I encountered the following question in complex analysis:
Let $P(z)$ be a polynomial such that if $P(z)\in \mathbb{D}$ then $z\in \mathbb{D}$ and $P'(z)\neq 0$. Prove that there exists an analytic function $f(z)$ on $\mathbb{D}$ such that $P(f(z))=z$ for all $z\in \mathbb{D}$.
The question seems innocuous but I am not sure where to start with. It is supposed to be solved using some analytic continuation and Riemann surface knowledge. Any help would be greatly appreciated.
On
Let's give a slightly different solution than the ones above - note that if $0<r<1, C$ the unit circle, $P^{-1}(rC)$ is a union of $n=\deg P$ Jordan curves $J_1(r),...J_n(r) \subset \mathbb D$ and then $P^{-1}(r\mathbb D)$ is the union of their interiors
(since $P$ is a proper map the preimage is compact and by hypothesis, it is included in the open unit disc, hence each component is traversed only once as $P' \ne 0$ inside and on it, but $P$ is a $\deg P$ to $1$ map etc - more generally any level curve that doesn't contain a critical value of $P$, where $P$ is a proper analytic map on some domain, is a Jordan curve traversed $m+1$ times, where $m$ is the number of critical points of $P$ inside it).
In particular, if we fix $a \in P^{-1}(0)$ and take $J_1(r)$ the Jordan curve containing $a$ inside and $U_r$ its interior, we get a Riemann map $g_r:\mathbb D \to U_r$, hence $P\circ g_r$ is a conformal bijection of $\mathbb D$ onto $r\mathbb D$ and by composing with an automorphism of the unit disc we get $f_r:\mathbb D \to U_r$ with $P \circ f_r(z)=rz$. But $f_r$ is a normal family on the unit disc so taking a normally convergent subsequence of $f_{r_n}, r_n \to 1$ we get our $f$ for which $P\circ f(z)=z, z \in \mathbb D$
Since $P$ is a polinomial, it is surjective, and thus $\forall z \in \mathbb{D}\ f^{-1}(\{z\})\neq \emptyset$. Thanks to the hypotesis, we obtain that $f^{-1}(\{z\})\in \mathbb{D}$. Thus $P: P^{-1}(\mathbb{D})\twoheadrightarrow \mathbb{D}$. Also, since $P'(z)\neq 0$, $P$ is locally invertible, i.e. given a point $z\in \mathbb{D}$ there exists a function $f$ defined in a neighbourhood of $z$ which satisfies $P\circ f=\text{Id}$.
Let us define $f_0$ as a local inverse of $P$ at $0$ (i.e. $P(f(0))=0$), defined on the open ball $U_0$. Given a path $\gamma:[0,1]\to \mathbb{D}$, we can extend $f_0$ on this path: To prove it, let $\Omega$ be the subinterval of $[0,1]$ such that $\gamma_{|\Omega}$ admits an analytic continuation of $f_0$.
$\Omega$ is non-empty, as it contains $0$.
$\Omega$ is closed , since given a cluster point $t$, we can define the analytic continuation in a neighborhood of $\gamma(t)$ as a local inverse of $P$ compatible with the analytic continuations in $\gamma(t^*);t^*<t$
$\Omega$ is open, since for every $t$ the analytic continuation $(f_t,U_t)$ in $\gamma(t)$ is also valid in $\gamma^{-1}(U_t)$ .
Thus $\Omega=[0,1]$, and $f_0$ can be analytically extended along every path in $\mathbb{D}$. Since $\mathbb{D}$ is simply connected, by the monodromy theorem we have an analytic extension $f$ of $f_0$ defined on $\mathbb{D}$.
It remains to prove that $P\circ f=\text{Id}$. This follows from the identity principle, since it is true in $U_0$
$(P^{-1}(\mathbb{D}),P)$ is a cover of $\mathbb{D}$, and since $\mathbb{D}$ is simply connected, there exists a function $f:\mathbb{D}\to P^{-1}(\mathbb{D})$ that lifts the identity, i.e. such that $f\circ P=\text{Id}$. However, since $P$ is locally invertible, and left and right inverse must be equal if they both exists, we have that, on $U_0$, $f=f_0$. The fact that $P\circ f=\text{Id}$ follows by the identity principle as in (1).