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.
Best Answer
We will use the Riemann criterion for complex simply connected spaces.
Let $U$ be simply connected and $f$ be biholomorphic.
We have only to prove that if $g:U'\to \Bbb C$ holomorphic with $g(z)\neq 0$ for every $z$, then there is a holomorphic $g_1:U'\to \Bbb C:g(z)=g_1^2(z)$ for every $z$.
Let such a holomorphic function $g$.The composition $gof:U\to \Bbb C$ is holomorphic and $gof(w)\neq 0$ for every $w$. Because $U$ is simply connected ,there is a holomorphic $h:U\to \Bbb C:gof(w)=h^2(w)$ for every $w\in U$.
Let $hof^{-1}:U'\to \Bbb C$. Then it is holomorphic and for every $z\in U'$ we have that $(hof^{-1}(z))^2=(h(f^{-1}(z))^2=g(z)$ for every $z$.