Let $f: \Omega \rightarrow \Omega$ be holomorphic where $\Omega \subset \mathbb{C}$ is a bounded region containing 0. If $f(0)=0$ and $f'(0)=1$, does $f(z)=z$?
This is true if $\Omega$ is a disk centered at 0 but does it hold if $D$ is only bounded?
Let $f: \Omega \rightarrow \Omega$ be holomorphic where $\Omega \subset \mathbb{C}$ is a bounded region containing 0. If $f(0)=0$ and $f'(0)=1$, does $f(z)=z$?
This is true if $\Omega$ is a disk centered at 0 but does it hold if $D$ is only bounded?
There is a substantial generalization of Schwarz' Lemma valid for arbitrary maps $f:\ \Omega\to\Omega$ of a Riemann surface into itself. It says that any such map which is not a conformal automorphism of $\Omega$ actually decreases the hyperbolic distance between points. This implies that if $|f'(z)|=1$ at some fixed point $z$ then $f$ is a conformal automorphism. This can be proven by lifting the map $f$ to a map $\tilde f:\ \tilde\Omega\to\tilde\Omega$ of the universal cover of $\Omega$. The latter is ("in most cases") conformally equivalent to the unit disk, where Pick's invariant version of Schwarz' Lemma can be applied. See here, p. 27:
Huber, Heinz: Analytische Abbildungen Riemannscher Flächen in sich. Comm. Math. Helv. 27 (1953), 1–73.
Find a conformal map $h$ that maps $\Omega$ to the unit disc $D$, fixing $0$. Then $g:= h\circ f\circ h^{-1}:D\to D$ is holomorphic, with $g(0)=0$ and $g'(0)=1$.