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Could you help me to explain this argument:

Let $f: S^{n-1} \rightarrow A$ for $n \ge 1$, form

$$X= C(f) := \dfrac{A\coprod D^n}{f(x) \sim x, \forall x \in S^{n-1}}$$ "$(D^n,S^{n-1}) \rightarrow (X,A)$ induces isomorphisms in $H_q(D^n,S^{n-1}) \rightarrow H_q(X,A)$"

user69833
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2 Answers2

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Let $g : (D^n, S^{n-1}) \to (X, A)$ be the characteristic map of the cell $e = X - A$. There are homotopy equivalences $(D^n, S^{n-1}) \to (D^n, D^n - \{0\})$ and $(X, A) \to (X, X - \{g(0)\})$, which induce isomorphisms of relative homology groups. By excising the boundary $\partial D^n$, one sees $((D^n)^\circ, (D^n)^\circ - \{0\}) \hookrightarrow (D^n, D^n - \{0\})$ induces an isomorphism in homology, and the same for $(e, e - \{g(0)\}) \hookrightarrow (X, X - \{g(0)\})$ by excising $A$. But $g$ restricts to a homeomorphism $(D^n)^\circ \stackrel{\sim}{\to} e$, hence it induces an isomorphism $H_*((D^n)^\circ, (D^n)^\circ - \{0\}) \stackrel{\sim}{\to} H_*(e, e - \{g(0)\})$. It follows then that $g$ also induces an isomorphism $g_* : H_*(D^n, S^{n-1}) \stackrel{\sim}{\to} H_*(X, A)$.

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Note $H_q(D^n, S^{n-1}) = H_q(D^n/S^{n-1}) = H_q(S^n)$ since $D^n/S^{n-1}$ is homeomorphic to $S^n$. On the other hand $H_q(X, A) = H_q(X/A) = H_q(S^n)$ since $X/A$ is also homeomorphic to $S^n$.

Let me know if you don't understand any of these isomorphisms.

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    Thanks a lot. Can you explain more about $X/A \approx S^n$? How can you recognise it immediately? I often construct a bijection, and it is complicated. – user69833 Apr 11 '13 at 01:57
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    The characteristic map $D^n \to X$ induces a map $D^n/S^{n-1} \to X/A$ which is a homeomorphism onto $\bar{e}/\partial \bar{e}$, where $e = X - A$ is the cell being attached, and the inclusion $\bar{e} \hookrightarrow X$ also induces a homeomorphism $\bar{e}/\partial \bar{e} \to X/A$. Anyway, I just posted another proof which is technically simpler. –  Apr 11 '13 at 10:56