If we consider $A^ {fr} \subset A^ {fg} $, where $A^ {fr}$ is the subcategory of torsion free finitely generated abelian groups, and $A^ {fg} $ is the category of finitely generated abelian groups, in $A^ {fr}$ are there cokernels ? How do they look like ?

  • $\begingroup$ Triple post .. boris why do you do that? $\endgroup$ – Martin Brandenburg Apr 17 '13 at 6:50

I'm not a specialist, so please correct me if I'm wrong, or if I'm answering the wrong question entirely. I'll be using this definition of cokernels.

There are cokernels in $A^{fr}$. Suppose $f: X \to Y$ is a morphism of torsion free FGAGs. I claim that the cokernel of $f$ is the group $Q = (Y/f(X))/T(Y/f(X))$ together with the natural surjection $Y \to Q$. Here $T(G)$ means the torsion subgroup of $G$.

Suppose we have a morphism $g: Y \to Q'$ such that $g \circ f = 0$. Then $g$ factors through $Y/f(X)$, i.e. there is a unique homomprism $\beta: Y/f(X) \to Q'$ such that the composition $$ Y \to Y/f(X) \stackrel{\beta}{\to} Q' $$ is equal to $g$.

Since $Q'$ is torsion free, $\beta$ must send $T(Y/f(X))$ to zero. Then there is a unique homomorphism $\gamma: (Y/f(X))/T(Y/f(X)) \to Q'$ such that the composition $$ Y/f(X) \to \frac{Y/f(X)}{T(Y/f(X))} \stackrel{\gamma}{\to} Q' $$ is equal to $\beta$. This also means that $\gamma$ is the only homomorphism such that this composition: $$ Y \to Y/f(X) \to \frac{Y/f(X)}{T(Y/f(X))} \stackrel{\gamma}{\to} Q' $$ equals to $g$. Looks like this proves what we want.


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