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

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

  • 1
    $\begingroup$ 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. $\endgroup$ Jan 24 at 16:48

1 Answer 1


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 $0<\theta\leq 1$.

  • 5
    $\begingroup$ It's helpful to visualize this for connected sums of surfaces. Instead of removing a whole disk from each surface and attaching along the boundary, imagine only removing the two center points, and then deforming the punctured disks into cylinders sticking out of the surface, and then you attach the cylinders together. $\endgroup$
    – Pedro
    Jan 24 at 14:52
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    $\begingroup$ Indeed, one needs to identify collars rather than glue along the boundary in order to get a differentiable structure on the result. $\endgroup$
    – Jim Conant
    Jan 24 at 16:01

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