TITLE: Question about Relative Homology
QUESTION [5 upvotes]: I am reading through Hatcher and I came across the following statement and am having trouble making sense of it.  I am not sure why elements may be written this way.  Any help will be appreciated.
By
considering the deﬁnition of the relative boundary map we see:
Elements of $H_n(X, A)$ are represented by relative cycles: $n$ chains $α ∈ C_n(X)$
such that $∂α ∈ C_{n−1}
(A).$
A relative cycle $α$ is trivial in $H_n(X, A)$ iﬀ it is a relative boundary: $α = ∂β + γ$
for some $β ∈ C_{n+1}
(X)$ and $γ ∈ C_n(A).$

REPLY [5 votes]: The relative chain groups are defined by
$$C_n(X, A) = C_n(X)/C_n(A).$$
The usual boundary map on $C_\ast(X)$ descends to this quotient so that we have a chain complex $(C_\ast(X, A), \partial)$. The relative homology groups of the pair $(X, A)$ are the homology groups of this chain complex.
Now if $\alpha = \partial \beta + \gamma$ for $\beta \in C_{n+1}(X)$ and $\gamma \in C_n(A)$, we have that $[\alpha] = [\partial \beta]$ in the quotient group $C_n(X,A) = C_n(X)/C_n(A)$. Since the boundary map descends to the quotient, we have that
$$\tilde{\partial}[\beta] = [\partial \beta],$$
where $\tilde{\partial}: C_{n+1}(X,A) \longrightarrow C_n(X,A)$ is the boundary map on $C_\ast(X,A)$. Therefore $[\alpha]$ is a boundary in $C_n(X,A)$ with respect to the boundary map $\tilde{\partial}$.