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Let $A=\{z \in \mathbb{C}:|z|>1\}$ and $B=\{z\in \mathbb{C}:|z|\ne 0 \}$.
Then which of the following are correct??
1. There exists a continuous function $f:A\to B$ which is onto.
2.There exists a continuous function $f:B\to A$ which is one-one.
3. There exists a non constant analytic function from B to A.
4. There exists a non constant analytic function from A to B.
Since A and B are homeomorphic 1 and 2 are correct.. But what about 3 and 4 ?

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  • $\begingroup$ C is the set of complex numbers $\endgroup$
    – mahan
    Sep 19 '16 at 17:57
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Hints: For your question #3. 1) consider what $\frac{1}{z}$ does to $A$. 2) Recall what removable singularity is. For question #4 you probably are missing something as there is clearly a nonconstant function from $A$ to $B$. Maybe you mean onto? In which case, think of a function you know really really well (And think of the point 1) above).

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  • $\begingroup$ Its non constant analytic function $\endgroup$
    – mahan
    Sep 19 '16 at 18:10
  • $\begingroup$ Then just think that $A$ is a subset of $B$ and you should be able to get a nonconstant function from $A$ to $B$. $\endgroup$
    – Jiri Lebl
    Sep 19 '16 at 18:12
  • $\begingroup$ Why are the first 2 options correct ? 4 is true by open mapping theorem. $\endgroup$
    – creative
    Dec 8 '16 at 13:22
  • $\begingroup$ For #1: $z = re^{it}$, then let $f(z) = (r-1)e^{it}$. For #2, let $f(z) = (r+1)e^{it}$. These are in fact homeomorphisms, the second is the inverse of the first. $\endgroup$
    – Jiri Lebl
    Dec 9 '16 at 5:43

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