Let $T:\mathbb{C_{\infty}}\mapsto \mathbb{C_{\infty}}$ be Möbius transformation where $\mathbb{C_{\infty}}$ stereographic projection of $\mathbb{C}$. Now let $\mathbb{D}$ be simply connected domain i.e. the first fundamental group of $\mathbb{D}$ is trivial ($\pi_1(\mathbb{D})=0$), my question is $T(\mathbb{D})$ also simply connected or not.
More precisely any $\mathbb{D} \subset \mathbb{C_{\infty}} $ with $\pi_1(\mathbb{D})=0$, $ \Rightarrow $ $\pi_1(T(\mathbb{D}))=0$ , is it correct or not ?
What I am thinking\ If image of $\mathbb{D}$ is a non simply connected domain then there must be a hole in the image of $\mathbb{D}$, as $\mathbb{D}$ is simply connected then the boundary of $\mathbb{D}$ must be topologically a circle now by Jordan curve theorem its divides the domain into two parts the inverse image of that puncture point must be the complement of $\mathbb{D}$ as $T$ is Möbius transformation so this is not possible because if a sequence $z_n$ approaches to this unique inverse image the image of that sequence also approaches to the puncture which disturbed the continuity of $T$.
Thanks in advance