Let $ A={z\in \mathbb{C}: |z|>1 }$ and $ B={z \in \mathbb{C}: z \neq 0}$ then which of the following is true. Let $ A=\lbrace z\in \mathbb{C}: |z|&gt;1 \rbrace$ and $ B=\lbrace z \in \mathbb{C}: z \neq 0\rbrace $ then which of the following is true.
The options are
a. There exists a continuous onto function from A to B

b. There exists a continuous one one function from B to A

c. There exists a non constant analytic function from A to B

d. There exists a non constant analytic function from B to A

I Can prove that C is true by taking just an inclusion map. Also i can prove that d is not true. As if it is true then say there is some non constant analytic function f from B to A. then clearly the function has either pole or essential singularity at 0 but then by picards theorem(in first case) and by casorati Weirstrass theorem(in second case) we get a contradiction to function being non constant. So c option is false. 

Now topologicaly both the sets A and B are connected, non compact, open etc. so have same properties.So trying a few example it seems that 1 is true by considering $f(z)=e^z$. And i think option B is true as we can have f to be some specific branch of logarithm. or function can also be defined in such a way that it shifts all the points continuously 1 unit distance radially outward i.e. $f(z)=(r+1)e^{i \theta}$ where $z=re^{i\theta}$.Is this argument right?
For options a) and b), we may use the fact that the punctured plane is homeomorphic to the compliment of closed unit disc
1.$f(z)=$ \begin{cases}|z| &amp;\text{if} |z|&gt;1\\\dfrac{1}{|z|} &amp;\text{if}|z|\le 1\end{cases}

2..$f(z)=$ \begin{cases}e^z &amp;\text{if} |z|&gt;1\\\dfrac{1}{e^z} &amp;\text{if}|z|\le 1\end{cases}

3.You are correct 


$(d)$ ;


If $f:B\to A$ then $|f(z)|&gt;1\forall z$.

Hence define $g(z)=\dfrac{1}{f(z)}$ which is entire .Also $|g(z)|&lt;1$ Hence $g$ is bounded entire function and so constant and hence so is $f$