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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? –  Dylan Moreland Dec 19 '11 at 2:29
    
@Dylan Moreland: I guess so....these are actually from Schapira's notes:link Pg44 –  abundent Dec 19 '11 at 2:31
    
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.] –  Dylan Moreland Dec 19 '11 at 2:50
    
@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. –  Zhen Lin Dec 19 '11 at 3:50
    
For people like me who never remember what an equalizer is, the definition of a kernel on Wikipedia 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). –  Dylan Moreland Dec 19 '11 at 3:56

1 Answer 1

up vote 3 down vote accepted

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. –  abundent Dec 19 '11 at 5:10
2  
This type of examples should not be taken too seriously :D –  Mariano Suárez-Alvarez Dec 19 '11 at 5:29

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