Deriving the long exact sequence of a pair $(X,A)$ from the Mayer-Vietoris sequence It is an exercise in Hatcher's Algebraic Topology text to derive the long exact sequence of  a pair $(X,A)$ from the Mayer-Vietoris sequence applied to $X \cup CA$. We may use the isomorphism $H_n(X,A) \cong \tilde{H}_n(X \cup CA)$ for all $n$.
But, the Mayer-Vietoris sequence is applied under the assumptions that we have a space $Y$ such that $Y = \text{int}(C) \cup \text{int}(D)$ for subspaces $C$ and $D$ of $Y$. (And, if we add the assumption that $C \cap D \neq \emptyset$, we get the Mayer-Vietoris sequence for reduced homology.)
In our situation, does the problem aim to have $X \cup CA$ play the role of $Y$? If so, how are we to write $X \cup CA$ as a union of interiors of subspaces of $X \cup CA$? I don't see why it would be true that $X \cup CA = \text{int}(X) \cup \text{int}(CA)$, so is the goal to apply the Mayer-Vietoris sequence to $Y = \text{int}(X) \cup \text{int}(CA)$?
The wording of the problem confused me and prevented me from getting started.
Thank you!
 A: Let $CA = \frac{A \times [0,1]}{A \times \{1\}}$, and let $q:A \times [0,1] \rightarrow CA$ be the quotient map. Consider the space $X \cup CA$ identifying $a \in A$ with $[a,0] \in CA$. Consider the open sets $U := X \cup q(A \times [0,\epsilon))$ and $V := q(A \times (\frac{\epsilon}{2},1])$ for $0 < \epsilon < 1$. Note that $X \cup CA = \text{int}(U) \cup \text{int}(V) = U \cup V$, $U$ is homotopy equivalent to $X$, $V$ is homotopy equivalent to a point (the vertex of $CA$), and $U \cap V$ is homotopy equivalent to $A$. Since homotopy equivalences induce isomorphisms in reduced homology groups for all degrees, the reduced homology groups of a point are all trivial, and $H_n(X,A) = \tilde{H}_n(X,A)$ for all $n$ and nonempty $A$, by the Mayer-Vietoris sequence for reduced homology applied to $X \cup CA$ and $H_n(X,A) \cong \tilde{H}_n(X \cup CA)$ for all $n$, we have the long exact sequence
\begin{align*}
    \cdots \rightarrow \tilde{H}_n(A) \rightarrow \tilde{H}_n(X) \rightarrow \tilde{H}_n(X,A) \rightarrow \tilde{H}_{n-1}(A) \rightarrow \cdots \rightarrow \tilde{H}_0(X,A) \rightarrow 0.
\end{align*}
We've now derived the long exact sequence of reduced homology groups of a pair $(X,A)$. It only remains to go further to obtain the long exact sequence of ordinary homology groups.
