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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!

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    $\begingroup$ 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 $A$. $\endgroup$
    – Sumanta
    Nov 9, 2021 at 17:32
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    $\begingroup$ @SumantaDas You should transform your comment into an answer. $\endgroup$
    – Paul Frost
    Nov 13, 2021 at 9:51
  • $\begingroup$ @SumantaDas Excellent. Thank you! If you write what you've written here as an answer, I will gladly accept it. $\endgroup$
    – michiganbiker898
    Nov 14, 2021 at 20:06
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    $\begingroup$ @michiganbiker898 you can answer your own question also, probably then you learn something more. See this page math.stackexchange.com/help/…. Also accept your answer. $\endgroup$
    – Sumanta
    Nov 15, 2021 at 4:50

1 Answer 1

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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.

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