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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.

circle and box

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?

Adrien Vakili
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    It's enough to show that your map is a local homeomorphism - that might be easier since you just have to define a local inverse. – ajd Dec 12 '14 at 18:59

3 Answers3

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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 $\Bbb R^2$.

Hamou
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Hint. Open arc can be broken up into the union of "small" open arcs. Such small that each one can easily be represented in your way.

Jihad
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Hint: Suppose that $f: M\to N$ is a smooth map of smooth manifolds, such that the derivative $df: T_xM\to T_yN$ is surjective for all $x\in M$ ($y=f(x)$). Then $f$ is a submersion and, hence, an open map. For equidimensional manifolds, like in you case, you can, alternatively, verify that $df$ is injective at each point. Now, compute the derivative of your map ...

Moishe Kohan
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  • This is borderline ridiculous! =) (Of course, it is a valid generalization, but if one cannot prove by hand that $t\mapsto e^{it}$ is open, then one will be using the tool above blindly, not knowing what is going on.) – Pedro May 29 '16 at 00:52
  • @PedroTamaroff: Dear Pedro, I like giving ridiculous answers (I think you should have omitted "borderline"), you will probably enjoy another one here: http://math.stackexchange.com/questions/295371/a-simply-connected-closed-surface-is-a-sphere/1799361#1799361. – Moishe Kohan May 29 '16 at 01:09