Deduce that there are short exact sequences 
Show that for $n>0$ there is a short exact sequence of chain complexes $0\rightarrow C_i(X;\mathbb{Z})\stackrel{f}{\rightarrow} C_i(X;\mathbb{Z})\stackrel{g}{\rightarrow} C_i(X;\mathbb{Z_n})\rightarrow 0$. Hence deduce that there are short exact sequences $0\rightarrow H_i(X;\mathbb{Z})/nH_i(X;\mathbb{Z})\rightarrow H_i(X;\mathbb{Z_n})\rightarrow n-Torsion(H_{i-1}(X))\rightarrow 0$ where $n-Torsion(H_{i-1}(X))$ is the kernel of the $\times n$ map of $H_{i-1}(X)$.

My Try:
If I take $f$ as $\times n$ map and $g$ as $\mod n$ map then I am done for the first part. For second part since I have to deduce I thought of using snake lemma but was not helpful. After that no clue at all. Can somebody explain me how to deduce it?
 A: The long exact sequence of homology (which indeed comes from the snake lemma) reads $$H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})\to H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$$
where the maps $H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})$ and $H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$ are given by multiplication by $n$ (because taking homology is compatible with addition), and $H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})$ is the connecting map (precisely given by the snake lemma).
The fact that it's exact means that $$H_i(X,\mathbb{Z})\to H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})$$ is exact, with the kernel of the leftmost map given by the image of $H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})$, ie $nH_i(X,\mathbb{Z})$, and the image of the rightmost map given by the kernel of $H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$, ie the $n$-torsion of $H_{i-1}(X,\mathbb{Z})$.
Thus there is an exact sequence $$0\to H_i(X,\mathbb{Z})/nH_i(X,\mathbb{Z})\to H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})[n]\to 0$$
just by taking the quotient by the kernel on the left and restricting to the image on the right.
