Critical points go to critical points under automorphism I'm working on $S^2 \cong \mathbb{C} \cup \{\infty\}$. It's well known that mobius transformations are the only automorphisms of $S^2$. 
Suppose I have a holomorphic function $f$ and a mobius transformation $m$, and I define $g = m\circ f \circ m^{-1}$. Then I want to show that if $\beta$ is a critical point of $f$, then $m(\beta)$ will be a critical point of $g$.
I just have one hitch in the (elementary) proof, so far I have this:
For later use let $m = (az + b)/(cz+d)$, then $m' = (cz + d)^{-2}$. Letting mobius transforms have determinant 1 so I'm assuming that $ad - bc = 1$.
I use chain rule on $g$:
\begin{align}
g'(z) &= (m'fm^{-1}(z)) \times (f'm^{-1}(z)) \times (m^{-1})'(z)\\
g'm &= (m'f(z)) \times (f'(z)) \times (m^{-1})'m(z)\\
&= (cf(z) + d)^{-2} \times f'(z) \times (-cm(z) + a)^{-2}\\
&= \frac{f'(z)(cz + d)^2}{(cf(z) + d)^{2}}
\end{align}
Now assuming that the denominator isn't $0$, this works. However, I can't see what to do when it is $0$. I've tried using L'hopital, but without much luck.
Alternatively, if someone knows a more insightful proof, I would be very glad.
 A: $g\circ m = m\circ f$. Therefore $$(g'\circ m)m' = (m'\circ f)f'$$ If $f' = 0$, then either $g'\circ m = 0$, or $m' = 0$. The only place where $m' = 0$ is at $\infty$.
So this gets you that every critical point of $f$ except possibly for $z = \infty$ is also a critical point of $g\circ m$. To finish it, you need to examine what exactly it means for $\infty$ to be a critical point for $f$.
A: Instead of viewing a critical at $x$ of $f$ as $f'(x) = 0$, view it geometrically. Let $x$ be a critical point of $f$ if $f$ is not injective in any neighbourhood of $x$. (As defined by Beardon, as well as differentiation textbooks). Now the claim follows trivially since $m$ is an automorphism. 
The two statements $f'(\infty) = 0$ and $\infty$ critical point are unrelated. 
Take $f = \frac {z} {z^2 - 1}$. Then $f' = \frac{-z^2 - 1}{(z^2 - 1)^2}$. Indeed $f'(\infty) = 0$, yet $\infty$ is not a critical point.
Take $f = z^2$. Then $f' = 2z$. Indeed $\infty$ is a critical point, but $f'(\infty) = \infty \neq 0$.
However, for a finite point, $f'(z) = 0$ is a still a valid test for critical points.
