1
$\begingroup$

My aim is to show that the category of free abelian groups not an abelian category.

I read that I could fix $n \in \mathbb{N} \setminus \left\lbrace 0,1 \right\rbrace $ and consider the "multiplication by $n$ " map $ \mathbb{Z} \rightarrow \mathbb{Z} $ is a mono and an epi but not an iso.

Can anyone provide me with a hint as to why this is a contradiction?

Thanks!

$\endgroup$
2
$\begingroup$

In an abelian category every morphism which is mono and epi is an isomorphism (that is, has an inverse).

This is based on the fact that, by assumption, every morphism has a kernel and a cokernel, every monomorphism is a kernel and every epimorphism is a cokernel. It's quite easy to see that a morphism which is both a kernel and a cokernel is an isomorphism.

The “multiplication by $2$” in a free abelian group is monic (obviously) and epic because the cokernel (computed in the category of abelian groups) is torsion, so there's no non zero morphism from it to any free abelian group. However it's not an isomorphism, because it has no inverse.

$\endgroup$
  • $\begingroup$ I guess my problem is with the first statement, I can't seem to prove it... Any hints? $\endgroup$ – Mia Dec 31 '13 at 10:57
  • $\begingroup$ Nice observations. I see free abelian groups, all of them, love you egreg. ;-) +1 $\endgroup$ – mrs Dec 31 '13 at 11:37
  • 1
    $\begingroup$ @B.S. Somebody tells me there are also non abelian groups, but I'm quite dubious about this assertion. ;-) “All groups, apart for some unfortunate exceptions, are abelian”. $\endgroup$ – egreg Dec 31 '13 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.