TITLE: Verify that $x\mapsto (\cos(x),\sin(x))$ from the real line to the unit circle is an open map.
QUESTION [9 upvotes]: The setup:
Let $p:\mathbb{R}\to S^1$ be defined by $x\mapsto (\cos(x),\sin(x))$. Prove that $p$ is open, i.e. $p$ sends open subsets of $\mathbb{R}$ to open subsets of $S^1=\{ (\cos(x),\sin(x))\in \mathbb{R}^2 \ \vert \ x\in \mathbb{R}\}$.
What I have so far:
Since the image of maps preserves unions, it suffices to check on basic open subsets of $\mathbb{R}$, i.e. open intervals. If such an interval has length no less than $2\pi$, then its image under $p$ is the whole circle, which is open.
On the other hand, if the interval has length less than $2\pi$, then its image is an "open arc" on $S^1$, which can be realized as an intersection of $S^1$ with an appropriately chosen open box of $\mathbb{R}^2$, as illustrated below.

My question:

How can I formalize this last bit, so as to avoid arguing from a figure only? My current attempts at constructing the desired box precisely have relied on checking cases. Is there a more elegant way?

REPLY [3 votes]: Let $I$ be an open set of $\Bbb R$. Then the set $O=\{ r(\cos x, \sin x) \ , \ r>0 ; x\in I \}$ is an open subset of $\Bbb R^2$ , we have
$p(I)=O\cap S^1$ open subset of $S^1$.
EDIT: $O$ is open: Let $f:\Bbb R^2\to \Bbb R^2$ defined by $f(r,x)=r(\cos x,\sin x)$ it is clear that $f$ is continuous , since $\Bbb R_+^*\times I$ is an open of $\Bbb R^2$ (product of two open) then $O=f^{-1}(\Bbb R_+^*\times I)$ is open of $\Bbb R^2$.
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