Talk:Homology braid II (Ex)
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by $(\gamma,t)\mapsto\gamma(\tilde g(t))$. | by $(\gamma,t)\mapsto\gamma(\tilde g(t))$. | ||
Excision gives | Excision gives | ||
− | $$H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) | + | $$H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) \cong \Zz \pi H_i(D^{k+1},S^k).$$ |
− | \cong \ | + | |
Similarly, swapping $W$ is homotopy equivalent to $M'\cup_{S^{n-k}}D^{n+1-k}$. | Similarly, swapping $W$ is homotopy equivalent to $M'\cup_{S^{n-k}}D^{n+1-k}$. | ||
This gives the second assertion. | This gives the second assertion. |
Revision as of 10:01, 29 March 2012
1) The diagram is composed of the exact sequences of the pair , and of the triple , and of the analogous sequences with and swapped. Because and are disjoint, we have an isomorphism and another one with and swapped. Both are implicit in the braid.
The squares that are half visible on the left and right hand side obviously commute. The upper triangles commute because of the following diagram,
The lower triangles work similarly.
The middle square only commutes up to sign: let~ be a relative cycle in~, so . Then the upper path maps to , and the lower path maps to . To repair this, put an additional minus sign on all boundary arrows landing in .
2) We have learned that is homotopy equivalent to . Let be a lift (ok if ) If we pass to the universal cover of , we are attaching a -cell by . Excision gives
Similarly, swapping is homotopy equivalent to . This gives the second assertion.
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