MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f: \mathbb{D} \rightarrow \mathbb{D}$ is analytic. Furthermore, suppose $f \circ f = g$, and $g$ is a linear fractional map. Does this guarantee that $f$ is linear fractional? I know it would be true if the domain was the sphere instead of the disk, or if $g$ was onto the disk (as we already have that $f$ is injective because $g$ is). But I want this to be true in general. Any insight?

share|cite|improve this question

Yes. Consider the sequence of open disks $$\mathbb D\subset g^{-1}(\mathbb D)\subset g^{-2}(\mathbb D)\subset \dots . $$ Let $\mathbb D'$ be the union. This is either an open disk or the complement of a point in the Riemann sphere. You can extend $f$ to $\mathbb D'$ by defining $f(z)=g^{-n}(f(g^n(z)))$ for sufficently large $n$. Now $f$ maps $\mathbb D'$ to itself in a one to one and onto fashion, so it is a linear fractional map.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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