# connected sum of surfaces is well defined proof attempt

Suppose $S_1$ and $S_2$ are compact surfaces (connected 2-dimensional manifolds). If we cut out of them two closed disks, and glue the surfaces along disk boundaries we get new surface, their connected sum, denoted by $S_1\#S_2$. I have precise definition in Massey's Algebraic topology, but couldn't find proof that definition is well:

Take disks $D_1 \subset S_1$ and $D_2 \subset S_2$, and define $S_i' = S_i \setminus Int(D_i)$ for $i=1,2$. Choose homeomorphism $h$ sending boundary of $D_1$ to boundary of $D_2$. Then $S_1 \# S_2$ is defined as a quotient space of $S_1' \cup S_2'$ with $x$ and $h(x)$ being identified for $x \in \partial D_1$.

My attempt (not finished):

Take now $\hat{D}_1 \subset S_1, \hat{D}_2 \subset S_2$, $\hat{S}_i' = \hat{S}_i \setminus Int(\hat{D}_i)$ for $i=1,2$ and another homeomorphism $\hat{h}$ mapping $\partial \hat{D}_1$ onto $\partial \hat{D}_2$. Construct a quotient space of $\hat{S}_1' \cup \hat{S}_2'$ where $x$ and $\hat{h}(x)$ are identified for $x \in \partial D_1$.

We must be able to construct homeomorphism $g$ between those two quotient spaces. We know that there exist homeomorphism $f$ mapping boundary of $\hat{D}_2$ onto boundary of $D_2$.

I want to define some homeomorphism on certain subsets of first space and then proceed by gluing lemma. Now I propose $g_1(x) = (h^{-1} f \hat{h})(x)$ for $x$ in the equivalence class of $\partial D_1$ and $g_2(x) = (h^{-1} f) (x)$ for $x$ in the equivalence class of $\partial D_2$. And now I don't have idea what to do with other points and have concern since both of those boundaries are closed as sets and what remains in complement is open... So I'm stuck here, since gluing lemma couldn't be applied, even if I would manage to define anything on the rest of the space.

Any idea, comment?

• What are you trying to show? Is it that if you take different disks from the boundary, the connected sum i.e the resulting manifolds are homeomorphic? – Faraad Armwood May 7 '16 at 23:35
• Exactly. And consequently a different homeomorphism between boundaries. – user133929 May 7 '16 at 23:38

It is highly nontrivial that the connected sum of manifolds is well-defined. First, you actually want to assume that you can thicken your embeddings of discs (the term is "locally flat"); otherwise one of your discs might be one side of the Alexander horned sphere! You want to prove that every embedding of the disc $D^n \hookrightarrow M$ is isotopic (meaning that the two embeddings are homotopic through locally flat embeddings). Two reasonable exercises are to show that if $f$ and $g$ are embeddings, then $g$ is isotopic to an embedding whose image is contained in the interior of the image of $f$; and to prove that if all (locally flat) embeddings of discs are isotopic, then the connected sum is well-defined. You'll then want the "isotopy extension theorem" for locally flat embeddings.

Now what you want to invoke is the annulus theorem to actually construct the final isotopy. This is not easy! You'll note that the Wikipedia article implies that the connected sum of manifolds was only finally proved well-defined in 1982. (It's a bit easier to prove that the connected sum of smooth manifolds is well-defined.)

If you are cynical about such theorems and demand to see proofs before you use these results, the way you should interpret theorems about $M \# N$ is "For every connected sum of $M$ and $N$, ..."

• Mild nitpick: note that it is really meant that every equivalently oriented embeddings $D^n \hookrightarrow M$ are isotopic. E.g., an orientation reversing embedding and orientation preserving embedding are not in general isotopic. It's worth noting that taking connected sum by removing two equally oriented embeddings and gluing the boundaries together and by removing two oppositely oriented embeddings often results in topologically different objects: e.g., $\Bbb{CP}^2 \#\Bbb{CP}^2$ and $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}$. – Balarka Sen May 8 '16 at 1:32
• To clarify my last statement: in one case one glues by identity along the removed balls, and in the other case by the antipodal map. The point that's being tried to convey is that orientation is important for specifying the resulting object even in the unoriented category, i.e., topologically. Maybe this is not so relevant though. – Balarka Sen May 8 '16 at 1:47
• @BalarkaSen Thanks for catching the error. The statement actually proved here (more or less) is "Given embeddings $f, g$, then $f$ is either isotopic to $g$ or its reflection." (Actually, all I do is show that $f$ is isotopic to an embedding with the same image as $g$; what's left is to understand the space of embeddings of $D^n$ into itself, which one can easily see has precisely two components.) – user98602 May 8 '16 at 2:32

Mike Miller's answer brings up some really interesting points in the general topological case. For surfaces the topological category and smooth category coincide, so let's use some differential topology to give an intuitive picture.

It's enough to be able produce a homeomorphism $p: S_1 - D_1 \to S_1 - \hat{D}_1$ that is orientation-preserving on $\partial D_1 \to \partial \hat{D}_1$. (Note that such a map will automatically induce a homeomorphism $\partial D_1 \to \partial \hat{D}_1.$ Then we can just do the same for $S_2$, and glue the pieces together, and we'll have a homeomorphism on the whole deal.)

$S_1$ is path-connected, so take $d_1, \hat{d}_1$ the centers of $D_1, \hat{D}_1$ and $\gamma: [0,1] \to S_1$ a path with $\gamma(0) = d_1$ and $\gamma_1 = \hat{d}_1$. Now we'll apply the tubular neighborhood theorem to the image of $\gamma$, which must be compact. So we get some open set $U \subset S_1$ that's an open neighborhood of $\gamma([0,1])$, and hence contains small disks $U_1$ and $\hat{U}_1$ around $d_1$ and $\hat{d}_1$.

Then we simply construct our desired homeomorphism in three stages:

1. Construct a homeomorphism $S_1 - D_1 \to S_1 - U_1$. (Just shrink!)

2. Transport $U_1$ inside $U$ to $\hat{U}_1 \ni \hat{d}_1$, also contained in $U$, so that we have a homeomorphism $S_1 - U_1 \to S_1 - \hat{U}_1$. An explicit construction would be unilluminating: let's have a picture instead. Think of $U$ as soap stretched across a long, skinny, ellipsoidal ring, and $\overline{U_1}$ as a small ring inside $U$ that doesnt' contain any soap. Drag $\overline{U_1}$ to $\overline{\hat{U}_1}$; let the homeomorphism be the identity outside $U$, and let it describe how the soap particles move when we move the ring.

3. Construct a homeomorphism $S_1 - \hat{U}_1 \to S_1 - \hat{D}_1$. (Just expand!)

The composition of these three maps gives you the desired homeomorphism $S_1 - D_1 \to S_1 - \hat{D}_1.$

• Where in your answer do you use smooth topology? The subtlety here is in the "Just expand" line, and that statement is subtle; it was only actually proved by Palais in the 60s (though is still a good exercise, given sufficient pointers). – user98602 May 8 '16 at 2:30
• @MikeMiller Certainly the tubular neighborhood theorem is smooth topology. – Thurmond May 8 '16 at 3:15
• Or locally flat topology, if you like. :) The standard proof that manifolds are path-connected can be extended to prove that manifolds are path-connected by paths that have tubular neighborhoods. – user98602 May 8 '16 at 3:17
• @MikeMiller Fair enough. Regarding "just expand", can't you take a small neighborhood of $\overline{D_1}$ - I think you can make it topologically trivial using the compactness assumption - and then just do a "straight-line expansion" there using a map into $\mathbb{R}^2$, which you can assume has image some open ball centered at 0. Admittedly this is all inherently using the 2-dimensionality; I agree that in higher dimensions things get much scarier! – Thurmond May 8 '16 at 3:19
• What you would need to do that is for the smaller ball to be star-shaped with center 0 in your chart. How do you set this up? – user98602 May 8 '16 at 3:21