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

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.

• What math do you know? For example, do you know complex variables? The complex logarithm function, especially the principal value? – Rory Daulton Apr 15 '16 at 0:44
• Some Readers may be momentarily confused by the mention of $\mathbb S$ without definition. From the context it must denote the unit circle of the complex plane. – hardmath Apr 15 '16 at 2:04

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.
• some hint to prove that $f^{-1}\lbrace I \rbrace$ is open set? – Infinito307 Apr 19 '16 at 17:53