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


$$(W-f(S^p\times D^{q+1}))\cup (D^{p+1}\times S^q).$$



Regarding the surgery construction in "A procedure for killing homotopy groups of differentiable manifolds" by Milnor
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 $0<\theta\leq 1$.


