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Why is the category of free abelian groups not an abelian category?

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I don't think it has cokernels. –  Olivier Bégassat Jan 3 '13 at 20:46
    
@OlivierBégassat A concrete example? One may think that the FrAb cokernel is somehow the torsion free part of the Ab cokernel. So writing something like $\mathbb Q$ or $\mathbb Z^{\mathbb N}$ as quotient of free abelian groups may lead to a case of nonexisting cokernel ... –  Hagen von Eitzen Jan 3 '13 at 20:57
    
@HagenvonEitzen I was actually thinking of the map Mariano describes, but I didn't want to check explicitely wether it has no cokernel in the category of free abelian groups, instead I just assumed it would have to be the same as in the category of groups. I didn't bother to check. –  Olivier Bégassat Jan 3 '13 at 21:19
    
Although that looks like a non-example, as there are no non trivial maps from $\Bbb Z$ to a free abelian group that vanish on the even integers. –  Olivier Bégassat Jan 3 '13 at 21:25
    
@Olivier: the cokernel is trivial. –  Qiaochu Yuan Jan 3 '13 at 22:41

1 Answer 1

up vote 8 down vote accepted

The map $\mathbb Z\to\mathbb Z$ given by multiplication by $2$ is a mono and an epi, but not an iso.

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