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For $r >0$, let $K_r \subseteq \mathbb{C}$ be the closed subset, $K_r = \mathbb{C} \setminus D(0,r)$. Define $S_r$ to be the quotient of $K_r$ under the identification:

$$ z \sim -z, \hspace{1cm} z\in \partial D(0,r)$$

Then, $S_r$ is a topological surface, as it is locally homeomorphic to $\mathbb{R}^2$, Hausdorff and second countable. I think that it can also be given the structure of an abstract smooth surface with an appropriate atlas of charts.

Question: is there any meaningful sense in which the limit of $S_r$ can be defined as $r \to 0$?

I was thinking about this as I learnt about on how the Riemann sphere $\hat{\mathbb{C}}$, functions which are meromorphic over $\mathbb{C}$ can be considered as holomorphic thanks to the addition of the "point at infinity". I am curious about what could happen if instead of adding $\{\infty\}$ to $\mathbb{C}$, we remove $0$ -- resulting in a twice-punctured sphere -- then try to close off (or even glue together) the ends of the "cylinder" to which this twice-punctured sphere is homeomorphic.

legionwhale
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  • If I understand the construction, "yes": Each quotient is a Möbius strip whose central circle has length $\pi r$. Letting $r \to 0$ effectively collapses the central circle to a point, yielding the plane. The concept of blowing up in algebraic geometry is related; there are some diagrams here. – Andrew D. Hwang May 23 '22 at 14:45
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    @AndrewD.Hwang Thanks for the answer. I haven't yet studied algebraic geometry, so the link you gave looks quite intimidating, but I should have a better understanding of all this next year. Could I ask two more questions? I'm quite surprised that the construction in the question gives the (punctured?) plane. Doesn't the twist in the centre affect anything? I am also curious what would happen if I now defined a surface by quotienting $A_{r,R} = {r \le |Z| \le R}$, with $z \sim -z$ on both boundary circles, and letting $r \to 0, R \to \infty$. Does this also give the plane? – legionwhale May 23 '22 at 19:17

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Here's a concrete version of the linked post: In polar coordinates $z = |z|e^{i\theta}$, define a mapping from $K_{r}$ to the Cartesian product of the complex plane with itself by $$ f(z) = (z(1 - r/|z|), rz^{2}/|z|^{2}) = ((\rho-r)e^{i\theta}, re^{2i\theta}). $$ This mapping is continuous, identifies antipodal points (and no others) on the circle of radius $r$, and the first coordinate maps the interior of $K_{r}$ bijectively (in fact diffeomorphically) to the punctured plane. Bijectivity in the first coordinate corresponds to the surprising fact that the Möbius-band, despite its twist, collapses to a disk when the central circle is collapsed. In other words, a Möbius band without its central circle is diffeomorphic to a punctured disk.

To see what happens in the annulus $\{r \leq |z| \leq R\}$, we may glue two copies of the preceding picture (or write down an appropriate mapping), yielding a Klein bottle (connected sum of two Möbious bands, a.k.a., cross caps) that collapses to a sphere (connected sum of two disks) as $r \to 0$ and $R \to \infty$.

Andrew D. Hwang
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