Revision as of 21:24, 21 February 2017

Problem statement

Let [ilmath]n \geq 1[/ilmath]. We show that the map [ilmath]H_k(B^n, S^{n-1}) \to H_k(B^n, B^n \setminus \{ 0 \}) [/ilmath] by the inclusion [ilmath]i: S^{n-1} \to B^n \setminus \{0\} [/ilmath] is an isomorphism.

Tools

First we show that the map [ilmath]H_k(S^{n-1}) \to H_k(B^n \setminus \{0\})[/ilmath] induced by the inclusion [ilmath]i: S^{n-1} \to B^n \setminus \{0\} [/ilmath] is an isomorphism, for [ilmath] k \geq 0[/ilmath]. Note that [ilmath]S^{n-1} [/ilmath] is a retract of [ilmath] B^n \setminus \{0\} [/ilmath].

• Define the map [ilmath] r: B^n \setminus \{ 0 \} \to S^{n - 1} [/ilmath] by [ilmath] r: x \mapsto \frac{x}{\vert\vert x \vert\vert} [/ilmath], where [ilmath] \vert\vert x \vert\vert [/ilmath] denotes the norm of [ilmath]x[/ilmath].
• Then [ilmath] r \circ i \eq \mathrm{id}_{S^{n-1} } [/ilmath].
• So [ilmath] i_*: S^{n - 1} \to B^n \setminus \{ 0 \} [/ilmath] is a monomorphism.
• Also, [ilmath] i \circ r[/ilmath] is homotopy equivalent to the identity map on [ilmath] B^n \setminus \{ 0 \} [/ilmath]. [left as an exercise to the reader]

Final proof

Alec's formatting