Möbius transformations are differentiable Let's consider the Möbius transformation 
$$
f:\mathbb{C}\cup \{\infty\}\rightarrow\mathbb{C}\cup\{\infty\},z\mapsto \frac{az+b}{
-\overline{b}z+\overline{a}},
$$
with $a,b\in\mathbb{C}$ such that $|a|^{2}+|b|^{2}=1$.
I want to check that, considering $\mathbb{C}\cup \{\infty\}$ as a differentiable manifold, $f$ is a differentiable map. My first question is:
If I prove that $f$ is a holomorphic map considering $\mathbb{C}\cup \{\infty\}$ as a complex manifold, would this imply that $f$ is differentiable?
The next question is:
How can I prove that $f$ is holomorphic?
My problem is that I haven't found an atlas that allows me to prove $f$ is holomorphic.
 A: $\newcommand{\Cpx}{\mathbf{C}}$As you surely know, $z$ is a holomorphic local coordinate on $U_{\infty}$, the complement of $\{\infty\}$, and $w = 1/z$ is a holomorphic local coordinate on $U_{0}$, the complement of $\{0\}$. Your atlas for the sphere will be the maximal holomorphic atlas compatible with these two charts.
Pre- and post-composition of a Möbius transformation $T$ with either chart map is still a Möbius transformation. To prove a general Möbius transformation $T$ is holomorphic (hence smooth) as a mapping of the Riemann sphere, it therefore suffices to restrict $T$ to the complement of its pole in $\Cpx$, where the calculation is elementary.
In more detail, separately restrict $f$ to the following four sets, which cover the sphere:


*

*The intersection $U_{\infty} \cap f^{-1}(U_{\infty})$ (a.k.a., the complement of the pole of $f$ in $\Cpx$);

*The intersection $U_{0} \cap f^{-1}(U_{\infty})$ (a.k.a., the complement of the pole of $f$ in $U_{0}$);

*The intersection $U_{\infty} \cap f^{-1}(U_{0})$ (a.k.a., the complement of the zero of $f$ in $\Cpx$);

*The intersection $U_{0} \cap f^{-1}(U_{0})$ (a.k.a., the complement of the zero of $f$ in $U_{0}$);
In each case, pre- and post-compose $f$ with $z$ or $w$ to get a mapping from $\Cpx$ to itself.
