Prove that the given map is not a covering map.

Question

This problem is from Munkres' Topology section 53. I understand how p is not a covering map if U is connected. How is it not a covering map when U is not connected.

Answer 1

The (continuous) path $\gamma:[0,2\pi]\to S^1$ defined by $$\gamma(t)=e^{-it}$$ does not "lift" to any continuous path $\tilde{\gamma}:[0,2\pi]\to (0,\infty)$ starting at $\tilde{\gamma}(0)=2\pi$ (such a path would have the expression $\tilde{\gamma}(t)=2\pi-t$, and then $\gamma(2\pi)= 0\not\in (0,\infty)$).

Answer 2

Suppose that it was a covering, there exists an open connected subset $V$ which contains $p(2\pi)$ such that $p^{-1}(V)=\cup_{i\in I}U_i$ where $U_i\cap U_j$ is empty and the restriction of $p$ to $U_i$ is an homeomorphism onto $V$.

The sequence $p({1\over n})$ converges towards $p(2\pi)$, there exists $N$ such that for $n>N$, $p({1\over n})\in V$, denote $U_n$ such that ${1\over n}\in U_n$ and the restriction $p_{\mid U_n}$ is an homeomorphism onto $V$. There exists $x_n$ such that $p(x_n)=p(2\pi)$, we deduce that $x_n=2k_n\pi_n$ since $U_n$ is connected $2\pi\in U_n$.

Let $n,m>N$, $U_n=U_m$ since $U_n\cap U_m$ is not empty, we deduce that for $n>N, U_n=U_{N+1}$ and $({1\over n},2\pi]\subset U_{N+1}$, this implies that the restriction of $p$ to $U_{N+1}$ is surjective and is an homeomorphism $U_{N+1}\rightarrow p(U_{N+1})=V=S^1$ contradiction.