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Let $\mathcal{A}$ be a category. If $\mathcal{A}$ is pointed, i.e. has zero objects, then $f$ monic $\implies \operatorname{ker} f = 0$. If $\mathcal{A}$ is abelian, we have the equivalence $f$ monic $\iff \operatorname{ker} f = 0$. However this is true in $\mathsf{Gp}$, which is not abelian. When in general (or more generally) is this true?

It is true when $\mathcal{A}$ is conormal, or more generally when the coequaliser of any two morphisms is a cokernel, but do we need this?

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Related: math.stackexchange.com/questions/53405/… –  Juan S Jun 9 '12 at 10:01
    
Thanks @JuanS that is useful. –  Paul Slevin Jun 9 '12 at 10:04
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I think the forward implication is always true, and the reverse is true in normal categories –  Juan S Jun 9 '12 at 10:09
    
@JuanS do you mean conormal? In a normal category if $f$ is a mono, then indeed it is a kernel, but if I take any $f$ with kernel $0$ it won't necessarily be a mono. However conormality would ensure that the coequaliser of any two morhisms would be a cokernel, which would do it. –  Paul Slevin Jun 9 '12 at 10:11
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For categories of algebraic structures, you probably need to have some kind of Mal'cev operation in your structure. –  Zhen Lin Jun 9 '12 at 10:38
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