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Is the map $x\to e^{ix}$ from real line $\Bbb R$ to circle open?

If I take any closed or half closed subset instead of $\Bbb R$ then this is definitely not open. But I'm little bit confused when domain is whole $\Bbb R $.

Could you please give me some hint. Thank you

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3 Answers 3

up vote 2 down vote accepted

An open subset of $\mathbb{R}$ is a union of open intervals. The map $x \mapsto e^{ix}$ maps open intervals of $\mathbb{R}$ to open intervals of the circle.

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A map is said to be open if the image of ANY open set is open. You can't just pick and choose which open sets to test. You also cannot look at the image of closed or half closed subsets [There is a notion of a closed map]. In your case, to show that your map is open, it suffices to prove that the image of open intervals in $\mathbb{R}$ is open in the circle.

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To see that $$f:\Bbb R\to S^1,\ z\mapsto e^{2\pi i z}$$ is open, note that it maps an interval $J=(a,b),\ a<b<a+1$ to the image of the set $$J'=[0,1]\cap((a-\lfloor a⌋,b-⌊a\rfloor)\cup(a-⌊a⌋-1,b-⌊a⌋-1))$$ under the quotient map $q:[0,1]\to S^1$ which is the restriction of $ f $. We know that $q$ is a quotient map since $[0,1]$ is compact and $S^1$ is Hausdorff, making $q$ a closed map. Now $J'$ is open and $q$-saturated, and this implies that its image $f(J)$ is open in $S^1$.

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