Explicitly prove that \(\mathbb R\) is that universal cover of \(S^1\).
Solution
View \(S^1\) as \(\{z \in \mathbb{C} : |z| = 1\}\) and let \(p\colon \mathbb{R} \to S^1\), \(p(t) = e^{2\pi i t}\). The map \(p\) is continuous and surjective, and \(p(t) = p(s)\) if and only if \(t - s \in \mathbb{Z}\).
1. Every \(z_0 \in S^1\) has an evenly covered open neighborhood.
Write \(z_0 = e^{2\pi i t_0}\) and let \(U = S^1 \setminus \{-z_0\} = \{e^{2\pi i t} : t_0 - \tfrac12 < t < t_0 + \tfrac12\}\), an open arc. Then \[ p^{-1}(U) = \bigsqcup_{k \in \mathbb{Z}} V_k, \qquad V_k = \left(t_0 - \tfrac{1}{2} + k,\ t_0 + \tfrac{1}{2} + k\right), \] because \(p(t) = -z_0\) exactly when \(t \in t_0 + \tfrac12 + \mathbb{Z}\); the \(V_k\) are pairwise disjoint open intervals. Let \(\operatorname{Arg}\colon \mathbb{C} \setminus (-\infty, 0] \to (-\pi, \pi)\) be the principal argument, which is continuous. For \(z \in U\), \(z e^{-2\pi i t_0} \neq -1\), so \[ \sigma_k(z) = t_0 + k + \frac{1}{2\pi} \operatorname{Arg}\left(z e^{-2\pi i t_0}\right) \] is a continuous map \(U \to V_k\) with \(p \circ \sigma_k = \operatorname{id}_U\) and \(\sigma_k \circ p|_{V_k} = \operatorname{id}_{V_k}\). Hence each \(p|_{V_k}\colon V_k \to U\) is a homeomorphism.
2. \(\mathbb{R}\) is simply connected.
\(H(t, s) = (1 - s)t\) is a homotopy from \(\operatorname{id}_{\mathbb{R}}\) to the constant map \(0\), so \(\mathbb{R}\) is contractible, hence path-connected with \(\pi_1(\mathbb{R}, 0) = 0\).
3. Q.E.D.
By step 1, \(p\) is a covering map. By step 2, its total space is simply connected, and \(S^1\) is connected and locally path-connected, so \(p\) is the universal cover of \(S^1\), unique up to isomorphism of covering spaces.