Composition of a kernel with a cokernel has a normal image Let $C$ be a category with zero morphisms, kernels and cokernels.
Let $m$ be a kernel, $e$ be a cokernel and assume that is defined the composition $e\circ m$.
Assume that $e\circ m$ has image; it's a normal monomorphism?
The answer is always yes in category of groups: for let $H,K$ be two normal subgroup of group $G$;
let $m:H\to G$ be the inclusion and $e:G\to G/K$ be the projection; then the image of $e\circ m$ is $HK/K$ which is a normal subgroup of $G/K$.
The same holds in the category $C$ above?
 A: I asked a colleague and she came up with a different proof:

So if you have a normal subobject, it is the kernel of its cokernel. And you have another regular epi. If you take the pushout of these two regular epis,  you get what is called a regular pushout, a pushout square where all morphisms are regular epis. In certain contexts (e.g. semiabelian) this is then automatically a double extension (the thing I told you about: the comparison morphism to the pullback is also a regular epi). So if you take kernels "upwards", the comparison between the kernels is also a regular epi. This means that you do get an image factorisation of the composite of your original normal subobject and regular epi through that second kernel you've taken.

If I understand correctly, the diagram to keep in mind is the one shown below:
        
Here, $A \to B$ is the regular epimorphism we start with, $M \to A$ is a normal monomorphism, $A \to C$ is the cokernel of $M \to A$, $D$ is the pushout of $A \to B$ and $A \to C$, $E$ is the pullback of $B \to D$ and $C \to D$, $N \to B$ is the kernel of $B \to D$, and $P \to E$ is the kernel of $E \to C$.
A: Apparently this is true in any regular Mal'cev category (such as a semi-abelian category). The general case is Proposition 3.2.7 in [Borceux and Bourn], and the special case of a semi-abelian category is covered in Proposition 3.9 in [Borceux, A survey of semi-abelian categories].
