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If we take the connected sum of four closed disks $S = 4 \mathbb{\overline{D}} = \mathbb{\overline{D}} \# \mathbb{\overline{D}} \# \mathbb{\overline{D}} \# \mathbb{\overline{D}}$, what does $S$ look like and how do we describe the boundary? (This has been resolved) Is it just a $2$-sphere, since $ \mathbb{\overline{D}} \# \mathbb{\overline{D}} = \mathbb{S}^2$ and $ \mathbb{S}^2 \# \mathbb{S}^2 = \mathbb{S}^2?$

How do we write down a plane model for this surface? If $S = 4 \mathbb{\overline{D}}$ is an annulus, would it just have the same plane model as the cylinder? (Since they are homeomorphic)

St Vincent
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    It's the 2-sphere with four open discs deleted; equivalently a disc with three open discs deleted from the interior. –  Nov 09 '15 at 05:45
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    I'm sorry, I looked back and I'm a little bit confused by your statement $D^2 # D^2 = S^2$. I think you have a different definition of connected sum than I do. Can you clarify why this is true? –  Nov 09 '15 at 05:50
  • @Mike Miller I am probably just confused but, the connected sum of two closed disks is just the $2$-sphere, and in my class we defined the $2$-sphere to be the identity element of the connected sum operation – St Vincent Nov 09 '15 at 05:53
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    You're stating that to be true - I'm asking why it's true. Can you define connected sum for me? –  Nov 09 '15 at 05:55
  • My class hasn't gone over a formal definition yet, but we have a description (from my notes): "We introduce an operation to construct new surfaces from other surfaces. Take a point $x_1$ on a surface $S_1$ and a point $x_2$ on $S_2$, with $D_i$ a small disk neighborhood of $p_i$ in $S_i$. Cut out and discard the interiors of the disks $D_i$ and glue the resulting circle boundaries to each other. The new surface is called the connected sum of $S_1$ and $S_2$, which we denote $S_1 # S_2$." – St Vincent Nov 09 '15 at 06:06
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    That sounds pretty close to the formal definition to me! So one thing to keep in mind: when $S_1$ is the disc, that point $x_1$ should be in the interior. When you delete that small neighborhood, you get the annulus $S^1 \times I$. Now delete the small neighborhood (from the interior of) $S_2$. Gluing on the $S^1 \times I$ to the boundary does not change the homeomorphism type (convince yourself of this). So $D^2 # \Sigma$, where $\Sigma$ is some surface, is precisely what you get by deleting a small open ball from $\Sigma$. In particular, $D^2 # D^2 = S^1 \times I$. –  Nov 09 '15 at 06:09

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First of all, as Mike Miller points out in his comment, $D^2 \# D^2 \approx S_1 \times [0,1] $ which is just a cylinder. So what you get is the connected sum of two cylinders.

In terms of writing down the planar model for this surface, this question of mine will be useful: What is the word associated with the connected sum of two surfaces with boundary?. The current answer is right for manifolds without boundary but not for manifolds with boundary. I found the answer for manifolds with boundary and I'll post it soon, but the result is that the word associated to the connected sum of surfaces with boundary (with associated words $p_1$ and $p_2$) is $c^{-1}p_1cp_2$.

The cylinder has as an associated word $aba^{-1}c$, so the word associated to the connected sum of two cylinders is:

$$g^{-1}aba^{-1}cgded^{-1}f$$

With this you can create the planar model and classify the surface using the classification of compact manifolds.

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