Coim factorization of a morphism in a complete well-powered category

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?

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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. – Zhen Lin Sep 23 '13 at 8:35
@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" – Anton Lyubinin Sep 23 '13 at 10:29
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.) – Zhen Lin Sep 23 '13 at 11:16
@ZhenLin: My fault, edited. – Anton Lyubinin Sep 23 '13 at 11:41
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 – Michael Barr Oct 14 '13 at 18:44