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 bo 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?

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 $q$.