Difference between revisions of "User:Harold/AlgTop1"
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| <center><span style="font-size:1.2em;"><m>\xymatrix{ | | <center><span style="font-size:1.2em;"><m>\xymatrix{ | ||
| − | \ldots \ar[r] & H_k(S^{n-1}) \ar[r] & H_k(B^n) \ar[r] & H_k(D^n, S^{n-1}) \ar[r] & H_{k-1}(S^{n-1}) \ar[r] & H_{k-1}(B^n) \ar[r] & \ldots \\ | + | \ldots \ar[r] & H_k(S^{n-1}) \ar[r] \ar@2{->}[d]^{i_*} & H_k(B^n) \ar[r] \ar@2{->}[d]^{\text{Id}} & H_k(D^n, S^{n-1}) \ar[r] \ar@{.>}[d]^{i'_*}& H_{k-1}(S^{n-1}) \ar[r] \ar@2{->}[d]^{i_*} & H_{k-1}(B^n) \ar@2{->}[d]^{\text{Id} } \ar[r] & \ldots \\ |
\ldots \ar[r] & H_k(B^n-\{0\}) \ar[r] & H_k(B^n) \ar[r] & H_k(D^n, B^n-\{0\}) \ar[r] & H_{k-1}(B^n-\{0\}) \ar[r] & H_{k-1}(B^n) \ar[r] & ... | \ldots \ar[r] & H_k(B^n-\{0\}) \ar[r] & H_k(B^n) \ar[r] & H_k(D^n, B^n-\{0\}) \ar[r] & H_{k-1}(B^n-\{0\}) \ar[r] & H_{k-1}(B^n) \ar[r] & ... | ||
| + | }</m></span></center> | ||
| + | |- | ||
| + | ! Diagram | ||
| + | |} | ||
| + | <!-- | ||
| + | <harrynoob> ... -> H_k(B^n\0) -> H_k(B^n) -> H_k(B^n, B^n\0) -> H_k-1(B^n\0) -> H_k-1(B^n) -> ... | ||
| + | <harrynoob> ... -> H_k(R^n\0) -> H_k(R^n) -> H_k(R^n, R^n\0) -> H_k-1(R^n\0) -> H_k-1(R^n) -> ... | ||
| + | --> | ||
| + | {| class="wikitable" border="1" style="overflow:hidden;" | ||
| + | |- | ||
| + | | <center><span style="font-size:1.2em;"><m>\xymatrix{ | ||
| + | \ldots \ar[r] & H_k(B^n-\{0\}) \ar[r] \ar[d]^{\text{?} } & H_k(B^n) \ar[r] \ar[d]^{\text{?}} & H_k(B^n, B^n-\{0\}) \ar[r] \ar@{.>}[d]^{\text{?} }& H_{k-1}(B^n-\{0\}) \ar[r] \ar[d]^{\text{?} } & H_{k-1}(B^n) \ar[d]^{\text{?} } \ar[r] & \ldots \\ | ||
| + | \ldots \ar[r] & H_k(\mathbb{R}^n-\{0\}) \ar[r] & H_k(\mathbb{R}^n) \ar[r] & H_k(\mathbb{R}^n, R^n-\{0\}) \ar[r] & H_{k-1}(\mathbb{R}^n-\{0\}) \ar[r] & H_{k-1}(\mathbb{R}^n) \ar[r] & ... | ||
}</m></span></center> | }</m></span></center> | ||
|- | |- | ||
! Diagram | ! Diagram | ||
|} | |} | ||
Latest revision as of 22:57, 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]
