Verify $p _0 : [0,1] \mapsto S^1 , p_0(s)=(\cos(2 \pi s),\sin(2\pi s))$ is a covering map.

Question

I want to verify that the restriction to the interval $[0,1]$ of the map $p : \mathbb{R} \mapsto S^1 $ given by $ p(s)=(\cos(2 \pi s),\sin(2\pi s))$ is a covering map.

I tried as follows.

Take $s _0 \in S^1 $ and an open set $U \subset S^1 $ that contains $s _0 $ and is evenly covered by $p $. Let $\{V _\alpha \} $ be the partition of $p^{-1 } (U) $ into slices. Then $[0,1]\cap V _{\alpha} $ are disjoint open set in $[0,1] $. Now how can I show that each $[0,1]\cap V _{\alpha} $ is mapped homeomorphically to $U $?

Thanks in advance!

Answer 1

Here are some properties of covering maps

A covering map $p:X\to Y$ is a local homeomorphism
Proof: If $x\in X$, then there is an open neighborhood $U$ of $p(x)$ whose preimage is a disjoint union of open $U_i$ with $U_i\approx U$ via $p$. So $x$ has an open neighborhood $U_j$ which is mapped homeomorphically to its open image.

A local homeomorphism is an open map.
Proof: If $U$ is open in $X$ and $y=f(x), x\in U$, there is an open neighborhood $V$ of $x$ such that $V\subseteq U$ and $p(V)\subseteq p(U)$ is an open neighborhood of $y$.

But your map $p'$ is not open. The open set $[0,1/2)$ has image $A=\{(x,y)\in S^1\mid x=1\vee y>0\}$. If $A$ were open in $S^1$, then so would be $S^1\setminus A=-A$, which is not possible since $S^1$ is connected.