Surgery on $S^m$

Question

On page 4 of the book "ALGEBRAIC AND GEOMETRIC SURGERY" by Andrew Ranicki, after the definition of surgery has written:

Example View the $m$-sphere $S^m$ as $$S^m=\partial (D^{n+1} \times D^{m-n})=S^n \times D^{m-n} \cup D^{n+1}\times S^{m-n-1}$$

The surgery on $S^m$ removing $S^n\times D^{m-n}\subset S^m$ converts the $m$-sphere $S^m$ into the product of spheres $$D^{n+1}\times S^{m-n-1}\cup D^{n+1}\times S^{m-n-1}=S^{n+1}\times S^{m-n-1}$$

Question: why $S^m=\partial (D^{n+1} \times D^{m-n})$?

and how does one derive $D^{n+1}\times S^{m-n-1}\cup D^{n+1}\times S^{m-n-1}=S^{n+1}\times S^{m-n-1}$?

Thanks for your help.

Answer 1

Think of disks as products of intervals, so $$D^{n+1}\times D^{m-n} \cong D^{n+1+m-n} = D^{m+1}$$ and $S^m = \partial D^{m+1}$.

The surgery on $S^m$ removes $S^n\times D^{m-n}$ and glues in a new product of spheres and disks along its boundary. Since $\partial (S^n\times D^{m-n}) = S^n\times S^{m-n-1} = \partial(D^{n+1}\times S^{m-n-1})$, you can glue in a $D^{n+1}\times S^{m-n}$ in place of the $S^n\times D^{m-n}$ along $S^n\times S^{m-n-1}$. This gives:

\begin{align*}S^m &= S^n\times D^{m-n}\cup D^{n+1}\times S^{m-n-1} \\ &\xrightarrow{\mbox{surgery}} D^{n+1}\times S^{m-n-1}\cup D^{n+1}\times S^{m-n-1} \end{align*}

To see this last expression is a product of spheres, observe that we should more precisely write $$D^{n+1}\times S^{m-n-1}\cup_{S^n\times S^{m-n}} D^{n+1}\times S^{n-m-1}.$$ Since we're gluing by the identity on the boundary, the two $D^{n+1}$ factors glue to a $S^{n+1}$, giving $S^{n+1}\times S^{m-n-1}$.

Answer 2

$S^m=\partial (D^{n+1} \times D^{m-n})$ since $S^m=\partial D^{n+1}$ and $D^k\times D^j \simeq D^{k+j}$; for your second question remember the rule for boundary of product spaces: $\partial (X \times Y)=(\partial X)\times Y \cup X \times (\partial Y)$.