And so this week, our algebraic topology class starts with relative homology groups. But there are some (REALLY) basic parts of the definition of the relative homology group that I don't understand why...
Our class is currently using Hatcher's Algebraic Topology.
Given that $A$ is a subspace of a topological space $X$.
Hatcher defines the chain group $C_n(X,A)=C_n(X)/C_n(A)$.
I understand that $C_n(A) \subset C_n(X)$.
1) Why is $C_n(A)$ even a subgroup of $C_n(X)$ in the first place?
2) I understand why the boundary map $\partial$ takes $C_n(A)$ to $C_{n-1}(A)$, but why does it induce a quotient boundary map from $C_n(X,A)$ to $C_{n-1}(X,A)$?
I apologize if it is really simple and probably obvious to most but I really can't see why.
Ad 1), since $A\subset X$, every chain in $A$ is also a chain in $X$. Then $C_n(A)$ is the subgroup of chains such that the image of every singular simplex happens to lie in $A$. The empty chain $0$ is a chain in $A$, since it contains no (singular) simplex whose image is not contained in $A$. If $c_1,c_2 \in C_n(A)$, then every simplex in $c_1+c_2$ lies in $A$, hence $c_1+c_2\in A$, and finally, for $c\in C_n(A)$, the chain $-c$ also consists only of simplices in $A$, hence $-c \in C_n(A)$. So $C_n(A)$ is a nonempty subset of $C_n(X)$ that is closed under the group operations, hence a subgroup.
Ad 2), we have the composition
$$\varphi = \pi \circ \partial \colon C_n(X) \to C_{n-1}(X) \to C_{n-1}(X,A)$$
of the boundary map and the canonical projection. That is a homomorphism, and
$$C_n(A) \subset \ker \varphi$$
since $\partial C_n(A) \subset C_{n-1}(A) = \ker \pi$. Thus we have an induced homomorphism
$$\overline{\varphi} \colon C_n(X)/C_n(A) \to C_{n-1}(X,A).$$
This induced homomorphism is the quotient boundary map.