# Thread: Complex Analysis - simply connected domain

1. ## Complex Analysis - simply connected domain

Prove that if domain $\displaystyle \Omega$ conformaly maps to unit disc $\displaystyle D=\left \{ \left | z \right |<1 \right \}$then $\displaystyle \Omega$ it's simply connected domain.

Thanks!

2. ## Re: Complex Analysis - simply connected domain

Originally Posted by sinichko

Prove that if domain $\displaystyle \Omega$ conformaly maps to unit disc $\displaystyle D=\left \{ \left | z \right |<1 \right \}$then $\displaystyle \Omega$ it's simply connected domain.
You need to assume that the conformal map $\displaystyle f:\Omega\to D$ is bijective (and therefore invertible), otherwise the result is not true.

Suppose that S is a loop in $\displaystyle \Omega$. The image f(S) of S is a loop in the simply-connected space D. So there is a continuous transformation in D that contracts this loop to a point. The image in of that deformation under the map $\displaystyle f^{-1}$ will be a continuous transformation in $\displaystyle \Omega$ which contracts S to a point S to a point.

3. ## Re: Complex Analysis - simply connected domain

[QUOTE=Opalg;696240]You need to assume that the conformal map $\displaystyle f:\Omega\to D$ is bijective (and therefore invertible), otherwise the result is not true.

Thanks a lot!