2D global conformal transformations and the $z= \frac1w$ argument 
For instance in Blumenhagen's CFT, there is a standard argument which determines that globally defined conformal transformations on the Riemann sphere where 
$$l_n = -z^{n+1} \partial_z$$
  is an element of the Witt algebra. In this argument we note that $l_n$ is non-singular at $z=0$ only for $n\geq -1$. Also substituting $z=-\frac1w$, we find 
$$l_n = -\left(-\frac1w\right)^{n-1}\partial_w$$
  is non-singular at $w=0$ only for $n\leq +1$. Therefore the global transformations are generated by $\{l_{-1},l_0,l_1\}$.

Why is the substitution $z=-\frac1w$ special? For instance if I use $z = -\frac{1}{w^2}$ I could repeat the argument above and conclude $n \leq 1/2$. 
 A: The minus sign is not essential, so let's remove it for simplicity. An atlas of the Riemann sphere, say $M$ ( which is a 1-dimensional complex manifold), is given by two coordinate charts 


*

*$z:M\to\mathbb{C}$, which covers the whole $M$ except the "infinity" $i\in M$. 

*$w:M\to\mathbb{C}$,  which covers the whole $M$ except the "origin" $o\in M$.
There is a holomorphic transition map: $w=1/z$ in the overlap $M\backslash\{i,o\}$. In polar coordinates $z=re^{i\theta}$, we have $w=e^{-i\theta}/r$.
Therefore the monodromy $z\to e^{2\pi i}z$ leads to the correct corresponding opposite monodromy $w\to e^{-2\pi i}w$ on the other chart, i.e. if we circle around the origin $o$ one time in one chart, it corresponds to circle around the infinity $i$ one time in the opposite direction in the other chart.
In contrast, a transition map of the form $z =1/w^m$, where $m\in \mathbb{N}\backslash\{1\}$, as OP suggests, would lead to wrong monodromy properties $w\to e^{-2\pi i/m}w$, and would hence not correspond topologically to a Riemann sphere.
