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Let $H,K$ be two normal subgoups of a group $G$. We know that there exists a group isomorphism: $HK\diagup H\simeq H\diagup{H\cap K}$. I want to generalize this statement in the language of category theory.

Let $\mathscr C$ be a category with zero morphisms, kernels, cokernels, products, coproducts, equalizers and coequalizers; moreover, assume $\mathscr C$ to be conormal. Let $h,k$ be two kernels with same codomain, say $h:H\to G$ and $k:K\to G$. Let consider the morphism $a=h\text{ coker}(k)$. We know that there exists a unique morphism $u$ such that $a=\text{ coker}(\ker (a))u\ker(\text{coker}(a))$. Moreover, by conormality, we know that $\ker(u)=0$, thus $u$ is a monomorphism.

My question: Is $u$ an isomorphism?

In order to prove it, it's enoght to show that the image of $a$ is normal, that's a kernel. In particular, consider the following diagram: \begin{matrix} & \rightarrow & K & & \cr \downarrow & & \downarrow & & \cr H & \rightarrow & G & \rightarrow & G\diagup H \cr & & \downarrow & & \downarrow \cr & & G\diagup K & \rightarrow_q & \end{matrix} in which the upper square is a pullback while the other below is a pushfoward. Then it's enoght to show that the image of $a$ is the kernel of $q$.

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  • $\begingroup$ 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 $H$. $\endgroup$ – Mikael Vejdemo-Johansson Nov 12 '12 at 14:42
  • $\begingroup$ Yes, because $\text{coker}(k)$ has domain $G$. $\endgroup$ – Fabio Lucchini Nov 12 '12 at 15:23
  • $\begingroup$ 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? $\endgroup$ – Chris Heunen Nov 12 '12 at 18:30
  • $\begingroup$ 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? $\endgroup$ – Fabio Lucchini Nov 12 '12 at 18:57

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