Schwarz's lemma $\Rightarrow$ an analytic conformal map UHP$\to$UHP must be an FLT? I read a solution to a conformal mapping problem that made the claim, "Schwarz's lemma implies that any analytic conformal map taking the upper half-plane to the upper half-plane must be a fractional linear transformation." I had not heard this, and I wonder why this is true. 
If you don't care to write out the whole proof, could you perhaps provide me with a reference?
Apologies if this is a repeat, I wasn't able to find this question on the site.
 A: Let $\mathbb H$ be a halfplane, and $\mathbb D$ the unit disk. We know there is a Möbius map $\phi:\mathbb H\to\mathbb D$. Given a holomorphic map $f:\mathbb H\to\mathbb H$, we can form $F=\phi\circ f\circ \phi^{-1}$, which is a holomorphic map $F:\mathbb D\to\mathbb D$. This is a very common thing to do, called the conjugation of $f$ by $\phi$. 
There is a Möbius map of $\mathbb D$ that sends $F(0)$ to $0$. Compose it with $F$ to get  $G:\mathbb D\to\mathbb D$ such that $G(0)=0$. So far we did not need any special assumptions on $f$. 
If $f$ is a bijection, then so is $G$. Applying the Schwarz lemma to $G$ and $G^{-1}$, we see that $|G(z)|\equiv |z|$; consequently, $G$ is a rotation. Unwinding the transformations that led from $f$ to $G$, we see that $f$ is a Möbius map.
Remark. It suffices to assume that $f$ is a surjection with nonvanishing derivative. In fact, if $f:\Omega\to \Omega'$ is a surjective holomorphic map with $f'\ne 0$, and $\Omega'$ is simply-connected, then $f$ is a bijection. This is actually a topological fact: a covering map onto a simply-connected space is a homeomorphism. In terms of holomorphic functions, this is given by the Monodromy theorem: since the (potentially multivalued) function $f^{-1}$ can be analytically continued in simply-connected domain $\Omega'$, it is single-valued there.
A: As a reference, consider Bak and Newman, Theorem 13.17, which I summarize here.
Essentially, you need to prove that all automorphisms of the upper half-plane are of the form $$f(z) = \frac{az+b}{cz+d}, ad-bc > 0, a,b,c,d \in \mathbb{R}.$$
It is easy to see that if $z$ is real-valued, then so too must be $f$.
Next, using $\textrm{Im} z = \frac{z-\overline{z}}{2i}$, we have
$$\textrm{Im}(f(z)) = \frac{ad-bc}{c^2+d^2} > 0,$$
so $f$ must map $i$ into the upper half-plane. This is sufficient to show that $f$ is an automorphism of the upper half-plane.
To show uniqueness, consider an automorphism from the upper half-plane onto the unit disc:
$$g(z) = e^{i\theta}\left(\frac{z-\alpha}{1-\overline{\alpha}z}\right), |\alpha| < 1$$
and consider the most basic automorphism, just discovered,
$$h(z) = \frac{z-i}{z+i}.$$
Then, $h^{-1} \circ g \circ h$ is an automorphism of the upper-half plane (this is a very trivial lemma).
Simply carrying out the arithmetic, and you will find that any automorphism of the upper half-plane must be of the desired form.
