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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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    @ZhenLin: I $k$now! I've never seen assumptions of existence of all equalizers either. $A$nyway,$S$cha$p$ira 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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    This type of examples should not be taken too seriously :D2011-12-19