Clarification on relative homology definition this is probably an extremely simple question so I apologize in advance. I just wanted to check that I have understood the definition of relative homology correctly.
Given a topological space $X$ with subspace ${A\subset X}$, then the relative chain groups are given by
$$
S_n(X,A) := \frac{S_n(X)}{S_n(A)}
$$
where ${S_n(X)}$ represents the ${n^{th}}$ chain group of $X$ and likewise for ${A}$.
So
$$
S_n(X,A) = \{\beta + S_n(A)\ |\ \beta \in S_n(X)\}
$$
is the induced boundary homomorphisms between ${S_n(X,A) \to S_{n-1}(X,A)}$, ${\overline{\partial}_n}$, given like this?
$$
\overline{\partial}_n(\beta + S_n(A)) = \partial_n(\beta) + S_{n-1}(A) \in S_{n-1}(X,A)?
$$
most texts I read just say "it induces a homomorphism...." without specifying exactly what it is. Thanks!
 A: The meaning of this phrase is quite general when working with quotients. Suppose that one has a diagram of homomorphisms as below, such that $h_1$ and $h_2$ are surjective, and therefore are quotient homomorphisms with respective kernels $K_1 = \text{kernel}(h_1)$ and $K_2 = \text{kernel}(h_2)$ (actually for this you don't need that $h_2$ is surjective):
$\require{AMScd}$
\begin{CD}
B_1 @>f>> B_2\\
@V h_1 V V @VV h_2 V\\
Q_1 @. Q_2
\end{CD}
Now there is a theorem:

If $f(K_1) \subset K_2$ then there exists a unique homomorphism $g : Q_1 \to Q_2$ that makes the diagram commute:
$\require{AMScd}$
\begin{CD}
B_1 @>f>> B_2\\
@V h_1 V V @VV h_2 V\\
Q_1 @>g>> Q_2
\end{CD}

In this situation one says that $f$ induces $g$; many authors use this "induces" language to refer to this theorem, without ever telling you that's what they are doing.
The proof of uniqueness is obvious: given $\beta \in B_1$, one must have  the equation
$$g(h_1(\beta)) = h_2(f(\beta))
$$
Then one proves existence by checking that this equation actually gives a well-defined formula for $g$, by using the inclusion $f(K_1) \subset K_2$ to prove that $g(h_1(\beta \kappa))=g(h_2(\beta))$ for any $\kappa \in K_1$.
