Consider the extension of the map$$f \colon \Bbb C \setminus \{1,-1\} \to \Bbb C$$ $$f \colon \frac{1}{z-1} -\frac{1}{\bar{z}+1}$$ to a map $$\tilde{f} \colon \Bbb S^2 \to \Bbb S^2$$ By adding to $f$ the relations $(1,\infty),(-1,\infty),(\infty,0)$. I believe I proved that the map $\tilde{f}$ is continuous in this way, so it makes sense to try and compute the degree of $\tilde{f}$ as a map of spheres. It's easy to see that $\tilde{f}$ is smooth, using the local charts of the riemann sphere given by the identity and $z \mapsto \frac{1}{z}$.
My definition of degree is the homological one (sum of local degrees, given by the sign of the Jacobian)
1) First of all, if I work far from the points $\pm 1, \infty$, my map $\tilde{f}$ coincides with $f$, so I have a nice formula for it.
2) By trial and error I found that $2/3 \in \Bbb C \hookrightarrow \Bbb S^2$ is a regular value, whose preimage is $f^{-1}(2/3)=\{2,-2\}$. By computing the Jacobian in these two points it appears that they both have positive sign so by the local degree theorem, $\deg \tilde{f}=2$.
I'd like to have some feedback on this approach, I'm mostly interested in knowing if there is a quicker way to do this kind of computations, and if there are observations to help me decide if my result makes sense or no.