A Covering Space of the Circle

This note fills in details in Hatcher, §1.1, page 30.

We take S^{1} to be the unit circle in ℂ, the complex numbers, or equivalently, in the plane ℝ^{2}. Consider the map p: ℝ → S^{1} given by p(t) = e^{2πit} ∈ ℂ, or equivalently (as in Hatcher) p(t) = (cos 2πt, sin 2πt) ∈ ℝ^{2}. Informally, it wraps the real line around the circle. We note that p is periodic, p(t+1) = p(t) for all t, so that we need discuss in detail only values of t that lie in (or near) the unit interval, [0, 1].

[figure: …; …; •; •; •; •; •; •; •; •; ℝ; −\frac{1}{4}; 0; \frac{1}{4}; \frac{1}{2}; \frac{3}{4}; 1; \frac{5}{4}; \frac{3}{2}; p; S^{1}; •; (−1, 0); •; (0, −1); •; (0, 1); •; (1, 0); 0]

Definition 1 Given a map p: E → B, a local section of p is any map s: U → E, where U is open in B, such that p ^{∘} s: U → B is the inclusion, i. e. s(x) ∈ p^{−1}(x) for all x ∈ U. The map s is a (global) section if U = B.

Definition 2 Given a map p: E → B, the open set U ⊂ B is evenly covered by p if p^{−1}(U) is the disjoint union of open sets Ũ_{λ} in E with local sections s_{λ}: U → Ũ_{λ} that are bijections (and hence homeomorphisms, with inverses p|Ũ_{λ}).

Theorem 3 The map p: ℝ → S^{1} is a covering map, i. e. there exists a covering of S^{1} by open sets U_{i}, each of which is evenly covered by p.

Remark The map p admits no global section, as will be clear later.

Remark The terminology is unfortunately confusing; the word covering has two quite different meanings here. In French, there are two separate words: recouvrement for an open covering of a space, and revêtement for a covering space.

The proof of Theorem 3 is essentially Calculus I.

Proof We use four open sets, U_{0}, U_{1}, U_{2} and U_{3}, described in terms of ℝ^{2} rather than ℂ.

Take U_{0} = {(x, y) ∈ S^{1} : x > 0}. As x^{2} + y^{2} = 1, y determines x on U_{0} by x = \sqrt{1−y^{2}}. We have the local section s_{0,0}: U_{0} → Ũ_{0,0} = (−\frac{1}{4}, \frac{1}{4}) given by s_{0,0}(x, y) = \frac{1}{2π} sin^{−1} y; as y increases from −1 to 1, sin^{−1} y increases from −\frac{π}{2} to \frac{π}{2} and s_{0,0}(x, y) increases from −\frac{1}{4} to \frac{1}{4}. More generally, we have for any integer n the local section s_{0,n}: U_{0} → Ũ_{0,n} = (n− \frac{1}{4}, n+ \frac{1}{4}) given by s_{0,n}(x, y) = n + s_{0,0}(x, y). For other points of ℝ, e. g. n + \frac{1}{4} ≤ t ≤ n + \frac{3}{4}, cos 2πt = cos 2π(t−n) ≤ 0 and p(t) ∉ U_{0}. So U_{0} is evenly covered by the intervals (n− \frac{1}{4}, n+ \frac{1}{4}).

Take U_{1} = \{(x, y) ∈ S^{1} : y > 0\}. Here, x determines y = \sqrt{1−x^{2}}. We have the local section s_{1,0}: U_{1} → Ũ_{1,0} = (0, \frac{1}{2}) given by s_{1,0}(x, y) = \frac{1}{2π} cos^{−1} x; as x decreases from 1 to −1, cos^{−1} x increases from 0 to π and s_{1,0}(x, y) increases from 0 to \frac{1}{2}. As with U_{0}, U_{1} is evenly covered by the intervals Ũ_{1,n} = (n, n+ \frac{1}{2}).

Take U_{2} = \{(x, y) ∈ S^{1} : x < 0\}. Here, y determines x = −\sqrt{1−y^{2}}. We have the local section s_{2,0}: U_{2} → Ũ_{2,0} = (\frac{1}{4}, \frac{3}{4}) given by s_{2,0}(x, y) = \frac{1}{2} + \frac{1}{2π} sin^{−1}(−y); as y decreases from 1 to −1, −y increases from −1 to 1, sin^{−1}(−y) increases from −\frac{π}{2} to \frac{π}{2} and s_{2,0}(x, y) increases from \frac{1}{4} to \frac{3}{4}. Then U_{2} is evenly covered by the intervals Ũ_{2,n} = (n+ \frac{1}{4}, n+ \frac{3}{4}).

Take U_{3} = \{(x, y) ∈ S^{1} : y < 0\}. Here, x determines y = −\sqrt{1−x^{2}}. We have the local section s_{3,0}: U_{3} → Ũ_{3,0} = (\frac{1}{2}, 1) given by s_{3,0}(x, y) = \frac{1}{2} + \frac{1}{2π} cos^{−1}(−x); as x increases from −1 to 1, −x decreases from 1 to −1, cos^{−1}(−x) increases from 0 to π and s_{3,0}(x, y) increases from \frac{1}{2} to 1. Then U_{3} is evenly covered by the intervals Ũ_{3,n} = (n+ \frac{1}{2}, n+1).