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My question is if there are Cokernels in the category of free abelian groups. The answer is yes in the case of finitely generated free abelian groups since one has the structure theorem of finitely generated modules over a Principal Ideal Domain and one can take as Cokernel just the free part of the usual Cokernel and is routine to prove that is indeed a Cokernel for the category. Does anyone has any clue about what happens if we drop the finitely generation condition?

Ignacio Barros
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1 Answers1

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Let $\prod\mathbb{Z}$ be the direct product of countably many copies of $\mathbb{Z}$, and let $$F_1\stackrel{\alpha}{\rightarrow}F_0\rightarrow \prod\mathbb{Z}\rightarrow 0$$ be a free resolution. Then $\alpha$ doesn't have a cokernel in the category of free abelian groups.

If it did, then the natural map from $F_0$ to the cokernel $C$ would factor uniquely through $\prod\mathbb{Z}$ (the cokernel in the category of all abelian groups), and the resulting map $\prod\mathbb{Z}\stackrel{\beta}{\rightarrow}C$ would be a universal map from $\prod\mathbb{Z}$ to a free abelian group.

But for any non-zero element $x$ of $\prod\mathbb{Z}$, at least one of the projections onto $\mathbb{Z}$ does not have $x$ in the kernel, and by the universal property, it factors through $\beta$. Therefore $x$ can't be in the kernel of $\beta$, and so $\beta$ is injective.

But then $\prod\mathbb{Z}$ is isomorphic to a subgroup of $C$. But every subgroup of a free abelian group is free abelian, and $\prod\mathbb{Z}$ is not free abelian.

Jeremy Rickard
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  • Thanks for the reply, but I didn't understood by universal property of what does $\pi$ factor through $\beta$? – Ignacio Barros Apr 23 '14 at 11:05
  • Every map from $\prod\mathbb{Z}$ to a free abelian group factors uniquely through $\beta$ (since $C$ is the cokernel of $\alpha$ in the category of free abelian groups, and $\prod\mathbb{Z}$ is the cokernel of $\alpha$ in the category of all abelian groups). – Jeremy Rickard Apr 23 '14 at 11:52