Talk:Homology braid II (Ex)

1) The diagram is composed of the exact sequences of the pair $(\widetilde W,\widetilde M)$$<wikitex>; '''1)''' The diagram is composed of the exact sequences of the pair $(\widetilde W,\widetilde M)$, and of the triple $(\widetilde W,\widetilde M\cup\widetilde M',\widetilde M)$, and of the analogous sequences with $\widetilde M$ and $\widetilde M'$ swapped. Because $\widetilde M$ and $\widetilde M'$ are disjoint, we have an isomorphism $H_k(\widetilde M\cup\widetilde M',\widetilde M')\cong H_k(\widetilde M)$ and another one with $\widetilde M$ and $\widetilde M'$ swapped. Both are implicit in the braid. The squares that are half visible on the left and right hand side obviously commute. The upper triangles commute because of the following diagram, $$\xymatrix{ H_{i+1}(\widetilde W,\widetilde M) \ar[d] \ar[r]^{\partial} & H_i(\widetilde M) \ar[d]^{\cong} \ar[r] & H_i(\widetilde W) \ar[d] \ H_{i+1}(\widetilde W,\widetilde M\cup\widetilde M') \ar[r]^{\partial} & H_i(\widetilde M\cup\widetilde M',\widetilde M') \ar[r] & H_i(\widetilde W,\widetilde M')}$$ The lower triangles work similarly. The middle square only commutes up to sign: let~$a\in C_{i+1}(\widetilde w)$ be a relative cycle in~$W$, so $\partial a=b+c\in C_i(\widetilde M)\oplus C_i(\widetilde M')$. Then the upper path maps $[a]$ to $[b]$, and the lower path maps $[a]$ to $[c]=-[b+\partial a]=-[b]$. To repair this, put an additional minus sign on all boundary arrows landing in $H_\bullet(\widetilde M)$. '''2)''' We have learned that $W$ is homotopy equivalent to $M\cup_{g|}D^{k+1}$. Let $\tilde g|\colon S^k\to\widetilde M$ be a lift (ok if $k\ge2$) If we pass to the universal cover of $M$, we are attaching a $\pi$-cell $\pi\times D^{k+1}$ by $(\gamma,t)\mapsto\gamma(\tilde g(t))$. Excision gives $$H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) \cong \Zz \pi H_i(D^{k+1},S^k).$$ Similarly, swapping $W$ is homotopy equivalent to $M'\cup_{S^{n-k}}D^{n+1-k}$. This gives the second assertion. --~~~~ <wikitex>(\widetilde W,\widetilde M)$, and of the triple $(\widetilde W,\widetilde M\cup\widetilde M',\widetilde M)$$(\widetilde W,\widetilde M\cup\widetilde M',\widetilde M)$, and of the analogous sequences with $\widetilde M$$\widetilde M$ and $\widetilde M'$$\widetilde M'$ swapped. Because $\widetilde M$$\widetilde M$ and $\widetilde M'$$\widetilde M'$ are disjoint, we have an isomorphism $H_k(\widetilde M\cup\widetilde M',\widetilde M')\cong H_k(\widetilde M)$$H_k(\widetilde M\cup\widetilde M',\widetilde M')\cong H_k(\widetilde M)$ and another one with $\widetilde M$$\widetilde M$ and $\widetilde M'$$\widetilde M'$ 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,

$\displaystyle \xymatrix{ H_{i+1}(\widetilde W,\widetilde M) \ar[d] \ar[r]^{\partial} & H_i(\widetilde M) \ar[d]^{\cong} \ar[r] & H_i(\widetilde W) \ar[d] \\ H_{i+1}(\widetilde W,\widetilde M\cup\widetilde M') \ar[r]^{\partial} & H_i(\widetilde M\cup\widetilde M',\widetilde M') \ar[r] & H_i(\widetilde W,\widetilde M')}$

The lower triangles work similarly.

The middle square only commutes up to sign: let~$a\in C_{i+1}(\widetilde w)$$a\in C_{i+1}(\widetilde w)$ be a relative cycle in~$W$$W$, so $\partial a=b+c\in C_i(\widetilde M)\oplus C_i(\widetilde M')$$\partial a=b+c\in C_i(\widetilde M)\oplus C_i(\widetilde M')$. Then the upper path maps $[a]$$[a]$ to $[b]$$[b]$, and the lower path maps $[a]$$[a]$ to $[c]=-[b+\partial a]=-[b]$$[c]=-[b+\partial a]=-[b]$. To repair this, put an additional minus sign on all boundary arrows landing in $H_\bullet(\widetilde M)$$H_\bullet(\widetilde M)$.

2) We have learned that $W$$W$ is homotopy equivalent to $M\cup_{g|}D^{k+1}$$M\cup_{g|}D^{k+1}$. Let $\tilde g|\colon S^k\to\widetilde M$$\tilde g|\colon S^k\to\widetilde M$ be a lift (ok if $k\ge2$$k\ge2$)

If we pass to the universal cover of

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$M$, we are attaching a $\pi$$\pi$-cell $\pi\times D^{k+1}$$\pi\times D^{k+1}$

by $(\gamma,t)\mapsto\gamma(\tilde g(t))$$(\gamma,t)\mapsto\gamma(\tilde g(t))$. Excision gives

$\displaystyle H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) \cong \Zz \pi H_i(D^{k+1},S^k).$ $$\displaystyle H_i(\widetilde W,\widetilde M)\cong H_i(\widetilde M\cup_{\pi\times S^k}(\pi\times D^{k+1}),\widetilde M) \cong \Zz \pi H_i(D^{k+1},S^k).$$

Similarly, swapping $W$$W$ is homotopy equivalent to $M'\cup_{S^{n-k}}D^{n+1-k}$$M'\cup_{S^{n-k}}D^{n+1-k}$. This gives the second assertion.

--~~~~