# Regarding the surgery construction in "A procedure for killing homotopy groups of differentiable manifolds" by Milnor

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

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

• Kosinski's text "Differentiable Manifolds" goes into the rationale for using this version of surgery (and an analogous construction for handle attachments) in considerable detail, in case you are looking for another source. Jan 24 at 16:48

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 $$0<\theta\leq 1$$.