for the standard examples of abelian categories i.e. rings, modules, vectorspaces, (abelian) groups this holds)

$\endgroup$
  • $\begingroup$ Try to show that the kernel of $\phi$ is $0$. $\endgroup$ – Max Jun 22 '17 at 14:57
  • 1
    $\begingroup$ Unless there is something I'm missing, this $\varphi$ is the kernel of the cokernel of $\phi$, so it is trivially a mono. $\endgroup$ – Arnaud D. Jun 22 '17 at 15:58
  • $\begingroup$ @ArnaudD. Ok, thanks, I worked it out. When I have time tomorrow evening I will post a proper diagramm as answer, unless you are faster and want the points ;-) $\endgroup$ – ctst Jun 22 '17 at 22:17
  • $\begingroup$ Ok, seems this is in general wrong, i.e. cokernels and kernels are not enough (don't have a proper counterexample yet, but I think we need abelian category). For abelian category this easily holds, as coimage and image are isomorphic. $\endgroup$ – ctst Jun 23 '17 at 8:36
  • $\begingroup$ @ctst You don't need an abelian category : this is true for groups, and also for rings (without unit), algebras and Lie algebras if you add the condition that $\phi$ is an extremal epimorphism. In fact we can prove this in homological (or even sequentiable) categories. $\endgroup$ – Arnaud D. Jun 27 '17 at 8:38

Your Answer

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

Browse other questions tagged or ask your own question.