Example 3: $p:S^{1}\rightarrow S^{1}$ given by $p(z)=z^{2}$ where $S^{1}$ is considered as a subspace of the complex plane.

Consider a general map $p(z)=z^{n}$ . If $U=\{z=e^{i\phi}|\phi\in(a,b)\subset(0,2\pi)\}$ , then $p^{-1}(U)=\{z=e^{i\phi}|\phi\in(a/n+2\pi k/n,b/n+2\pi k/n),k\in\overline{0,n-1}\}$ is the union of $n$ open intervals in $S^{1}$ such that the restriction of $p$ onto each such interval is a homeomorphism of the interval with $U$ . Similarly, for $U=\{z=e^{i\phi}|\phi\in(a,b),-\pi\le a<0<b\le\pi\}$ ,$p^{-1}(U)=\{z=e^{i\phi}|\phi\in(a/n+2\pi k/n,b/n+2\pi k/n),k\in\overline{0,n-1}\}$ also is the union of $n$ open intervals in $S^{1}$ such that the restriction of $p$ onto each such interval is a homeomorphism of the interval with $U$ . Overall, every point of $S^{1}$ has such an open neighborhood.