Schwarz Lemma (Theorem 4.3.4.13 of Ahlfors Complex Analysis). Let $f$ be holomorphic on $B_1(0)$ such that $|f(z)|\leq 1$ and $f(0)=0$. Then, for all $z$, $|f(z)|\leq |z|$ and $|f'(0)|\leq 1$. If $|f(z)|=|z|$ for some $z\neq 0$, or if $|f'(0)|=1$, then $f(z)=cz$ with a constant $c$ of absolute value $1$.
Lemma. Let $\mu(z)=\frac{z-a}{\overline{a}z-1}$ for a fixed $a\in B_1(0)$ and each $z\in \bC$. Then $\mu(B_1(0))= B_1(0)$.
Proof. First note that $|\mu(0)|=|a|<1$. Furthermore,
$$|\mu(1)|=\left|\frac{1-a}{\overline{1-a}}\right|=1,|\mu(-1)|=\left|\frac{1+a}{\overline{a+1}}\right|=1$$
and
$$|\mu(i)|=\left|\frac{i-a}{\overline{a}i-1}\right|=\left|\frac{i-a}{-\overline{a}-i}\right|=\left|\frac{i-a}{\overline{i-a}}\right|=1.$$
So $\mu$ is a Möbius transformation sending $\partial B_1(0)$ to $\partial B_1(0)$. Since $0$ is sent into $B_1(0)$, then $\mu(B_1(0))=B_1(0)$.
Exercise. Let $g:B_1(0)\to B_1(0)$ be a holomorphic function such that $g(\frac{1}{2})=0$. Then $|g(0)|\leq \frac{1}{2}$.
Proof. We'll use part of the proof of the Schwarz-Pick theorem, along with the facts that the Möbius transformations used are involutions.
Let $$\mu(z)=\frac{\frac{1}{2}-z}{1-\frac{z}{2}}.$$
Take $f=g\circ\mu$. Then $f(B_1(0))\subseteq B_1(0)$
and $f(0)=g\left(\frac{1}{2}\right)=0.$
Applying Schwarz Lemma, $|f(z)|\leq |z|$ for all $z$. But then,
$$\left|g\left(\frac{\frac{1}{2}-z}{1-\frac{z}{2}}\right)\right|\leq |z|,$$
which for $z=\frac{1}{2}$ gives
$$\left|g(0)\right|\leq \frac{1}{2},$$
the desired result.
Note: In the proof of the Schwarz-Pick Theorem the inverse $\mu^{-1}$ is used instead of $\mu$, and there is a second Möbius transformation which in this case is $\lambda(z)=-z$. So in this case $\mu=\mu^{-1}$ and $\lambda$ can be removed.
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