# Does zero-kernel imply monic in Abelian categories?

I'm trying to learn how to perform diagram-chasing in abstract Abelian categories. Instead of an approach with some elements one have to use universal properties somehow in the proof. But I reckon the need of some lemmas...

A category is Abelian if:

• it has a zero object
• it has all binary products and binary coproducts
• it has all kernels and cokernels
• every monomorphism is a kernel to some morphism
• every epimorphism is a cokernel to some morphism

The first "lemma" that coming into my mind (though I don't really know if it is true) deals with the connection between being monic and have a kernel equal to zero.

From the universal properties:

(1)$\quad f$ is a monomorphism if it given morphisms $m,n$ holds that $\beta m=\beta n\implies m=n$ $\require{AMScd}$ \begin{CD} X @>m>n> B@>\beta>> B' \end{CD} (2)$\quad k$ is a kernel to $\beta:B\to B'$ if $\beta k=0$ and for each $k'$ with $\beta k'=0$ there is a unique morphism $\phi$ such that $k\phi=k'$ \begin{CD} K'@>k'>>B\\ @V\exists!\phi VV\# @|\\ K @>k>> B@>\beta>> B' \end{CD}

How to prove that $\operatorname{ker}\beta=0\implies \beta$ is mono, using (1) and (2)?

My own approach, unfortunately, consists of staring on the diagrams above.

• @QiaochuYuan Actually, I don't see how we get enrichment over commutative monoids. That's the only problem, though – once we have biproducts, there's a trick for getting negatives. – Zhen Lin Feb 8 '15 at 19:50
• Hmmm. It turns out there's a trick for getting biproducts too. See Q6 here. – Zhen Lin Feb 8 '15 at 19:53
• @Zhen: oh, that's curious. Well, in any case, the OP needs either a stronger definition or some lemmas to get an enrichment over abelian groups. – Qiaochu Yuan Feb 8 '15 at 19:59
• Pity! This definition from Wikipedia seemed so gentle, but obviously it's a rather long way to prove the lemma from it. But Q6 might be a good exercise. – Lehs Feb 8 '15 at 20:26
• The exercise that Zhen Lin links to uses the notion of a pseudomonomorphism which in this context probably means a morphism $m$ such that $mf=0$ implies $f=0$. – tcamps Feb 8 '15 at 23:57

First prove that $f$ is a monomorphism if and only if $fg = 0$ implies $g = 0$ for all $g$.
Now assume $f : X \to Y$ has kernel $(0, 0 \to X)$ and suppose $g : W \to X$ with $fg = 0$. Then by (2), there is a morphism $h : W \to 0$ such that $g = 0h = 0$.