Are quotients of chain groups $C_n(X)/C_n(A)$ still free? Suppose you have a topological space $X$ and a subspace $A$. Their chain complexes are made up of free abelian groups $C_n(X)$ and $C_n(A)$ are the free abelian groups on the $n$-simplexes on $X$ and $A$, respectively. 
If we take the quotient complex we get another chain
$$
\cdots\to C_2(X)/C_2(A)\to C_1(X)/C_1(A)\to C_0(X)/C_0(A)\to\cdots
$$
of abelian groups. 
Are these quotients still free abelian groups though? I know in general that the quotient of a free abelian need not be free, but is it any different in this case however?
 A: Yes, they are free. A basis for $C_n(X)/C_n(A)$ consists of all $n$-simplices in $X$ not contained in $A$ (they may intersect $A$, so long as they are not entirely contained in $A$ in which case they will be in $C_n(A)$ and hence $0$ in the quotient $C_n(X)/C_n(A)$).
Hope this helps!
A: I was going to add this as a comment, but oh well.
$C_n(X)$ is freely generated over $\mathbb{Z}$ by the singular simplexes $\sigma$ in $X$. These can be partitioned into those such that $\sigma(\Delta^n) \subset A$ and those for which $\sigma(\Delta^n) \not\subset A$. The former freely generate $C_n(A)$ and the latter generate another free abelian group we can call $G$. We have, as a result of this decomposition, that $C_n(X) \cong C_n(A) \oplus G$. Each element of $C_n(A)$ and no nonzero element of $G$ perishes under the quotient map $C_n(X) \to C_n(X)/C_n(A)$. It follows from the first isomorphism theorem that $C_n(X)/C_n(A) \cong G$ is free abelian.
A: This has very little to do with algebraic topology. The general result is :

If $S' \subseteq S$ is an inclusion of sets, then 
  $$ \mathbb Z[S] \mathop{/} \mathbb Z[S'] \simeq \mathbb Z[S \setminus S'] .$$

Here you apply it with $S = \operatorname{Sing}_n(X)$ and $S' = \operatorname{Sing}_n(A)$. ($\operatorname{Sing_n}(T)$ being the set of $n$-singular complexes of the topological space $T$.)
A: Yes, they're still free. The reason is that if some $n\sigma$ is in $C_2(A)$ then so is $\sigma$. In other words, $C_2(A)$ is a direct summand in $C_2(X)$.
