# Kernel of Zero Morphism in Abelian Categories

Consider an abelian category $C$. Let $f:M \rightarrow N$ be a zero morphism, i.e. the zero element of the abelian group $Mor_C(M,N)$. What is the kernel of $f$? Applying the definition, i get that it must be a morphism $p: P \rightarrow M$, such that whenever we have a morphism $q:Q \rightarrow M$, then there exists unique morphism $\bar{q}: Q \rightarrow P$ such that $q = p \circ \bar{q}$. So the kernel is a morphism, such that any other morphism between any object and $M$ must factor through. Is that precise? Can we characterize the kernel more specifically?

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The kernel is $0\rightarrow M$, the identity element of $\mathrm{Hom}_C(0,M)$ (where $0$ is a zero object of $C$). Your definition of kernel is not correct. A kernel of $M\rightarrow N$ is an arrow $K\rightarrow M$ such that $K\rightarrow M\rightarrow N=0$ and for any $K^\prime\rightarrow M$ with this property, there is a unique arrow $K^\prime\rightarrow K$ such that $K^\prime\rightarrow K\rightarrow M=K^\prime\rightarrow M$. – Keenan Kidwell Nov 11 '12 at 18:08
As Qiaochu points out in his answer, the first sentence in my comment is totally wrong. When I started writing it I somehow got it in my head that your morphism was supposed to be a monomorphism. – Keenan Kidwell Nov 11 '12 at 18:33
Check that every isomorphism $k \colon M' \to M$ is a kernel of $f$. Notice that you can factor $f$ as $M \to 0 \to N$. Maybe this makes it a bit clearer. – commenter Nov 11 '12 at 19:59
@KeenanKidwell: The definition of the kernel boils down to what i say in my question for the particular case of the zero morphism, right? This is because whenever there is a morphism $q:Q \rightarrow M$, then clearly $f \circ q=0$ and so there must exist $\bar{q}: Q \rightarrow M$ such that $q = p \circ \bar{q}$. As Qiaochu and Makoto mention, $id_M$ does the job because we can take $\bar{q}=q$. – Manos Nov 12 '12 at 1:19

A kernel $g\colon K \rightarrow M$ of $f\colon M \rightarrow N$ is characterized by the following conditions.

1) $f\circ g = 0$

2) Let $p\colon P \rightarrow M$ be a morphism such that $f\circ p = 0$. Then there exists a unique $q\colon P \rightarrow K$ such that $p = g\circ q$.

If $f = 0$, clearly $g = id_M \colon M \rightarrow M$ satisfies 1) and 2). Hence $id_M$ is a kernel of $f$.

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The kernel is the identity map $\text{id}_M : M \to M$. This agrees with the set-theoretic answer in, say, abelian groups. Keenan's comment contains the correct definition but not the correct answer.

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Thanks for correcting me Qiaochu. I don't know why I was thinking $M\rightarrow N$ was supposed to be a monomorphism instead of the zero morphism. – Keenan Kidwell Nov 11 '12 at 18:32