Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Remove the closure of simply connected region from the interior of a simply connected region. Is it true that the resulting domain can be mapped conformally to some annulus?

share|improve this question
1  
Another "counter-example", take whole C^1 and remove a disk from it... –  Dmitri Jan 8 '10 at 23:46
add comment

3 Answers

up vote 4 down vote accepted

The answer is yes. This is a special case of theorem 10 in Ahlfors' Complex Analysis, section 5, chapter 6. (Special in that the theorem more generally says that if the complement of the domain has $n$ connected components not reduced to points in the extended plane, then the domain is equivalent to an annulus from which $n-2$ concentric slits have been removed. In your case $n=2$.)

share|improve this answer
add comment

See Wikipedia. The third entry of the list gives an affirmative answer.

share|improve this answer
2  
Thanks for the lesson, Mariano and Scott. Henceforth I will look in standard references before asking. –  Anweshi Jan 8 '10 at 20:51
add comment

No. The resulting set need not even be connected. And if it is, it need not be doubly connected, as the interior region may have boundary points in common with the original region. Aside from these crude objections, however, the answers of Mariano and Scott are OK.

share|improve this answer
    
Thanks for the pointers towards complete rigor. –  Anweshi Jan 8 '10 at 20:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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