# 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

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Open subsets of $\mathbb R$ are unions of open intervals, and $f(\cup I_i)=\cup f(I_i)$, so it suffices to show that $f(I)$ is open in $S^1$ when $I$ is an open interval. – Jonas Meyer Nov 22 '12 at 5:19
@JonasMeyer why $f(\cup I_i)=\cup f(I_i)$? – user42912 Nov 22 '12 at 5:22
Rafael: That is a property that always holds for images and unions, regardless of the function. You can prove it. – Jonas Meyer Nov 22 '12 at 5:33
@JonasMeyer ha, yes of course, sorry – user42912 Nov 22 '12 at 5:40

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 $S^1$.
when you said "Let $I=(a,b)$. Then the image $f(I)$ is precisely those points on $S^1$ with angle between $a$ and $b$." I think it's not write. Thank you for your answer. – user42912 Nov 22 '12 at 5:57
another problem is how we can prove that $f(I)=S^1\cap O$? – user42912 Nov 22 '12 at 5:59
Recall that I assumed (without loss of generality) that $I$ was contained in $[0,1)$. Note that $f$ wraps $[0,1)$ around the circle in such a way that $f(\theta)$ is precisely the point on $S^1$ with angle $\theta$. Then $f$ maps $(a,b)$ to the set of points with angles between $a$ and $b$. Of course, you can think of this as the set of all nonzero points in $\mathbb{R}^2$ with angle between $a$ and $b$ (this was the set $O$) restricted to $S^1$. Is that any clearer? – Isaac Solomon Nov 22 '12 at 6:20
take for example $\theta = 1/2$ the angle of $f(1/2)$ is $1/2$? – user42912 Nov 22 '12 at 7:20
Sorry, I edited it, the angle of $f(\theta)$ should be $2\pi \theta$. So if $\theta = 1/2$, we should have an angle of $\pi$, but $f(1/2) = (\cos \pi, \sin \pi) = (-1,0)$, which indeed has angle $\pi$. – Isaac Solomon Nov 22 '12 at 7:34
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.