I just proved the following: Let $U=S^1-\{(-1,0)\}$ and $V=S^1-\{(0,1)\}$. Let $Y$ be a topological space, $\alpha:Y\to S^1$ be a path continuum. Let $y_0\in Y$ and $\theta_0\in\mathbb R$ be such that $\alpha(y_0)=(\cos\theta_0,\sin\theta_0)$. If $\alpha(Y)\subset U$ or $V$, then there exists $\theta:Y\to\mathbb R$ continues such that $\alpha(y)=(\cos\theta(y),\sin \theta(y))$ and $\theta(0)=\theta_0$.
But now I am faced with the following problem that I don't know how to attack clearly: Let $Y$ be a topological space, $\alpha:Y\to S^1$ be a continuous path, then the following statements are equivalent:
- There exists $\theta:Y\to\mathbb R$ continuous such that $\alpha(y)=(\cos\theta(y),\sin\theta(y))$, $\forall y\in Y $
- $\alpha:Y\to S^1$ is homopotic to a constant.
On the other hand, it would help me to prove that under hypothesis 2, $\alpha$ cannot be surjective. Is there a way to proof that?
It is well-known that $p : \mathbb R \to S^1, p(x) = (\cos x, \sin x) = e^{ix}$, is a covering projection.
Assume there exists a continuous $\theta : Y \to \mathbb R$ such that $\alpha = p \circ \theta$ (i.e. of a lift of $\alpha$). Since $\mathbb R$ is contractible, $\theta$ is homotopic to a constant map. Thus also $p \circ \theta$ is homotopic to a constant map.
For example, you may take $\phi : Y \times I \to \mathbb R, \phi(y,t) = t \cdot \theta(y)$. Then $(p \circ \phi)(y,t) = (\cos (t \cdot \theta(y)), \sin (t \cdot \theta(y)))$, is a homotopy from $\alpha$ to a constant map.
Clearly each constant map $c : Y \to S^1$ has a lift $\tilde c :Y \to \mathbb R$. Since covering projections have the homotopy lifting property, you can take a homotopy $H$ from $c$ to $\alpha$ and lift it to a homotopy $\tilde H$ such that $H(-,0) = \tilde c$. The map $\theta = H(-,1) : Y \to \mathbb R$ is then a lift of $\alpha$.
Note that under hypothesis 2. the map $\alpha$ can be surjective. As an example take $p : \mathbb R \to S^1$. Since $\mathbb R$ is contractible, $p$ is homotopic to a constant map.