Exact sequence of finitely generated abelian groups Let $$0\rightarrow G_{1}\xrightarrow{f_{1}} G_{2}\xrightarrow{f_{2}} G_{3}\rightarrow 0$$ is a short exact sequence of finitely generated abelian groups. We call $\overline{G_{i}}$ the quotient of $G_{i}$ by its torsion. Then, we consider :
$$0\rightarrow \bar{G_{1}}\xrightarrow{\bar{f_{1}}} \bar{G_{2}}\xrightarrow{\bar{f_{2}}} \bar{G_{3}}\rightarrow 0.$$

Prove that $\bar{f_{2}}$ is surjective and that $Ker(\bar{f_{2}})/Im(\bar{f_{1}})$ is a torsion group.

My answer is (I'm not sure, that's why it looks like questions) : Is it true that $\bar{f_{2}}$ is surjective since $f_{2}$ is ?
Is it true that $Ker(\bar{f_{2}})/Im(\bar{f_{1}})$ is a torsion group since $Im(f_{1})=Ker(f_{2})$ ?
Thank you.
 A: Writing $G_i = \mathbb{Z}^{r_i} \oplus T_i,$ where $r_i \geq 0$ is an integer and $T_i$ is a torsion group, we have the following exact sequence:
$$
0 \rightarrow \mathbb{Z}^{r_1} \oplus T_1 \xrightarrow{f_1} \mathbb{Z}^{r_2} \oplus T_2 \xrightarrow{f_2} \mathbb{Z}^{r_3} \oplus T_3 \rightarrow 0.
$$
From this, taking quotient by trosion subgroups, we get the following complex:
$$
0 \rightarrow \mathbb{Z}^{r_1} \xrightarrow{\bar{f_1}} \mathbb{Z}^{r_2} \xrightarrow{\bar{f_2}} \mathbb{Z}^{r_3} \rightarrow 0.
$$
This is not exact in general. First we want to show that $\bar{f_2}$ is surjective. This follows from the fact that $f_2$ is surjective, and $f_2(t) \in T_3$ for every $t \in T_2.$ Now we want to show that $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}$ is a torsion group. Let $x \in \text{ker}{\bar{f_2}}.$ Then $\bar{f_2}(x) = 0 \Rightarrow f_2(x) \in T_3.$ But $T_3$ is a torsion group. So there exists $n \in \mathbb{N},$ such that $f_2(nx) = nf_2(x) = 0.$ Hence there is $y \in G_1$ such that $f_1(y) = nx.$ Note that $y \notin T_1.$ So $nx = 0$ in $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}.$ This shows that, every element of $\text{ker}{\bar{f_2}}/ \text{Im}{\bar{f_1}}$ is a torsion element.
