2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ CSIR NET 2016 june question. I think it is duplicate/somewhat related-math.stackexchange.com/questions/1833661/… $\endgroup$
    – vidyarthi
    Nov 26 '16 at 7:35
  • $\begingroup$ @vidyarthi that question is somewhat related to first 2 options but i need reverse way function. $\endgroup$
    – Meow
    Nov 26 '16 at 7:39
0
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ ok. But is my argument right? $\endgroup$
    – Meow
    Nov 26 '16 at 7:43
  • $\begingroup$ @Prajakta yes, since the question asks only for existence, therefore examples may suffice. But, the general property in the background is, I think, homeomorphism. $\endgroup$
    – vidyarthi
    Nov 26 '16 at 7:45
  • $\begingroup$ ok. right. thanks. can you share a link of proof of homeomorphism? $\endgroup$
    – Meow
    Nov 26 '16 at 7:48
  • $\begingroup$ @Prajakta This is somewhat related --math.stackexchange.com/questions/1920002/… $\endgroup$
    – vidyarthi
    Nov 26 '16 at 7:57
-1
$\begingroup$

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 $f$

$\endgroup$
3
  • $\begingroup$ I am not getting how it answers b $\endgroup$
    – Meow
    Nov 26 '16 at 7:41
  • $\begingroup$ Sorry it will be d $\endgroup$
    – Learnmore
    Nov 26 '16 at 7:41
  • $\begingroup$ ya then its fine. $\endgroup$
    – Meow
    Nov 26 '16 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.