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!)

  • $\begingroup$ Finitely generated abelian groups of rank 0 are torsion! $\endgroup$ – 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.

  • $\begingroup$ 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 $im(i_{*})=dom(f)$? $\endgroup$ – beroal Mar 24 '11 at 9:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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