3
$\begingroup$

In a complete, cocomplete and well-powered category with zero object consider the canonical factorization of a morphism $f=k\circ \mathrm{Coim}(f)$. Does cocomplete+complete+well-powered guarantee that $k$ is a monomorphism?

$\endgroup$
5
  • $\begingroup$ How are you defining $\operatorname{coim} f$ (or $k$)? Given only the data of a complete and well-powered category, the only reasonable definition is to take $k$ to be the intersection of all subobjects of the codomain of $f$ through which $f$ factors, in which case $k$ is necessarily a monomorphism. $\endgroup$
    – Zhen Lin
    Sep 23, 2013 at 8:35
  • 1
    $\begingroup$ @ZhenLin: what you are talking about is $\mathrm{im}(f)$, which is indeed always monic. $\mathrm{coim}(f)$ is defined as $\mathrm{coker}(\mathrm{ker}(f))$ and $k$ is obtained from the universal property of coker (which is a coequalizer). One always has a canonical map $\mathrm{coim}(f)\to\mathrm{im}(f)$, which in general is neither epic nor monic. However when the category is complete+wellpowered, it has (ExtrEpi, Mono) and (Epi, ExtrMono) factorization structures. So I would like to know if coim-factorization belongs to one of them. A good reference is AHS book "Abstract and concrete cat-s" $\endgroup$ Sep 23, 2013 at 10:29
  • $\begingroup$ Then you should specify that your category has coequalisers or cokernels. (Also, in my view, that construction is not the right definition of coimage. The coimage of $f$ is supposed to be the universal epimorphism through which $f$ factors.) $\endgroup$
    – Zhen Lin
    Sep 23, 2013 at 11:16
  • $\begingroup$ @ZhenLin: My fault, edited. $\endgroup$ Sep 23, 2013 at 11:41
  • $\begingroup$ Yes, how do you define coimage. Consider the dual situation. Would you need co-well-powered to define image? I don't think so. That would allow to define coimage. To define image, you want well-powered. And the map to the image is, in a complete category, epic. If not use the equalizer of two maps out of the coimage. As stated, I don't think there is a good answer $\endgroup$ Oct 14, 2013 at 18:44

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.