Homeomorphism between $\Bbb S$\ $\lbrace-1\rbrace$ and $\left(-1,1\right)$

Question

I Know that $\Bbb S$\ $\lbrace-1\rbrace=\lbrace e^{i\theta}:\theta\in \left(-\pi,\pi\right)\rbrace=\lbrace e^{i\pi t}: t\in\left(-1,1\right)\rbrace$

Let $f:\Bbb S$ \ $\lbrace-1\rbrace\longrightarrow\left(-1,1\right)$ be a map where $e^{i\pi t}\mapsto t$

Why $f$ is continous function?

Can You help me please? or give me an Hint.

Answer 1

Let $I\subset (-1,1)$ and open interval. Note that $f^{-1}(I)=\{e^{\pi t} : t\in I\}$ is an open subset of $\mathbb{S}\setminus\{-1\}$ (indeed it's an arc in $\mathbb{S}$). In general, si $U$ is an open set it $(-1,1)$, there exists a family $\{I_a\}_{a\in A}$ of open intervals such that $U=\bigcup_{a\in A}I_a$ (this is beacuse the open intervals form a basis for the topology over $(-1,1)$). Then

$$f^{-1}(U) = f^{-1}\left( \bigcup_{a\in A}I_a \right)=\bigcup_{a\in A} f^{-1}(I_a),$$

as each $f^{-1}(I_a)$ is open, and the arbitrary union of open sets is open, then $f^{-1}(U)$ is open. This proves that $f$ is continuous.