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 $A=\lbrace z\in \mathbb{C}: |z|>1 \rbrace$ 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?

• CSIR NET 2016 june question. I think it is duplicate/somewhat related-math.stackexchange.com/questions/1833661/… Nov 26 '16 at 7:35
• @vidyarthi that question is somewhat related to first 2 options but i need reverse way function.
– Meow
Nov 26 '16 at 7:39

2 Answers

For options a) and b), we may use the fact that the punctured plane is homeomorphic to the compliment of closed unit disc

• ok. But is my argument right?
– Meow
Nov 26 '16 at 7:43
• @Prajakta yes, since the question asks only for existence, therefore examples may suffice. But, the general property in the background is, I think, homeomorphism. Nov 26 '16 at 7:45
• ok. right. thanks. can you share a link of proof of homeomorphism?
– Meow
Nov 26 '16 at 7:48
• @Prajakta This is somewhat related --math.stackexchange.com/questions/1920002/… Nov 26 '16 at 7:57

1.$f(z)=$ \begin{cases}|z| &\text{if} |z|>1\\\dfrac{1}{|z|} &\text{if}|z|\le 1\end{cases}

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

3.You are correct

1. $(d)$ ;

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

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

• I am not getting how it answers b
– Meow
Nov 26 '16 at 7:41
• Sorry it will be d Nov 26 '16 at 7:41
• ya then its fine.
– Meow
Nov 26 '16 at 7:42