This is really bothering me, as I thought it an easy exercise. Let $A,B,C$ be abelian groups and $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ be exact. I want to show that
$$A/2A\rightarrow B/2B\rightarrow C/2C\rightarrow 0$$ is exact.
I have already shown exactness of the top line, surjectivity of $\bar{g}$, and that $\text{im}(\bar{f})\subseteq \ker(\bar{g})$. I tried to show $\ker(\bar{g})\subseteq \text{im}(\bar{f})$. However, if I pick a $\bar{b}=b+2B$ in $\ker(\bar{g})$, all I know is that I can take $b\in B$ and $g(b)+2C = \bar{g}(\bar{b})=0$, so $g(b)\in 2C$. Now, I can get something in $2B$ that maps to $g(b)$, but that is not helping me show that $\bar{b}\in \text{im}(\bar{f})$. What else can I do to finish showing exactness at $B/2B$?
If $g(b-2b')=0$ then
$b-2b'\in\operatorname{im}(f)$ hence
$\bar{b}\in \text{im}(\bar{f}).$