Talk:Homology braid II (Ex)
(Created page with "<wikitex>; '''1)''' The diagram is composed of the exact sequences of the pair $(\widetilde W,\widetilde M)$, and of the triple $(\widetilde W,\widetilde M\cup\widetilde M',\...") |
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To repair this, put an additional minus sign on all boundary arrows | To repair this, put an additional minus sign on all boundary arrows | ||
landing in $H_\bullet(\widetilde M)$. | landing in $H_\bullet(\widetilde M)$. | ||
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+ | '''2)''' | ||
+ | We have learned that $W$ is homotopy equivalent to $M\cup_{g|}D^{k+1}$. | ||
+ | Let $\tilde g|\colon S^k\to\widetilde M$ be a lift (ok if $k\ge2$) | ||
+ | If we pass to the universal cover of $M$, we are attaching a $\pi$-cell $\pi\times D^{k+1}$ | ||
+ | by $(\gamma,t)\mapsto\gamma(\tilde g(t))$. | ||
+ | Excision gives | ||
+ | $$H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) | ||
+ | \cong \Z\pi H_i(D^{k+1},S^k).$$ | ||
+ | Similarly, swapping $W$ is homotopy equivalent to $M'\cup_{S^{n-k}}D^{n+1-k}$. | ||
+ | This gives the second assertion. | ||
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Revision as of 08:58, 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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