A question regarding complex valued function Let $A=\{z\in \mathbb C:|z|>1\}$ and $B=\{z\in \mathbb C:z\neq 0\}$. Then which of the following is/are true?


*

*There exists a continuous onto function $f:A\to B$.

*There exists a continuous one-one function $f:B\to A$.

*There exists a non-constant analytic function $f:A\to B$.

*There exists a non-constant analytic function $f:B\to A$.
3 is correct as $f(z)=e^z$ is one such function. What about others? I am confused.
 A: for (4) : 
if $f : B \to A$ is analytic then $g(z) = (f(\exp z))^{-1}$ is analytic from $\Bbb C$ to $\{z \in \Bbb C \mid 0 < |z| < 1 \}$.
Since $g$ is entire and bounded it has to be a constant $c$. Then since $\exp$ is onto $B$, $f$ has to be constant too (to $c^{-1}$).
A: $4.$ The answer is false. 
Consider the Laurant series of $f(z) = \sum_{n\in \mathbb{Z}} a_n z^n$. Suppose that there is an integer $M_0$ such that $a_n= 0$ for all $n\leq M_0$. If there is an integer $N_0$ and $a_n\neq 0$ for infinitely many $n\geq N_0$, then $f(1/z)$ has an essential singularity at $0$. Then by Casorati-Weierstrass Theorem, $f(B)$ is dense   in $\mathbb{C}$. 
Suppose that there is an integer $M_0$ such that $a_n= 0$ for all $n\leq M_0$. And, if there is an integer $N_0$ such that $a_n\neq 0$ for only finitely many $n\geq N_0$. Then $z^kf(z)$ is a polynomial for some $k\in \mathbb{Z}$. By the Fundamental Theorem of Algebra, the image of $f$ contains $\mathbb{C}-\{z_0\}$ for some $z_0\in\mathbb{C}$. 
Suppose $f(z)$ has an essential singularity at $z=0$, then by Casorati-Weierstrass Theorem, $f(B)$ is   dense  in $\mathbb{C}$. 
We have a stronger result if we apply the Great Picard Theorem instead of Casorati-Weierstrass. For some $z_0\in\mathbb{C}$, 
$f(B)$ contains $\mathbb{C}-\{z_0\}$.  
