The problem is this:
Let $V\subset\mathbb{C}$ be a simply connected domain with $V\neq\mathbb{C}$ and let $a,b\in V$ with $a\neq b$. Let $g:V\rightarrow V$ be analytic with $g(a)=a$ and $g(b)=b$. Prove that $g(z)=z$ for every $z\in V$.
I am assuming this involves applying the Riemann Mapping Theorem (probably the uniqueness part) since it uses an analytic map defined on a proper simply connected domain and concludes with uniquely identifying the function with required properties. But one problem with this is that the function $g$ is not necessarily a bijection.
Also, could be expecting to use Schwarz' Lemma after composing with appropriate Mobius map(s)?
Any help would be greatly appreciated.
Best Answer
Let $\phi: V \to \Bbb C$ be a conformal mapping from $V$ onto $\Bbb D$ with $\phi(a) = 0$, this exists according to the Riemann mapping theorem.
Then $f = \phi \circ g \circ \phi^{-1}$ is a holomorphic function from the unit disk into itself, with $f(0) = 0$. Apply the Schwarz Lemma to $f$ and note that $f$ has another fixed point, different from zero.