Is this a valid way to show a sequence does not split?

Question

Suppose you have $$ 0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to 0 $$ viewed as additive groups. Let $i$ be the inclusion of $\mathbb{Z}$ into $\mathbb{Q}$. I want to say this is not a split sequence, since there is no retract $f:\mathbb{Q}\to\mathbb{Z}$ such that $f\circ i=id_\mathbb{Z}$. If there were, then $f(n)=n$ for all integers $n$. Then $$ nf(1/n)=f(1/n)\cdots+f(1/n)=f(1/n+\cdots+1/n)=f(1)=1 $$ so $f(1/n)=1/n$, which doesn't make any sense if $n>1$, since $f(1/n)\in\mathbb{Z}$. So no such $f$ exists. Is that all there is to do?

Answer 1

Yes! That is good. You're basically observing that $\text{Hom}(\mathbb{Q},\mathbb{Z})=\{0\}$ and so you clearly can't have a retraction.

You could also note that if it split then $\mathbb{Q}\cong \mathbb{Z}\oplus(\mathbb{Q}/\mathbb{Z})$, but $\mathbb{Q}$ has no torsion :)

Answer 2

Yes; this is the extension lemma, and it holds in Abelian categories:

http://en.wikipedia.org/wiki/Splitting_lemma