The exponential function $f:\left(-\frac{1}{2}, \frac{1}{2}\right)\to \mathbb{S}^1\backslash\{ -1\}$ defined by $f(x)=e^{2\pi i t}$ is a homeomorphism.
I have already proved that $f$ is continuous and injective. But can't show the inverse is continuous or that $f$ is onto.
Can anybody help for continuity of inverse part? Thanks in advance.
Best Answer
The exponential function is $e : \mathbb R \to S^1, e(t) = e^{2\pi i t}$. It is well-known that this a continuous surjection with the property
$$e(s) = e(t) \Leftrightarrow s - t \in \mathbb Z .\tag{1}$$
This implies that $e$ maps $(- \frac 1 2, \frac 1 2]$ bijectively onto $S^1$. Since $e(\frac 1 2) = -1$, we see that $e$ maps $(- \frac 1 2, \frac 1 2)$ bijectively onto $S = S^1 \setminus \{-1\}$. This shows that the restriction $f: (- \frac 1 2, \frac 1 2) \to S$ of $e$ is a continuous bijection.
$f$ is an open map (and thus a homeomorphism):
Let $U \subset (- \frac 1 2, \frac 1 2) $ be open. Then $C = [- \frac 1 2, \frac 1 2] \setminus U$ is compact, hence $e(C)$ is a compact subset of $S^1$ and $V = S^1 \setminus e(C)$ is open in $S^1$. Clearly $V \subset S$ since $\frac 1 2 \in C$ which implies $-1 \in e(C)$. Thus $V = S \cap V$ is open in $S$. We have $[- \frac 1 2, \frac 1 2] = C \cup U$ and therefore $S^1 = e([- \frac 1 2, \frac 1 2]) = e(C) \cup e(U)$. The sets $e(C)$ and $e(U)$ are disjoint because points $s \in C$ and $t \in U$ cannot have the same image under $e$ (recall that $- \frac 1 2, \frac 1 2$ are the only two points of $[- \frac 1 2, \frac 1 2]$ mapped by $e$ to the same point of $S^1$, and both of them belong to $C$ whereas $t \notin C$). We conclude $f(U) = e(U) = S^1 \setminus e(C) = V$.