Unique Lifting Property of the Circle Proof

Question

I am trying to understand a line in the proof of this theorem as given by Lee in Introduction to Topological Manifolds.

The theorem states: "Let $B$ be a connected topological space. Suppose $\varphi:B\rightarrow\mathbb{S}^1$ is continuous, and $\widetilde\varphi_1,\widetilde\varphi_2$ are lifts of $\varphi$ that agree at some point of $B$. Then $\widetilde\varphi_1$ is identically equal to $\widetilde\varphi_2$.

The proof method is to show that the set $A=\{b\in B\mid \widetilde\varphi_1(b)=\widetilde\varphi_2(b)\}$ is both open and closed. The open part of the proof I understand. I am confused about the beginning of the closed part, which is as follows (note that $\epsilon$ denotes the exponential map from $\mathbb{R}$ to $\mathbb{S}^1$).

So we suppose $b\not\in A$. Set $r_1=\widetilde\varphi_1(b)$ and $r_2=\widetilde\varphi_2(b)$, so $r_1\neq r_2$. Let $z=\epsilon(r_1)$=$\epsilon(r_2)=\varphi(b)$, and let $U$ be an evenly covered neighborhood of $z$. If $\widetilde U_1$ and $\widetilde U_2$ are the components of $\epsilon^{-1}(U)$ containing $r_1$ and $r_2$, respectively, then $\widetilde U_1\cap \widetilde U_2=\emptyset$.

My question is the following: Why can't $r_1$ and $r_2$ be contained in the same component of $\epsilon^{-1}(U)$?

Answer 1

I presume you have some covering map $\epsilon:\Bbb R\to S^1$ and you want $\varphi=\pi\circ\widetilde\varphi$ etc. Probably your $\epsilon$ maps $x$ to $\exp(2\pi ix)$, and for definiteness' sake I'll assume this.

You have $z=\epsilon(r_1)=\epsilon(r_2)$. Then $r_2=r_1+k$ where $k$ is a nonzero integer. Take say, $\widetilde U_1=(r_1-\frac12,r_1+\frac12)$ and $\widetilde U_2=(r_2-\frac12,r_2+\frac12)$ and $U =\epsilon(\widetilde U_1)=\epsilon(\widetilde U_2)$. Then each of $\widetilde U_i$ is a component of $\epsilon^{-1}(U)$ and the argument goes through.

Answer 2

Lord Shark the Unknown has completely answered your question by constructing a special $U$.

Lee, however, takes an arbitrary $U$ which is evenly covered. We have $\epsilon^{-1}(U) = \bigcup_{\alpha \in A} V_\alpha$ with pairwise disjoint open $V_\alpha$ which are mapped by $\epsilon$ homeomorphically onto $U$. Let $\widetilde U_i$ be the connected component of $\epsilon^{-1}(U)$ containing $r_i$. Then $\widetilde U_i \cap V_\alpha \ne \emptyset$ only for a single $\alpha = \alpha_i \in A$; otherwise we could represent $\widetilde U_i$ as the disjoint union of nonempty open subsets. Hence $\widetilde U_i \subset V_{\alpha_i}$. Assume $\alpha_1 = \alpha_2$. Then $r_1, r_2$ are distinct points of $V_{\alpha_1}$ having the same image under the homeomorphism $\epsilon : V_{\alpha_1} \to U$, a contradiction. Hence $\alpha_1 \ne \alpha_2$ which implies $\widetilde U_1 \cap \widetilde U_2 = \emptyset$.