# Connected sum of surfaces

I'm looking for a detailed reference about connected sums. I'd like it to contain a proof that a connected sum of connected surfaces is independent - up to homeomorphism - of the various choices involved in the process. There are several books in which it is stated, but I cannot find one in which it is proved. In particular, I do not see a simple argument implying that changing the orientation of the circle in the glueing has no influence on the homeomorphism class of the surface. I'm aware that for compact surfaces, this point more or less follows from the classification, but then what about non compact ones?

Any suggestion?

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What do you mean by "non-compact"? Finite type with punctures, or something more general? –  Igor Rivin Feb 6 '12 at 19:54
Maybe I should have said that I take the word "surface" in the topological sense, i.e. a topological space that is separated and locally homeomorphic to $\mathbb{R}^2$. Thus, by non compact, I simply mean a surface in the above sense, that is not compact as a topological space. There is a well-known classification theorem for compact surfaces (they have a finite number of connected components and these are all connected sums of (a sphere and) tori and projective spaces). –  Baptiste Calmès Feb 6 '12 at 21:08
True or false: every orientable surface has an orientation-reversing homeomorphism? –  Tom Goodwillie Feb 6 '12 at 22:38
Baptiste -- a small remark: did you in fact mean connected sum of connected surfaces? –  algori Feb 6 '12 at 23:34
This follows from the classification of non-compact surfaces by Richards: ams.org/journals/tran/1963-106-02/S0002-9947-1963-0143186-0/… –  Ian Agol Feb 7 '12 at 3:55

For surfaces, smooth and topological classification is the same, so let me argue in the smooth category.

In Bröcker and Jänichs book "Einführung in die Differentialtopologie" (I have seen references to an English translation), the following result is shown in full detail. Let $M_0$ and $M_1$ be two connected smooth manifolds and $D_i \subset M_i$ be two embedded discs. The diffeomorphism type of $M_0 \sharp M_1$ (formed along the two discs) only depends on the orientation-behaviour of the discs $D_i$ (if $M_i$ is orientable) and it does not depend on the choice of the discs if $M_i$ is nonorientable. Moreover, if both manifolds admit orientation-reversing self-diffeomorphisms, then the diffeomorphism type of $M_0 \sharp M_1$ does not depend on any choices. As indicated by Toms comment, the question is whether any surface admits an orientation-reversing diffeomorphism.

Theorem: ''Every smooth connected orientable surface $M$ admits an orientation-reversing involution.''

Proof: the closed case is easy (by a picture), so let us assume that $M$ is open. Take a handlebody decomposition of $M$, we can arrange it so that $M$ has only one $0$-handle and no $2$-handles. Now $M$ is the union $D^2=M_0 \subset M_1 \subset M_2 \subset \ldots$. $M_{n+1}$ is obtained from $M_{n}$ by gluing in a copy of $D^1 \times D^1$ along an orientation-preserving embedding $S^0 \times D^1 \to \partial M_{n}$.

Now construct the orientation-reversing diffeomorphism $f$ inductively. On $M_0 =D^2$, take a reflection. Assume that we have constructed $f_n:M_n \to M_n$ with the extra property that $f_n$ induces the identity on $\pi_0 (\partial M_n)$ (i.e., $f_n$ does not permute the components of the boundary) and each boundary component has a parametrization by $S^1 \subset R^2$ so that $f_{n+1}$ is given by $(x,y) \mapsto (x,-y)$ in these coordinates. There are two cases to distinguish in the induction step.

1st case: The attaching embedding $S^0 \times D^1 \to \partial M_n$ takes the two copies of the interval into two different components. We can find an isotopic embedding $S^0 \times D^1 \to \partial M_n$ that is $Z/2$-equivariant with respect to the already constructed $f_n$ and the self-map $(x,y) \mapsto (x,-y)$ of $D^1 \times D^1$. This is possible by the form of $f_n$ on the boundary. The involution extends to an involution $f_{n+1}$ of $M_{n+1}$ and because the two components where the embedding lands in are joined together, the new $f_{n+1}$ does not permute the boundary components.

2nd case: The attaching embedding takes both components of $S^0 \times D^1$ into the same component of $\partial M_{n+1}$. This time we isotope the attaching embedding so that it becomes equivariant with respect to $f_n$ and $(x,y) \mapsto (-x,y)$. This is possible by taking a non-fixed point of $f_n$ on the boundary. The component is split into two parts, but the way I have arranged the gluing guarantees that the new $f_{n+1}$ still does not permute the components.

Note that a straightforward adaption of that argument also settles the closed case.

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Thanks for this answer. However, although I did not mention it in the question, I would like to avoid any argument using a differentiable structure. Somehow, the spirit of my question is: "How come at the beginning of various introductory textbooks on algebraic topology (ex: Massey), this topological fact is stated as true and intuitive, although there seems to be no proof avoiding some kind of classification result, itself quite non-trivial in the non compact case? Am I missing some elementary argument, or is there really a lot of work swept under the carpet at that point?". –  Baptiste Calmès Feb 7 '12 at 10:48
I would say that there is a lot of work swept under the carpet. Unless the involution exists, things can go completely wrong. For example, $CP^2 \sharp CP^2$ is not even homotopy equivalent to the manifold $CP^2 \sharp \bar{CP^2}$ that you obtain if one of the embedded discs is negatively oriented. So the surface case is special. If you make a picture of my construction, you will find it ''intuitive''. P.S: I really think that the smooth structure makes everything easier. –  Johannes Ebert Feb 7 '12 at 14:57

A relatively clean and intuitive proof is given in Kosinski's "Differential Manifolds," which works in the topological setting and essentially boils down to the following:

If $M$ is path-connected and $i_1,i_2:D\rightarrow M$ are isotopic embeddings (smooth or topological), then by the so-called "Cerf-Palais disk theorem" (a consequence of the Isotopy Extension Property) there is an ambient isotopy $\Phi:M\times I\rightarrow M$ (smooth or topological) such that for all $t$, $\Phi_t$ is identity outside a contractible compact set, and $\Phi(i_1(x),1)=i_2(x)$. Intuitiely, $\Phi$ translates the image of $i_1$ to the image of $i_2$, and tries hard to not effect anything else.

So if $M, i_1, i_2$ are as above, $N$ is another topological (or smooth) manifold and $i:D\rightarrow N$ is another embedding, then let $M\\#_1N$ be formed by attaching $N\setminus i(0)$ to $M\setminus i_1(0)$, and form $M\\#_2N$ using $M\setminus i_2(0)$. Since these objects are actually pushouts, we can define a homeo(diffeo)morphism in pieces: If $y\in N\setminus i(0)$, send $y$ to itself; if $y\in M\setminus i_1(0)$, send $y$ to $\Phi(y,1)$. These will assemble to give the required equivalence from $M\\#_1N$ to $M\\#_2N$. (then you could repeat the argument on the $N$ side, or just say it follows from commutativity)

The fact that the connected sum is associative and commutative follows naturally from the fact that it is actually a pushout (if you're careful, it does make a pushout in the smooth category). Then to show that it doesn't depend on the attaching disk, I think you need something equivalent to the "Cerf-Palais" theorem I mentioned.

Edit: because of what was mentioned in the comments above, it was necessary for me to assume that the embeddings were isotopic to begin with

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Whoops, I may have made a huge gaff: does Isotopy Extension Property apply to Topological manifolds? I know the original question is just about surfaces, but I'm wondering more generally. –  William Feb 9 '12 at 0:32
Thanks for your answer. In my mind, though, the important case is the one where the maps are not isotopic (i.e. the surface is orientable, and the disks are glued in opposite ways). –  Baptiste Calmès Feb 9 '12 at 9:33
Ah, well then in that case it is as Johanns said: there's no guarantee that the connected sums are even homotopy equivalent without some restriction on the embeddings or on the manifolds. –  William Feb 9 '12 at 22:29
@William. I am talking about surfaces. I know about the classical counter example in dimension 4 that Johannes Ebert was mentioning (captured by homology, by the way). However, it has to be true for surfaces, even non-compact ones, in view of the classification result mentioned by Agol. But this is definitely not a simple argument, and given no one has come up with an obvious trick, it certainly isn't intuitive before classification that connected sums of oriented surfaces are homeomorphic, whatever glueing you use. –  Baptiste Calmès Feb 11 '12 at 12:19
It seems like it comes down to the case where you are embedding the disk into an oriented surface with embeddings who have different orientation parity (reversing/preserving), because in every other case you get an isotopy of the embeddings. Maybe there is an elementary cut-and-paste argument you could give for this case. After all, being able to prove with cut-and-paste techniques is what makes the theory of surfaces so elementary. –  William Feb 11 '12 at 23:23