If $h$ is analytic and non-zero at $e^{it}$ then (since $h(z)=\sum_k c_k (z-e^{it})^k$) so is $g(z)=1/\overline{h(\overline{z}^{\ -1})}$
If $h(e^{it}) = e^{i\theta}$ then $g(e^{it}) = e^{i\theta}$.
Thus $h$ is meromorphic $\Bbb{C\to C}$ and $|h(z)|=1$ on $|z|=1$ means that $g-h$ is analytic around $e^{it}$ and vanishes on $e^{ix},x-t\in (-\epsilon,\epsilon)$ ie. $g=h$.
Next since $h$ is analytic $\Bbb{P^1(C)\to P^1(C)}$ then it is a rational function $ h(z)=C\frac{\prod_j(z-a_j)}{\prod_i(z-b_j)}$, that $h$ sends each hemisphere to itself gives $|h(z)|=1$ on $|z|=1$ so that $h=g$ and hence $$h(z)=e^{i\theta}\prod_{j=1}^J \frac{z-a_j}{1-\overline{a_j}z}$$
Finally each $|a_j|<1$ since otherwise $h$ would have some zero on the $\infty$-hemisphere.
To restate the question more precisely: when can a meromorphic function $f:\mathbb{C}\to\mathbb{C}$ be extended to a function $\mathbb{C}^*\to\mathbb{C}^*$ that is holomorphic as a map between Riemann surfaces?
This is equivalent to asking if $f(\infty)$ can be defined so that this extended $f$ is meromorphic on the whole Riemann sphere.
This is equivalent to:
- The original $f$ is "bounded" in a neighborhood of $\infty$. That is, either (i) for some $r,b\in\mathbb{R}^+$, we have $|f(z)|<b$ whenever $|z|>r$, or else (ii) for some $r,b\in\mathbb{R}^+$, we have $|f(z)|>b$ whenever $|z|>r$.
(i) is bounded in the usual sense: the values lie in some disk around 0, whenever $z$ is large enough. (ii) is "bounded" in the sense that the values lie in some "disk around $\infty$", for large enough $z$. (People sometimes say "bounded away from 0" for this.)
Proof: We can move the domain to a neighborhood of 0 by looking at $g(z)=f(1/z)$. If this is bounded in that neighborhood, then the Riemann removable singularities theorem says that $g(z)$ can be extended to a function holomorphic at 0, by setting $g(0)=\lim_{z\to 0}g(z)$ ($=a$, say). (The theorem asserts that the limit exists.) Then set $f(\infty)=a$ and $f$ is holomorphic at $\infty$.
If $f(z)$ is "bounded" in sense (ii) above, then look at $g(z)=1/f(1/z)$: this moves both the domain and range to a neighborhood of $0$. Now we have $g(z)$ bounded near 0, so we set $g(0)=\lim_{z\to 0}g(z)=a$ as before. If $a\neq0$, then $\lim_{z\to 0}f(1/z)=1/a$ and we have the previous case. Otherwise $\lim_{z\to 0}g(z)=0$ and so $\lim_{z\to 0}f(1/z)=\infty$, so set $f(\infty)=\infty$ and this extended $f$ is meromorphic at $\infty$.
On the other hand, if we can extend $f$ meromorphically to include $\infty$ in its domain, then the original $f$ is obviously bounded near $\infty$ if $f(\infty)$ is finite, and "bounded" in sense (ii) near $\infty$ if the new $f$ has a pole at $\infty$.
It is well-known that a function meromorphic on the whole sphere must be rational. So that's another equivalent condition.
By one of Weierstrass's theorems, if $f(1/z)$ has an essential singularity at 0, then it cannot be bounded there; likewise for $1/f(1/z)$. So another equivalent condition is that $f$ does not have an essential singularity at $\infty$.
These all imply that the original $f$ has only finitely many poles. But by itself, that condition is not enough. For example, $e^z/z$ has only one pole plus an essential singularity at $\infty$. Or more simply still, $e^z$ has no poles and an essential singularity at $\infty$.
(Note that if $f$ with domain $\mathbb{C}^*$ has infinitely many poles, then these will have an accumulation point where $f$ will have an essential singularity. By initial assumption, $f$ has no essential singularities in the finite plane.)
Finally, all this "moving the domain (or range) from a neighborhood of $\infty$ to 0" is just an instance of the use of charts, as you noted: near $\infty$ we use the chart $z\mapsto 1/z$.
Best Answer
Existence of $\varphi_1:U_1 \rightarrow V_1$: Forster's wording is a little confusing. Since $Y$ is a Riemann surface, there is a chart $(\psi, U')$ around $b$. Since $f$ is continuous, there is an open set $W$ with $a \in W$ and $f(W) \subset U'$. There is also a chart $(\varphi, U)$ around $a$. Set $U_1 = W \cap U$ and $\varphi_1 = \varphi|U_1$. Then replacing $(U, \varphi)$ with $(U_1, \varphi_1)$ satisifes (i) and (ii).
The identity theorem: Pretty much what you said. Since $f$ is non-constant, by the identity theorem it is non-constant on the open set $U_1$. Since $\varphi_1$ is surjective from $U_1$ to $V_1$, $f \circ \varphi_1^{-1}$ is non-constant on $V_1$. And since $f \circ \varphi_1^{-1}(V_1) \subset U'$ and $\psi$ is 1-1 on $U'$, $f_1 = \psi \circ f \circ \varphi_1^{-1}$ is non-constant on $V_1$.
The function $h$: since $g(0) \neq 0$, $g$ is nonzero in a disk containing $0$, and therefore has a well-defined logarithm there, so set $h(z) = e^{\log(g(z))/k}$.