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Let $f:D(0,1) \to D(0,1)$ be a holomorphic map where $f$ has two distinct fixed points. Could anyone advise me how to prove $f$ is identity map? Do I use Schwarz-Pick somewhere?

Let $g(z)=f(z)-z.$ Then $ |zg(z)|\leq \dfrac{a-z}{1-\overline{a}z}, \dfrac{b-z}{1-\overline{b}z},$ where $a,b$ are the distinct fixed points. How should I proceed from here?

Thank you.

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Denote the fixed points by $\mu$ and $\eta$ ($\mu\ne\eta$) and define $$\varphi_{\mu}(z)=\frac{\mu-z}{1-\bar\mu z},$$then $\varphi_\mu$ is an automorphism of $D(0,1)$ such that $\varphi_{\mu}(0)=\mu$, $\varphi_{\mu}(\mu)=0$ and $\varphi_{\mu}\circ \varphi_{\mu}=\rm{Id}_{D(0,1)}$. (You can check this or see it in Stein's book.)

Now, it follows that $$\varphi_{\mu}\circ f\circ\varphi_{\mu}(0)=0\quad \text{and}\quad\varphi_{\mu}\circ f\circ \varphi_{\mu}\circ\varphi_{\mu}(\eta)=\varphi_{\mu}(\eta).$$

Since $\varphi_{\mu}\circ f\circ\varphi_{\mu}$ is a holomorphic map from $D$ to $D$ and $\varphi_{\mu}(\eta)\ne 0$, by Schwarz Lemma, it follows that $\varphi_{\mu}\circ f\circ\varphi_{\mu}=\rm{Id_{D}}$, hence $f=\rm{Id_{D}}$.

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Hint: By composing with suitable automorphisms of $D(0, 1)$, you can construct a map which has two fixed points, one of which is $0$. Now see what the Schwarz Lemma tells you.

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  • $\begingroup$ I am sorry. I still cant come up with that automorphism now $\endgroup$ – Alexy Vincenzo Sep 27 '14 at 8:29

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