# Riemann mapping for doubly connected regions

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?

• Another "counter-example", take whole C^1 and remove a disk from it... – Dmitri Panov Jan 8 '10 at 23:46

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 $n=2$.)