I am stuck with an exercise which is concerned with a proof that various open subsets of the complex plane are not conformally equivalent.
Show that no two of the following open subsets of $\mathbb{C}$ are conformally equivalent.
$U_1=\mathbb{C}, \ U_2=\{z \in \mathbb{C} \ | \ |z|<1 \} ,U_3=U_2 \setminus \{0\}, \ U_4=\{z \in \mathbb{C} \ | 1 < |z| <2 \}$
(Hint: In the case of $U_3,U_4$ you may want to consider what kind of singularities a holomorphic mapp on $U_3$ can have at $z_0=0$.)
This is what I have so far:
I know that any conformal mapping will also be biholomorphic and vice versa.
Suppose that there is a map $f: U_3 \to U_4$. Because $f$ is bounded on a punctured neighborhood of $0$ it follows that $0$ is a removable singularity. So there is a holomorphic map
$$\bar{f}: U_2 \to \mathbb{C}, \ \bar{f}= \begin{cases} f(z), & z \ \neq 0 \\ w_0, & \text{otherwise} \end{cases}$$
Case 1: $w_0 \in U_4$
Because f is bijective there is $z_1$ such that $\bar{f}(z_1)=\bar{f}(0)=w_0$. So there are neighborhoods $A_0, A_1$ such that $A_0 \cap A_1=\varnothing$. Let $g$ be the inverse of $f$. Then $\bar{f}(A_0)=g^{-1}(A_0)$ and since $g$ is holomorphic and therefore continous it follows that $\bar{f}(A_0)$ is a neighborhood of $w_0$. If I can show that $\bar{f}(A_1)$ is also a neighborhood of $w_0$ then $\bar{f}(A_0) \cap \bar{f}(A_1)$ is also a nonempty neighborhood of $w_0$. Then there is $w' \in \bar{f}(A_0) \cap \bar{f}(A_1)$ and $z \in A_0,z' \in A_1$ such that $f(z)=w=f(z')$ which contradicts the injectivity of $f$.
Case 2: $w_0 \notin U_4$
Because $\bar{f}$ is holomorphic and nonconstant and $U_2$ is a domain (connected and open set) it follows from the open mapping theorem that $\bar{f}(U_2)=U_4 \cup \{f(w_0)\}$ is also a domain which is a contradiction since the set is not connected.
In the case of $U_3, U_2$ I can use the same argument as above.
In the case of $U_1,U_i \ i \in \{2,3,4\}$ I can use Liouville's theorem.
Consider $U_2, U_4$. Suppose there is a conformal map $f: U_2 \to U_4 $.
I know that $U_2$ is contained in the inner ring of $U_4$.
I have two problems.
1) In the case of $U_3,U_4$ I do not see how to show that $\bar{f}(A_1)$ is a neighborhood of $w_0$.
2) I do not see how to proceed in the case of $U_2, U_4$. I think it is possible to conclude that $f$ is not injective. But I fail to see how this follows.
\Edit: I am not allowed to use arguments involving homeomorphisms or the fundamental group.
