Timeline for A generalization of a group isomorphism.

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Nov 12, 2012 at 18:57	comment	added	Fabio Lucchini		Yes. Indeed it's suffice to show that the image of a (which is a monomorphism) is a kernel. This is true in group category because a is the composition of a kernel with a cokernel. So, in general, the question can be stated as follows: ker(h′) coker(k) as a normal image?
Nov 12, 2012 at 18:52	history	edited	Fabio Lucchini	CC BY-SA 3.0	added 671 characters in body
Nov 12, 2012 at 18:30	comment	added	Chris Heunen		For abelian categories, this is known, right? (See e.g. 2.67 in Freyd's book "Abelian categories".) Are you asking if it still holds without assuming that every monomorphism is a kernel?
Nov 12, 2012 at 15:23	comment	added	Fabio Lucchini		Yes, because $\text{coker}(k)$ has domain $G$.
Nov 12, 2012 at 14:42	comment	added	Mikael Vejdemo-Johansson		Does $a=h\text{coker}(k)$ even exist? It seems to me that $\text{coker}(k)$ should be a subobject of $G$, while $h$ takes input from $H$.
Nov 12, 2012 at 14:05	history	asked	Fabio Lucchini	CC BY-SA 3.0