This seems like such an egregious error in such an otherwise solid book that I felt I should ask if anyone has noticed to be sure I'm not misunderstanding something basic.

In his section on connect sums, Kosinski does not seem to acknowledge that, in the case where the manifolds in question do not admit orientation reversing diffeomorphisms, the topology (in fact homotopy type) of a connect sum of two smooth manifolds may depend on the particular identification of spheres used to connect the manifolds.

His definition of connect sum is as follows. For embeddings $h_i: \mathbb{R}^m \rightarrow M_i, i=1,2$ (if the $M_i$ are oriented, choose $h_1$ to be orientation preserving and $h_2$ orientation reversing) and orientation reversing diffeomorphism $\alpha : (0, \infty) \rightarrow (0,\infty)$, define $\alpha_m : \mathbb{R}^m-0 \rightarrow \mathbb{R}^m-0 , v \mapsto \alpha (|v|) v/|v|$ (I.e. perform $\alpha$ along radial lines). Then form $M_1 \# M_2$ by gluing along the map $h_2 \alpha_m h_1^{-1}$.

His theorem 1.1 on page 90 reads

"$M_1 \# M_2$ is a smooth manifold, connected if $m>1$ and oriented if both $M_i$ are oriented. It does not depend - up to diffeomorphism- on the choice of $\alpha$ and the embeddings $h_i$."

The mistake in the proof seems to come at the bottom of page 91 when he claims: "if $h'$ and $h_1$ both embed $\mathbb{R}^m$ as a proper tubular neighborhood of a point $p$ in $M_1$ then there is a diffeomorphism $g: M_1 \rightarrow M_1$ such that $gh_1 = h'$."

Clearly if $h'$ and $h_1$ disagree in orientation then such a $g$ can only exist if $M_1$ admits an orientation reversing diffeomorphism.

Later on page 95 he claims in Theorem 2.2, "The set of connected oriented and closed m-dimensional manifolds is, under the operation of connected sum, an associative and commutative monoid with identity [...] of course, 2.2 holds without assuming the manifolds are oriented"

Am I missing something?

  • $\begingroup$ As the textbook says on the bottom of pg 91 (at least in my 1993 edition), the existence of your g comes from Theorem 3.3.5 (pg 51) applied here with M={p}, N=M1, and F0, F1 your embeddings of Rm as tubular nbds of p. Then take g=H1 (as in the theorem statement). This has nothing to do with orientations. $\endgroup$ – user31714 Jan 11 '13 at 23:14
  • $\begingroup$ @user31714 : Yes but as I read theorem 3.3.5 it only guarantees a diffeo $g$ of $M_1$ such that $gh_1$ is a diffeomorphism onto $h(\mathbb{R}^m)$. It does not guarantee that you can make this composition any embedding $h$ that you have in mind. Maybe I'm misreading or misunderstanding. $\endgroup$ – Tim kinsella Jan 12 '13 at 0:26
  • $\begingroup$ ...in particular $h$ must be isotopic to $h_1$ relative to $h(0)=h_1(0)$! $\endgroup$ – Tim kinsella Jan 12 '13 at 0:33

I think there is no conceptual difficulty at here. For his definition of connected sum we have:

  1. Two manifolds $M_{1},M_{2}$ with the same dimension in the gluing spot;
  2. Two embedding $h_{1},h_{2}$ from the Euclidean space to $M_{1},M_{2}$;
  3. An orientation reversing differeomorphism of the real line which we use to induce an orientation reversing differeomorphism of the Euclidean space minus a point;

  4. A resulting manifold generated by glueing $M_{1},M_{2}$ along the embedding Euclidean space (or sphere).

The objection you raised that $h_{1}$ or $h_{2}$ is only well-defined up to the orientation of the underlying manifold is not really justified. By assumption if both spaces are oriented already, then the orientation of $h_{1}$ and $h_{2}$ are pre-determined. So two different $h_{2}$'s, for example should not give rise to an orientation reversing differeomorphism you claimed. Clearly this definition was proposed with objects like $\mathbb{C}\mathbb{P}^{2}$ at mind.

I disagree that Kosinski's book is solid though. There are numerous difficult proofs and various errors found out by our topology group last semester (we used this one to understand $h$-corbordism theorem). So if you feel really confused you should consult other sources or even the original paper in some of the topics.

  • $\begingroup$ Do you maybe have an erratum of the book? $\endgroup$ – Marko Karbevski Sep 27 '16 at 20:58

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