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Identify $\mathbb{R}^2$, with coordinates $x, y$, with $\mathbb{C}$, with coordinate $z = x + iy$. Likewise, identify a copy of $\mathbb{R}^2$ with coordinates $u, v$ with $\mathbb{C}$ with coordinate $w = u+iv$. Let $f : \{\mathbb{R}^2 - (1, 0) - (-1, 0)\} \to \mathbb{R}^2$ be the function $f(z) = \frac{1}{z-1} - \frac{1}{\bar{z}+1}$. Show that $f$ extends to a smooth map $\tilde{f} : S^2 \to S^2$, where $S^2$ (or $\mathbb{CP}^1$ is the one-point compactification of $\mathbb{R}^2$ (or $\mathbb{C})$.

Following FBD's comment below, I want to show:

  • 1, $f$ is continuous, which is obvious;

  • 2, $f$ is surjective;

  • 3, $f^{-1}$ maps compact set to compact set

Correct?

Thank you very much.

WishingFish
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  • Do you want something more than "Hints: Define $f(\infty)$ to be what you want it to be--zero. Define the image of the poles to be $\infty$. Then, just work in charts."? – Alex Youcis Aug 13 '13 at 02:38
  • Thanks @AlexYoucis, but no, I don't have a direction - at all. Do you mind givingme some.....? Thank you! – WishingFish Aug 13 '13 at 02:56
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    Where are you having trouble? I told you what you need to do, which part of it seems hard? – Alex Youcis Aug 13 '13 at 03:58
  • To start with, I don't know how to show $f$ extends (I mean this type of question), in general. @AlexYoucis.. – WishingFish Aug 13 '13 at 04:00
  • There is a result that continuous surjective maps extend iff the inverse image of a compact set is compact. You can check first that the extended map is continuous in the topology of the compactification, and then , like Alex said, you can use charts on $S^2$ to show the differentiability of the extension; show that $(phi)ofo(\csi^{-1})$ is a differentiable as a map from $R^2$ to itself, where $\phi$, $\csi$ are charts, but you need to make sure your charts cover every point in $S^2$ – FBD Aug 13 '13 at 04:51
  • So maybe you can start by, like Alex said, defining the image at the two poles; introduce $\infty$ the "point at infinity" , and also define the value of $f(\infty)$, then show the map is continuous ( you need to know what is the topology of the compactified space), by, e.g., showing inverse image of open sets is open. Then find manifold chart maps for $S^2$, and show the extended map is differentiable. Remember that differentiability is independent of the choice of chart; if the map is differentiable in one chart , it is differentiable under any change of charts. – FBD Aug 13 '13 at 04:56
  • Thank you very much for your comment @FBD, it is very helpful. Just to confirm - so I want to show, 1, $f$ is continuous, which is obvious; 2, $f$ is surjective; 3, $f^{-1}$ maps compact set to compact set - right? – WishingFish Aug 13 '13 at 19:44
  • @WishingFish, yes correct. I don't think you're ready for these types of problems; I honestly think (based on all of your previous questions) that your main problem that you have to overcome first is understanding basic Multivariable Calculus and then basic Point-Set Topology. Please heed this message. – Chris Gerig Aug 13 '13 at 20:38
  • @ChrisGerig As I told you before, it is way too late for me to drop the course. And I can not just give up and stop showing up for classes and exam! But I appreciate your help and advice. I'll try to go to classes you mentioned in the fall. – WishingFish Aug 13 '13 at 20:42

1 Answers1

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O.K, I think the comments are not enough to give a good suggestion, so let me expand a bit; sorry, I don't have too much time, but I'll do my best:

Re my suggestion of showing the map is a proper map, while this is true, I think it will not be helpful here, because we would have to show surjectivity, etc. which may be too hard.

So, let's get started by extending the map from a map $f: \mathbb R^2 \rightarrow \mathbb R^2$ , into a map $f^ :S^2 \rightarrow S^2 $, seeing $S^2$ as the 1-pt. compactification of $\mathbb R^2$ , i.e., $S^2= \mathbb R^2 \cup$ {$\infty$}. My suggestion is that we define $f(1,0)=f(-1,0)=f(\infty)=\infty$.

Now, we need to check that , as defined, this map is continuous, as a map from $S^2$ into $S^2$ , but, notice that the 1-pt compactification has a specific topology you can find, e.g., in Munkres; I think it is called the Alexandrov compactification, but I am not sure. You can do this, e.g., by showing that the inverse image of an open set is open.

Once you have done that, you need to find manifold charts for this one-point compactification; i.e., charts that make this particular topological space into a manifold. I think the charts for the Riemann sphere will do.

Given the charts, you need to show that , for every x in $S^2$ (as the domain), and for f(x) in , say $S^{2'}$ (to distinguish the domain from the codomain ), you have a chart $ (U_x,\phi_x)$ containing x, and a chart $( V_{f(x)}\beta_{f(x)}$ containing f(x) , so that the composition $\beta \circ f \circ \phi^{-1} $ is smooth, as a map from $R^2$ to itself.

At some point, it would be a good idea to show that a continuous, surjective map between spaces X,Y extends continuously into the respective compactifications iff it is a proper map, i.e., if the inverse image of every compact set is compact.

I'll be busy , but I'll try to get back in as soon as I can. Good Luck!

Pedro
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FBD
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  • Hi FBD, thank you so much for your generous help! I think at this point, I get most of the part, except for the local chart. Do you mind showing me a bit more on how to show $\beta \circ f \circ \phi^{-1}$ is smooth? Thank you so much! – WishingFish Aug 14 '13 at 06:09
  • No problem, WishingFish; glad to help. You need to find the actual chart maps that make $S^2$ a manifold under the topology of 1-pt. compactification. I think if you see $S^2$ as a Riemann surface, these charts will make $S^2$ with the guven topology into a manifold. Notice that the composition $\beta o f o \phi^{-1} $ is a map from a subset of $R^2$ into a subsetof $R^2$. Then you can apply the standard rules and results for differentiability/smoothness of maps between subsets of $\mathbb R^n $. Let me know of any other question. – FBD Aug 14 '13 at 06:16
  • Sorry, I gotta go sleep now, but will check back later. – FBD Aug 14 '13 at 06:17
  • Same here, just trying to put some of your words into my dream before I go to sleep. Thank you so much FBD. – WishingFish Aug 14 '13 at 06:18
  • Hi FBD - thanks for your great answer. May I request for a more explicit form of the local charts? Perhaps they are two patches covers $S^2$ by stereographic projection covering $\infty, (\pm 1, 0)$? – 1LiterTears Aug 16 '13 at 02:54
  • Hi,yes, I'm pretty sure that it is the stereographic projection. – FBD Aug 27 '13 at 04:19