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I encountered the following HW level problem:

Assume $\mathcal{C}$ is a category which admits a zero object and kernels (by this word I think the author means equalizers). Prove that a morphism $\mathbb{}f:X \rightarrow Y$ is a monomorphism iff $\mathbb{} Kerf \simeq 0$.

So $\Rightarrow$ is obvious, while I have no idea why $\Leftarrow$ should work. Any ideas? General comments are also welcome!

P.S. According to Mitchell's "Theory of categories" pg15, $\Leftarrow$ is not true in general ... so seeing counterexamples would be delightful. Seems that we should start from a category which is not normal....

P.S.2 These are actually from Schapira's notes:link Pg44. Alas, Mitchell vs Schapira!

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    I'm confused: are you being asked to prove something that isn't true?2011-12-19
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    @Dylan Moreland: I guess so....these are actually from Schapira's notes:[link](http://people.math.jussieu.fr/~schapira/lectnotes/HomAl.pdf) Pg442011-12-19
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    Ah, it seems like he would know what he's talking about! This seems clear if $\mathcal C$ is additive, but otherwise I, like you, don't see what to do. [It would be good to include that reference in the question, by the way.]2011-12-19
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    @abundent: There's a perfectly good definition of kernel in any category with zero morphisms: $\ker f$ is the equaliser of $f$ and the zero morphism. A category with a zero object a fortiori has zero morphisms.2011-12-19
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    For people like me who never remember what an equalizer is, the [definition of a kernel on Wikipedia](http://en.wikipedia.org/wiki/Kernel_(category_theory)) is simple and requires nothing more than a zero object (or even just a "zero morphism", which I hadn't come across as an isolated concept before).2011-12-19
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    @ZhenLin: I know....But it seems that the existence of equalizers is stronger than that of kernels.2011-12-19
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    @abundent: A kernel is an equaliser. You don't have to assert that all equalisers exist before you can talk about them.2011-12-19
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    @ZhenLin: I know! I've never seen assumptions of existence of all equalizers either. Anyway,Schapira denotes it ker(f,g) and refers to these as kernels throughout. I do hope this stronger(?) condition is useful. Please inform me otherwise.2011-12-19

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Consider the graph

$$a \underset{g}{\stackrel{f}\rightrightarrows} b \stackrel{h}\to c$$

next consider the category freely generated by it, add to it a zero object $0$, all the zero arrows, and mod out arrows so that $hf=hg$. In the resulting category, $h$ is not a monomorphism, yet the kernel of $h$ is $0$. A short list of checks shows that this category has all kernels.

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    :Thanks! I guess my concrete thinking has become a hinderance.2011-12-19
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    This type of examples should not be taken too seriously :D2011-12-19