How do I prove that $f(x)=(\cos2\pi x,\sin2\pi x)$ is an open map? How do I prove that $f:\mathbb R\to \mathbb S^1$, $f(x)=(\cos2\pi x,\sin2\pi x)$ is an open map?
I'm thinking about a simple solution.
Maybe it's a hard question
I need help here
Thanks
 A: As Jonas Meyer pointed out in the comments, it suffices to show that $f(I)$ is open in $S^1$ for an open interval. Moreover, it should not be hard to convince yourself that we may assume $I \subset [0,1)$ by $2\pi$-periodicity and the fact that if $I$ contains an interval of the form $[a,b)$ with $b-a = 1$, then its image is all of $S^1$. Let $I = (a,b)$. Then the image $f(I)$ is precisely those points on $S^1$ with angle between $2\pi a$ and $2\pi b$. Consider the open set $O$ in $\mathbb{R}^2$ given in polar coordinates by $O = \{(r,\theta): r > 0, 2\pi a < \theta < 2 \pi b\}$. Then $f(I) = S^1 \cap O$, which is open in the relative topology on $S^1$.
A: We identify $\mathbb{S}^1$ with $\mathbb{T} = \{z \in \mathbb{C}|\ |z| = 1\}$.
Then $f$ becomes $f(x) = exp(i x)$.
Hence $f$ is a continuous homomorphism of topological groups.
$f$ factors to $\mathbb{R} \rightarrow \mathbb{R}/2\pi \mathbb{Z} \rightarrow \mathbb{T}$.
Since $\mathbb{R}/2\pi \mathbb{Z}$ is compact, $\mathbb{R}/2\pi \mathbb{Z} \rightarrow \mathbb{T}$ is an isomorphism. Since $\mathbb{R} \rightarrow \mathbb{R}/2\pi \mathbb{Z}$ is open, $f$ is open.
