Let $A,B$ be domains in $\mathbb{C}$, with $A$ simply-connected, and $f:A\rightarrow B$ be analytic and invertible, with analytic inverse. I need to prove that $B$ is also simply-connected.
I have tried proof by contradiction:
Suppose that $B$ is not simply-connected. Then there exists a cycle $\gamma$ in $B$ and $a\in\mathbb{C}\setminus B$ with $n(\gamma,a)\neq0$.
$f^{-1}\circ\gamma$ is a cycle in $A$ and $A$ is simply-connected, so $n(f^{-1}\circ\gamma,w)=0$ $\forall w\in \mathbb{C} \setminus A$.
Then by the general form of Cauchy's Theorem, we have that $\int_{f^{-1}\circ\gamma} f=0$.
Not sure if I'm heading in the right direction. Any hints would be much appreciated.
There are many characterizations of simply connectedness. One of them says that a region is simply connected iff every analytic function on it wit o zeros has an analytic logarithm primitive. Now let $g$ be analytic in $B$ and suppose $g$ has no zeros. Then $g\circ f$ is analytic in $A$ and it has no zeros. So $g(f(z))=e^{h(z)}$ for some analytic function $h$ on $A$. Since $h\circ f^{-1}$ is analytic on $B$ and $e^{(h\circ f^{-1})(z)}=g(z)$ we are done.