Is $S^1 \to [-1,1]$ a covering?

Question

Let $p \colon S^1 \to [-1,1]$, $(x,y) \mapsto x$ be the vertical projection. Is this a cover? It seems to be one.

Edit: Why it seems to be one: For every $x \in (-1,1)$, we find some $\varepsilon > 0$ such that $U := B_\varepsilon(x) \subseteq (-1,1)$. Then $p^{-1}(U) \cong U \times S^0$ since the preimage of~$U$ is a disjoint union of two open subsets of $S^1$. For $x \in \{-1,1\}$, we could choose $U := B_1(x) \cap [-1,1]$. Then $p^{-1}(U) \cong U \times \{*\}$ since the preimage of $U$ is one open subset of $S^1$.

Answer 1

Let $x\in\{-1,1\}$ and $U$ (small enough) open neighbourhood of $x$. $p^{-1}(U)$ is an open arc of $S^1$ and in particular, it is connected. If we restrict $p$ to $p^{-1}(U)$, it is not injective because $(x_0,\pm y_0)$ are both mapped to $x_0$, but connectedness of $p^{-1}(U)$ prevents us from forming disjoint union $V_1\cup V_2 = p^{-1}(U)$ such that $p|_{V_i}$ is homeomorphism.

Note that we know that $p^{-1}(U)$ should be split into the union of two connected open sets due to the fact that this happens for any $x\in(-1,1)$.

Answer 2

No. The preimage of either $\pm 1$ under $p$ is a singleton, while all the other points have fibres of cardinality $2$.

Answer 3

Other answers correctly address your specific function, but let me give you a bit more general results:

No continuous map $S^1\to [-1,1]$ is a covering map. In fact, no map $X\to Y$ is a covering map, given both $X$, $Y$ are path connected, $\pi_1(X)\neq 0$ and $\pi_1(Y)=0$. That's because a covering map $p:X\to Y$ induces an injective group homomorphism $\pi_1(p):\pi_1(X)\to\pi_1(Y)$, which is the important result in covering maps theory.

Another reason that no map $S^1\to [-1,1]$ is a covering map, is because covering maps are local homeomorphisms. But these spaces are not locally homeomorphic. In fact no open neighbourhood of $-1$ in $[-1,1]$ is homeomorphic to an open subset of $S^1$. Because it has an "end", which formally can be seen as non cut-point (i.e. a point such that its removal doesn't disconnect the space). This of course gives us another, different generalization: no map $X\to Y$ between non-locally homeomorphic spaces is a covering map.