Question about Automorphism group of $\mathbb{H}$

Question

I already know/proved, that $Aut(\mathbb{H})$ consists of all Mobius-transformations $\Phi_M=M\tau$ where $M \in SL_2(\mathbb{R})$.

I know that they are bijective since $\Phi_M^{-1}=\Phi_{M^{-1}}$, i.e there exists an inverse function. However, I am not sure why these functions are holomorph because afaik there is a Pole of 1. order at $\tau=-d/c$.

EDIT: Since the first question was answered fairly quickly. I'd like to ask another one: How can we know that there is no other function in $Aut(\mathbb{H})$ other then the Mobius-transformations.?

Answer 1

With $\tau \mapsto 1-\frac{1}{1/2-i\tau}$ you can send $H = \{Im(\tau) > 0\}$ to the unit disk $D$.

Then with Schwarz's lemma show the automorphisms of $D$ are the Möbius transformations $c\frac{z-a}{\bar{a}z-1}$ where $|c|=1$ and $|a| < 1$ :

Let $\phi(z)$ being an arbitrary automorphism of $D$. Choosing the Möbius automorphism of $D$ : $f(z) = \frac{z-\phi(0)}{\bar{\phi(0)}z-1}$ we have $g(z) = f \circ \phi (z)$ an automorphism of $D$ satisfying $g(0) = 0$.

Hence by Schwarz lemma, $g(z) = c z = f \circ \phi (z)$ for some $|c| =1$, i.e. $ \phi(z) = f^{-1}(cz)$ was a Möbius automorphism of $D$.

Finally, let $W(\tau) = 1-\frac{1}{1/2-i\tau}, T(z) = W^{-1}(z) = \frac{i}{1-z}-i/2$,

you get that if $\varphi(z)$ is an automorphism of $H$ then $\phi(z) = W \circ \varphi \circ T(z)$ is a Möbius automorphism of $D$ so that $\varphi(z)$ was indeed a Möbius automorphism of $H$.