Conformal quivalence

Question

Let's say there are two conformally equivalent domains $A,B$ in the complex plane.

If A is simply connected, can we infer that B is also simply connected?

Thank you in advance

Answer 1

Given a closed loop in $B$, tranport it to a closed loop in $A$ via the equivalence, contract it to a point in $A$ and transport this homotopy back to $B$.

Answer 2

Yes.

DEFINITION: A domain $D$ is simply connected if $n(\gamma,z)=0$ for every closed curve $\gamma$ in $D$ and $\forall z \in D^c$,where $n(\gamma,z)$ is the $\text{rotation number}$ or $\text{index}$ of $\gamma$ with respect to $z$

Now let $f:A \to B$ a holomorphic and one-to-one and onto function(i.e a conformal mapping from $A$ onto $B$)

Assume that $B$ is not simply connected.

Then exist a closed curve $\gamma$ in $B$ and $w_0 \in B^c$ such that $\frac{1}{2 \pi i} \int_{\gamma}\frac{1}{w-w_0}dw=n(\gamma,w_0) \neq 0$

Thus the function $h(w)=\frac{1}{w-w_0}$ does not have a primitive in $B$.

Let $g$ the holomorphic function in $A$ such that $g(z)=h(f(z))f'(z)$.

Since $A$ is simply connected,exists a primitive $G$ of $g$ in $A$, i.e $G'(z)=g(z), \forall z \in A$

Then define the holomorphic function $H$ in $B$ by $H(w)=G(f^{-1}(w))$ and we have: $$H'(w)=G'(f^{-1}(w))(f^{-1})'(w)=g(f^{-1}(w))(f^{-1})'(w)=h(w)f'(f^{-1}(w))(f^{-1})'(w)=h(w)$$ for every $w \in B$.

This a contradiction because $h$ does not have a primitive in $B$

So $B$ is simply connected.