Open subsets of $S^1$ I am trying to prove that the projection mapping: 
$$p:S^1 \rightarrow [-1,1]:p(x,y)=x$$
is an open map but not a covering map. My initial idea was to let $A \subset S^1$ be open. Then we could write $A$ as the union of open sectors of the circle. I tried formalizing this as $A = \bigcup_{i \in I} A_{i}$, with 
$$A_{i} = \{(x,y):x^2+y^2=1, a_{i}< x < b_{i}\}$$
and from there go to prove that $p(A_{i})$ is open and that the restriction of $p$ to $A_{i}$ is not necessarily a homeomorphism. But the definition above seems wrong as it would not properly define an open sector that includes the point $(1,0)$ for example. Though one option would be to change the definition of $A_{i}$ to set the constraint on $y$, this would introduce the same problem for the open set containing $(0,1)$ for example. 
Do you have any suggestions on how to correct this? Or another simpler idea to prove $p$ is open and not a covering map?
 A: Let $S^1_\pm = \{ (x,y) \in S^1 \mid (-1)^{\pm1}y \ge 0 \}$ denote the upper and the lower half of the circle. Then the projection maps $p_\pm : S^1_\pm \to [-1,1]$ are homeomorphisms. Let $U \subset S^1$ be open. Then $U_\pm = U \cap S^1_\pm$ is open in $S^1_\pm$ and $p_\pm(U_\pm)$ is open in $[-1,1]$. Hence $p(U) = p(U_+ \cup U_-) = p(U_+) \cup p(U_-) = p_+(U_+) \cup p_-(U_-)$ is open in $[-1,1]$.
$p$ is not a covering projection because $p^{-1}(1)$ has one element, but $p^{-1}(0)$ has two.
A: What if we define $A_{i}=\{(\cos(x), \sin(x)): x \in (a_{i}, b_{i}) \subseteq [0,2\pi)\}?$ 
Then let $U \subset S^1$ be open. We can write $U= \bigcup_{i \in I} A_{i}$. Thus: 
$$p(U)=\bigcup_{i \in I} p(A_{i})=\bigcup_{i \in I} \{\cos(x):x\in(a_{i}, b_{i})\}$$
As each $\{\cos(x):x\in(a_{i}, b_{i})\}$ is open in $[-1,1]$, $p(U)$ is open and $p$ is an open map.
By contradiction, assume $p$ is a covering map. Then for the point $(0,1)$, we can find an open neighborhood $V \subset S^1$ such that $p$ restricted to $V$ (call it $q$) is a homeomorphism. But then $V=\bigcup_{i \in I} A_{i}.$ Thus, we can find $j \in I$ such that:
$$(0,1) \in A_{j}=\{(\cos(x), \sin(x)): x \in (a_{j}, b_{j})\subseteq [0,2\pi)\}.$$ 
But then, $\pi/2 \in (a_{j}, b_{j})$, thus $a_{j}<\pi/2<b_{j}$. Then $\exists \epsilon > 0$ such that:
$$\pi/2+\epsilon, \pi/2-\epsilon \in (a_{j}, b_{j}).$$ 
Then, $(\cos(\pi/2+\epsilon), \sin(\pi/2+\epsilon)), (\cos(\pi/2-\epsilon), \sin(\pi/2-\epsilon)) \in A_{j}$. This gives:
$$q(\cos(\pi/2+\epsilon), \sin(\pi/2+\epsilon)) \neq q(\cos(\pi/2-\epsilon), \sin(\pi/2-\epsilon)).$$
As $\cos(\pi/2+\epsilon) > 0$ and $\cos(\pi/2-\epsilon) < 0$. Thus $q$ is not injective and thus not a homeomorphism.
