In the first section of "A procedure for killing homotopy groups of differentiable manifolds", Milnor gives the surgery construction as follows. Let $W$ be an $n=p+q+1$ dimensional manifold. Given a smooth, orientation preserving embedding:
$$f:S^p\times D^{q+1}\to W$$
we may obtain a new manifold as the disjoint sum
$$(W-f(S^p\times 0))\cup (D^{p+1}\times S^q)$$
modulo some equivalence relation.
My question is: why in this construction is Milnor deleting $S^p\times 0$ from $W$ and not $S^p\times D^{q+1}$? I expected this disjoint sum to be
$$(W-f(S^p\times D^{q+1}))\cup (D^{p+1}\times S^q).$$
Best Answer
There is no error in the paper as far as I can tell. One thinks of the surgery as taking out the interior of the image of $S^p\times D^{q+1}$, and glueing in a copy of $D^{p+1}\times S^q$ along the common boundary $S^p\times S^q$. However, that does only define a topological space and not an orientable differentiable manifold. That's why one takes $W'=W\setminus f(S^p\times\{0\})$ and glues in $D^{p+1}\times S^q$ on the overlap $$ f(S^p\times (D^{q+1}\setminus\{0\}))\cong (D^{p+1}\setminus\{0\})\times S^q $$ the identification being $f(u,\theta v)\leftrightarrow (\theta u,v)$, for $(u,v)\in S^p\times S^q$ and $0<\theta\leq 1$.