Let $f:U\rightarrow U$ be analytic, where $U \subsetneqq \mathbb{C}$ is a simply connected domain. Suppose that $f$ has at least two fixed points. Prove that $f$ is the indentity map.

I am able to prove the theorem in the case where $U = \mathbb{D}$ is the unit disc (the proof uses Schwarz lemma, and is discussed in this question).

The proof however skips the (rigorous) details of how the Riemann Mapping Theorem allows us to reduce from $U$ to $\mathbb{D}$. Any details are appreciated.


By the Riemann mapping theorem, there is a biholomorphism $\phi : U \to \mathbb{D}$. Now consider the map $\phi\circ f\circ\phi^{-1} : \mathbb{D} \to \mathbb{D}$. If $a, b \in U$ are fixed points of $f$, then $\phi(a), \phi(b) \in \mathbb{D}$ are fixed points of $\phi\circ f\circ\phi^{-1}$. By the result you have proved, $\phi\circ f\circ\phi^{-1} = \operatorname{id}_{\mathbb{D}}$, and hence

$$f = \phi^{-1}\circ\operatorname{id}_{\mathbb{D}}\circ\phi = \phi^{-1}\circ\phi = \operatorname{id}_U.$$

  • $\begingroup$ Thanks @MichaelAlbanese! This makes perfect sense. $\endgroup$ – GaussTheBauss Dec 20 '15 at 1:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.