TITLE: Simply Connected domains.
QUESTION [7 upvotes]: If $U$ and $U'$ be two domains in $\Bbb C$, and $f$ be a homeomorphism in $U$ and $U'$ then domain $U$ is simply connected $\iff$ $U'$ is simply connected. I found this problem in complex analysis. So I would prefer to know its proof from complex point of view rather using topological propositions. Thanks.
There are few properties which are equivalent for a domain $D$ in Complex plane. 
a)$D$ is simply connected.
b)for each $z_0\in \Bbb C$\ $D$ there is a analytic branch of $log(z-z_0)$ defined on $D$.
c)The compliment of $D$ in the extended complex plane $\Bbb C^*$ is connected.

REPLY [2 votes]: The question is quite weird ; the notion of simply connectedness is a topological notion and the property you are asking is a lot simpler than complex analysis. If $U$ and $U'$ are homeomorphic , then any topological notion on $U$ can be carried to $U'$ by the homeomorphism.
For your question, this can take the following form. Assume that $U$ is simply connected. Choose a point $y$ in $U'$ and take a loop $\gamma(t)$ with base-point $y$ in $U'$. Then $f^{-1} \circ \gamma(t)$ is a loop in $U$ with base-point $x = f^{-1}(y)$. Since $U$ is simply connected, you have an homotopy $F(s,t)$ from $f^{-1} \circ \gamma$ to the constant loop based at $x$. Then $f \circ F(s,t)$ is a homotopy from $\gamma$ to the constant loop based at $y$.