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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?

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Interesting question. Here is an idea (not yet a proof): Assume that pushouts exist in $C$. Write $m : H \to G$ as the kernel of $G \to G/H$ and $e : G \to G/K$ as the cokernel of $K \to G$. Now let $P$ be the pushout of $G/H \leftarrow G \to G/K$. Then the kernel $L$ of $G/K \to P$ has the property that $em$ factors as through $L$. Unfortunately, I cannot show that it is the image. However, remark that this is the case for the category of groups (and probably also abelian categories). – Martin Brandenburg Jan 6 at 20:30
@Martin The pushout $P$ can also be characterised as the cokernel of $e \circ m$, so your claim is true in an abelian category simply because the image of any morphism is the kernel of its cokernel. But then again in an abelian category any monomorphism is already normal... – Zhen Lin Jan 6 at 22:42

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up vote 1 down vote accepted

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:

        commutative diagram

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 $E \to C$.

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Why $P\to N$ is epic? – user54738 Jan 19 at 14:08
The square $PNBE$ is a pullback square, and the axioms for regular categories say that the pullback of a regular epi is regular epi. – Zhen Lin Jan 19 at 14:50
If $PNBE$ is a pull-back square, then $P\to E$ is the kernel of $E\to B\to D$, thus $E\to B\to D$ and $E\to C$ have same kernel? – user54738 Jan 19 at 15:15
Hm, no, that doesn't sound right. Anyway $PNBE$ is not a pullback square either, but $P \to N$ is an isomorphism because $EBDC$ is a pullback square. (Use the pullback pasting lemma and pullback cancellation lemma.) – Zhen Lin Jan 19 at 16:00
Ok, the pullback pasting lemma is enough, do you think? – user54738 Jan 19 at 23:54
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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].

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