# Torsion and torsion-free abelian groups

I am missing some knowledge about torsion and torsion-free groups that I need to understand an example (let's say I have not seen these expression before). We have the exact sequence of abelian groups:

$$0 \to H_2(\mathbb{R} P^2) \to \mathbb{Z} \stackrel{2}{\to} H_1(\mathbb{R} P^1) \stackrel{i_*}{\to} H_1(\mathbb{R} P^2) \to \mathbb{Z}$$

We can see that $H_2(\mathbb{R} P^2) = 0$. We also know that the rank of $H_2(\mathbb{R}P^2)$ and $H_1(\mathbb{R}P^2)$ are equal (and hence 0).

So I know from here that abelian groups of rank 0 are torsion. (i.e. each element has finite order).

The next statement is that exactness shows that $i_*$ is surjective (because $H_1(\mathbb{R}P^2)$ is torsion and $\mathbb{Z}$ is torsion free. Can anyone shed some light on this statement? (Or just a wiki link!)

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Finitely generated abelian groups of rank 0 are torsion! – user641 Mar 25 '11 at 18:26

The image of the map $H_1(\mathbb{R}P^2)\rightarrow \mathbb{Z}$ is a quotient of $H_1(\mathbb{R}P^2)$ and thus torsion. It must then be $0$ since it is a subgroup of a torsion-free group. And since the sequene is exact, the kernel of this map (which is then $H_1(\mathbb{R}P^2)$) is the image of $i_{*}$ so this map is surjective.
Is the shorter proof correct: $f: H_1(\mathbb{R}P^2)\to \mathbb{Z}$, $dom(f)$ is torsion and $cod(f)$ is torsion-free, then $f=0$, then $ker(f)=dom(f)$, the sequence is exact, then $im(i_{*})=dom(f)$? – beroal Mar 24 '11 at 9:52