Talk:Homology braid II (Ex)
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 central 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.
3) By part 2, we get the following diagram.
This proves part (a).
To prove part (b), we decompose using the above braid as
Now let be a Morse function on with and such that there is only one critical point of index . Fix a gradient vector field for such that the stable and unstable disks are exactly the core disk of the -handle and its dual. We can multiply this vector field with a positive function such that the flow moves points in to in time except in a small neighbourhood of the image of . All this lifts to .
Now, let be a smooth cycle in that is transversal to , which is the image of , for all . The gradient flow moves from to except in a small neighbourhood of the image of . Hence, only depends on the intersections of with the various~ for . This is the main step in the proof of (b).
$ except in a small neighbourhood of the image of $g|$. All this lifts to $\widetilde W$. Now, let $y$ be a smooth cycle in $\widetilde M$ that is transversal to $\gamma x$, which is the image of $\gamma \circ \tilde g$, for all $\gamma\in\pi$. The gradient flow moves $y$ from $M$ to $M'$ except in a small neighbourhood of the image of $\gamma\circ\tilde g$. Hence, $\alpha(y)$ only depends on the intersections of $y$ with the various~$\gamma x$ for $\gamma\in\pi$. This is the main step in the proof of (b).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 central 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.
3) By part 2, we get the following diagram.
This proves part (a).
To prove part (b), we decompose using the above braid as
Now let be a Morse function on with and such that there is only one critical point of index . Fix a gradient vector field for such that the stable and unstable disks are exactly the core disk of the -handle and its dual. We can multiply this vector field with a positive function such that the flow moves points in to in time except in a small neighbourhood of the image of . All this lifts to .
Now, let be a smooth cycle in that is transversal to , which is the image of , for all . The gradient flow moves from to except in a small neighbourhood of the image of . Hence, only depends on the intersections of with the various~ for . This is the main step in the proof of (b).