Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
Triple post .. boris why do you do that? – Martin Brandenburg Apr 17 '13 at 6:50
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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