I need to prove that the map $$f \colon \Bbb R^2 \setminus \{(1,0)\} \cup \{(-1,0)\} \to \Bbb R^2$$ $$f(z)= \frac{1}{z-1} -\frac{1}{\bar{z}+1}$$ extends to the one-point compactification $\Bbb S^2$ of $\Bbb R^2$ where we identified $\Bbb R^2$ with $\Bbb C$ in the usual way.
Since I have to define $f$ on $\infty, 1,-1$ in order to extend it to the one-point compactification (so not only on the infinity point I add), do I proceed as always? I.e. I'd try to show that $f$ is proper.
1) Assuming my approach is correct, how can I prove that such map is proper? Are there any clever way to do this? My approach is to try and prove that the preimage of a ball of radius $r<\infty$ is contained in a ball of radius $R<\infty$, but I'm not able to control its behaviour near the critical points.
2) I'm asked to compute its degree: since it's not surjective ($0$ is not hit) then it should be $0$ am I right?
If you want to show that the map extends, then you have to see if there is a limit for the case that the argument converges against $\pm 1$ or $\infty$.
You can do this quite comfortably using simple instruments from real analysis, considering that convergence of a sequence against $\infty$ (in the one-point compactification) is equivalent to the absolute value converging against $\infty$ in $\mathbb{R} \cup \{\infty\}$.
Long story short:
$$ \forall \left(z_n\right)_n \in \left(\mathbb{C}\cup\{\infty\}\right)^\mathbb{N} \left( \lim_n z_n = \infty \iff \lim_n |z_n| = \infty \right) $$ Now, argue with the limit definition for continuity...
You will find out that the map is surjective; so even though the implication is right, you can't apply it.