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I'm trying to show that the following map is a covering map:

$ p_{n}: S^{1} \rightarrow S^{1},\space(\cos(t), \sin(t)) ↦ (\cos(nt), \sin(nt))$

I've taken the approach of dividing $S^{1}$ into four overlapping sections: 1) positive first coordinate, 2) negative first coordinate, 3) positive second coordinate, 4) negative second coordinate.

The only thing I'm stuck on is that the 'slices', or $V_{i}$'s, that make up the pre-image of each of these sections are not a disjoint union. In fact, it seems that all the 'slices' are identical and so it only takes one 'slice' to cover each section. Is this okay? Every definition I can find describes the 'slices' as a union of disjoint open sets or a 'collection', but none specify whether this union can be finite, infinite, or whether it matters at all.

Am I already done with this proof and I'm just overthinking things, or have I taken the wrong approach?

Jos van Nieuwman
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bones_mccoy
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2 Answers2

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Hint

Use that $p_n$ restricted to $\left(t_0,t_0+\frac{2\pi}{n}\right)$ is a homeomorphism with its image and show that the preimage of any point consists of $n$ points since $p_n(t_0)=p_n\left(\frac{2kn\pi t_0}{n-1}\right)$ for all $k\in\{0,\dots, n-1\}.$

mfl
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  • Could you clarify your notation more? It would seem to me that $$p_{n}(t_{0}) \neq p_{n}(n(t_{0} + \frac{2k\pi}{n}))$$ – bones_mccoy Dec 03 '16 at 21:34
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I think it's better to approach this question.... $pictorially$.
So take the case n=4.
Hence the set $p^{-1 } \{1\}$ contains 4 elements. enter image description here Now let A be the points with first coordinate positive. enter image description here Now divide the circle into n=4 equal parts with each part having boundary the pre-images of 1. Now divide each of these parts into 4 equal parts. Then $p^{-1}(A)$ is shown as the red section.enter image description here. So if you're taking any n, then the preimage of A consists of those sections lying close to the pre-images of 1. Hence these are you're slices. Hope this helps you

Ivin Babu
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