proof of conformal Möbius circle map theorem

Let $f$ be a conformal map from the unit disk $\mathrm{\Delta }$ onto itself.Let $a=f(0)$. Let $g_a(z)=\frac{z-a}{1-\overline{a}z}$.Then $g_a\circ f$ is a conformal map from $\mathrm{\Delta }$ onto itself,with $g_a\circ f(0)=0$.Therefore, by Schwarz’s Lemma for all $z\in \mathrm{\Delta }$ $|g_a\circ f(z)|\le |z|$.

Because $f$ is a conformal map onto $\mathrm{\Delta }$,$f^{-1}$ is also a conformal map of $\mathrm{\Delta }$ onto itself.${(g_a\circ f)}^{-1}(0)=0$ so that by Schwarz’s Lemma$|{(g_a\circ f)}^{-1}(w)|\le |w|$ for all $w\in \mathrm{\Delta }$.Writing $w=g_a\circ f(z)$ this becomes $|z|\le |g_a\circ f(z)|$.

Therefore, for all $z\in \mathrm{\Delta }$ $|g_a\circ f(z)|=|z|$.By Schwarz’s Lemma, $g_a\circ f$ is a rotation.Write $g_a\circ f(z)=e^{i\theta }z$, or $f(z)=e^{i\theta }{g}_a^{-1}$.

Therefore, $f$ is a Möbius Transformation.