Show that $f$ extends to a smooth map. 
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.
 A: 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!
