Question on chain complex Let $X$ be a topological space. Here, $S_n(X)$ denotes the free abelian group generated by all singular simplexes $\sigma: \Delta^n \to X$, let $S_{\ast}(X)$ denote the chain complex d
$$\ldots\xrightarrow{\partial_{n+1}} S_n(X) \xrightarrow{\partial_n}\ldots \xrightarrow{\partial_3} S_2(X) \xrightarrow{\partial_2} S_1 \xrightarrow{\partial_1} S_0.$$
How can I prove that there is a short exact sequence of chain complexes $$0 \to S_{\ast}(X) \xrightarrow{\cdot m} S_{\ast}(X) \to S_{\ast}(X ; \mathbb{Z}/ m\mathbb{Z})\to 0$$
I totally get the construction of $S_{\ast}(X)$ and the map $\cdot m$, but what is the definition of $S_{\ast}(X ; \mathbb{Z}/ m\mathbb{Z})$? Considering this as a lemma, I also have to prove in the exercise that there is a short exact sequence $$0 \to H_n(X)/mH_n(X) \to H_n(X; \mathbb{Z} / m \mathbb{Z}) \to T_m(H_{n-1}(X)) \to 0$$ where $T_m(A)$ denotes the $m$-torsion free subgroup of $A$.
I must admit I am a little lost, any help is appreciated.
 A: The definition of $S_*(X;A)$, singular chains with coefficients in abelian group $A$, is usually just $S_*(X;A)=S_*(X)\otimes A.$
So there is a short exact sequence of abelian groups $0\to\mathbb{Z}\overset{m}\to \mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0.$ In general tensoring with other abelian groups is right-exact; it only preserves the right-hand end of short exact sequences. But tensoring with flat modules does preserve the whole S.E.S., and free $\mathbb{Z}$-modules are flat, and $S_*(X)$ is a free $\mathbb{Z}$-module, by definition.
So tensoring with $S_*(X)$ gives $0\to S_*(X)\overset{m}\to S_*(X)\to S_*(X;\mathbb{Z}/m\mathbb{Z})\to 0.$
If you're not familiar with these facts of homological algebra about flatness and free modules, it is also straightforward to verify directly that this short exact sequence is exact. Let me know if you need those details.
Then to get your result, we take homology of these chain complexes. Taking homology doesn't preserve short exact sequences, but it does turn short exact sequence into long exact sequences, which is more or less its raison d'être. So we have 
$$\ldots\to H_n(X)\overset{m}\to H_n(X)\to H_n(X;\mathbb{Z}/m\mathbb{Z})\overset{\partial}\to H_{n-1}(X)\overset{m}\to H_{n-1}(X)\to\ldots.$$ 
Now in general, any morphism $A\overset{f}\to B$ can be extended to a short exact sequence $0\to \ker(f) \to A \to \text{coker}(f)\to 0.$ 
Also any long exact sequence $\ldots \to A\overset{f}\to B\overset{g}\to C\to\ldots$ can be truncated to $0\to \ker(f)\to B\to \text{im}(g)\to 0.$ Intuitively, you can lop off everything to the left of any point and retain exactness, as long as you replace the first group you cut $A$ with $\ker(f),$ the kernel of the map out of it to the right. Similarly you can lop off everything to the right of any point if you replace the cut group with $\text{im}(g),$ the map into it from the left.
Now let's apply the latter option to truncate this long exact sequence in homology. By the first isomorphism theorem, we have that $\text{im}(H_n(X)\to H_n(X;\mathbb{Z}/m\mathbb{Z})) = H_n(X)/\ker(H_n(X)\to H_n(X;\mathbb{Z}/m\mathbb{Z})).$ Then by exactness, we have that $\ker(H_n(X)\to H_n(X;\mathbb{Z}/m\mathbb{Z})) = \text{im}(H_n(X)\overset{m}\to H_n(X)) = mH_n(X).$ Hence we have 
$$
0\to H_n(X)/mH_n(X)\to H_n(X;\mathbb{Z}/m\mathbb{Z})\overset{\partial}\to H_{n-1}(X)\overset{m}\to H_{n-1}(X)\to\ldots.$$
Then by exactness the image of $H_n(X;\mathbb{Z}/m\mathbb{Z})\overset{\partial}\to H_{n-1}(X)$ is the kernel of $H_{n-1}(X)\overset{m}\to H_{n-1}(X),$ which is the $m$-torsion subgroup $T_m(H_{n-1}(X))$ of $H_{n-1}(X).$ Thus we have 
$$
0\to H_n(X)/mH_n(X)\to H_n(X;\mathbb{Z}/m\mathbb{Z})\to T_m(H_{n-1}(X))\to 0
$$
which was to be proved.
