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I want to prove that inclusion $i:(B^n,S^{n-1})\to (B^n,B^n\setminus\{0\})$ induces isomorphism in relative homology.

Now, $i$ obviously induces isomorphism of $H_p(B^n)$ with itself for every $p$. Restriction of $i$ to $S^{n-1}$ induces isomorphism of $H_p(S^{n-1})$ with $H_p(B^n\setminus\{0\})$ for every $p$ because $S^{n-1}$ is deformation retract of $B^n\setminus\{0\}$. Hence, by the naturality of long exact homology sequence of pair and five lemma it follows that $i$ induces isomoprhism of $H_P(B^n,S^{n-1})$ with $H_p(B^n,B^n\setminus\{0\})$ for every $p$.

Is my argument ok?

Jean Marie
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Madara Uchiha
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  • I don't know what you mean by "$i$ induces a homotopy equivalence". Inclusion $i$ is not homotopy equivalence, as it was shown here http://math.stackexchange.com/questions/1911324/inclusion-is-not-homotopy-equivalence. – Madara Uchiha Sep 03 '16 at 12:34

1 Answers1

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Notice that $S^{n-1}$ is homotopic to $B^n \setminus \{0\}$ (in fact a deformation retract.

From this, we have the following short exact sequences:

$\dots H_n(S^{n-1}) \to H_n(B^n)\to H_n(B^n,S^{n-1})\to H_{n-1}(S^{n-1})\to H_n(B^n) \dots$

and

$\dots H_n(B^n \setminus \{0\}) \to H_n(B^n)\to H_n(B^n,B^n \setminus \{0\})\to H_{n-1}(B^n \setminus \{0\})\to H_n(B^n) \dots$

since $S^{n-1}$ and $B^n \setminus \{0\}$ are homotopic, there is an isomorphism $\phi_*$ between the first and last written elements, and likewise, the second and fourth have isomorphisms induced by the identity map.

Now apply the five lemma

Andres Mejia
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  • Sorry, I just read the body of your question. For the sake of having an answer, I'll leave this here, but: yes, your argument is perfect. – Andres Mejia Apr 07 '18 at 21:07