Conformal quivalence 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
 A: 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 $B$.
A: 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.
