We work in Singular homology. This exact sequence can be found in pg97 of Intro. Alg. Top. by Rotman.
Let $A \subseteq X$ be subspace of space $X$. In category of topological pairs, the inclusions then induces SES $$0 \rightarrow S_*(A, \emptyset) \rightarrow S_*(X, \emptyset) \rightarrow S_*(X)/S_*(A)\rightarrow 0$$ By diagram chasing in homological algebra, there is a LES, where $H_n(A,\emptyset)=H_n(A)$, $$\cdots \rightarrow H_n(A) \rightarrow H_n(X) \rightarrow H_n(X,A) \xrightarrow{d} H_{n-1}(A)\rightarrow \cdots $$
My question is , what exactly does the map $d$ do? It seems weird and so simple:
Take a cycle in $S_n(X)/S_n(A)$, which must be of form $[\sum m_i \tau_i ]$. Pull back to $S_n(X)$, $\sum m_i \tau_i$, of which choice of representative is relevant, and $\tau_i$ are the basis maps $\Delta^n \rightarrow X$.
Push to $S_{n-1}(X)$, $\partial \sum m_i \tau_i = \sum m_i \partial \tau_i$. But by diagram chase, this is can be seen as an element of $S_{n-1}(A)$.
So we have that $$d:[\sum m_i \tau_i ] \mapsto [ \sum m_i \partial \tau_i ] $$
Am I doing this right?
You can think of $S_n(X)/S_n(A)$ as the free Abelian group on the singular $n$-simplices in $X$ which do not lie in $A$. If one does that, the boundary of a cycle $\alpha$ in $S_n(X)/S_n(A)$ is a chain $\partial\alpha\in S_{n-1}(A)$, since those are the chains that become zero in $S_{n-1}(X)/S_{n-1}(A)$. Then $\partial\alpha$ is a cycle in $A$ (obviously $\partial(\partial\alpha))=0$), so induces an element of $H_{n-1}(A)$. The map $\alpha\mapsto\partial\alpha$ therefore induces a map $d:H_n(X,A)\to H_{n-1}(A)$.