# 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!

• Draw Picture: Let $CA=\frac{A\times [0,1]}{A\times \{1\}}$, consider the space $X\cup CA$ identifying $a\in A$ with $[a,0]\in CA$. Let $q\colon A\times [0,1]\to CA$ be the quotient map. Consider $\mathcal U:= X\cup q\big(A\times [0,\varepsilon)\big)$ and $\mathcal V:=q\left(A\times\left(\frac{\varepsilon}{2},1\right]\right)$ for $0<\varepsilon<1$. Then, the open set $\mathcal U$ homotopy equivalent to $X$ and the open set $\mathcal V$ homotopy equivalent to a point(vertex of the cone). Now, $\mathcal U\cap \mathcal V$ is homotopy equivalent to $A$. Nov 9, 2021 at 17:32
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.