A path from $S^1$ to $\mathbb{C} \setminus \{0\}$ can be expressed in "polar" form if and only if, the winding number of the normalized path is zero I'm trying to prove the following result:

Let $f: \mathbb{S}^1 \to \mathbb{C} \setminus \{0\}$ be a continuous map. Then there exists a continuous $g: \mathbb{S}^1 \to \mathbb{R}$ such that $f(z) = |f(z)|e^{ig(z)}$ for every $z \in \mathbb{S}^1$ if, and only if, the path $c: [0, 1] \to \mathbb{C} \setminus\{0\}$, defined by $c(s) = f(e^{2\pi i s})$, satisfies $n(c) = 0$.

Here $n(c)$ is the winding number of the path $[0, 1] \ni s \mapsto \alpha(s) =  \dfrac{f(e^{2\pi i s})}{|f(e^{2\pi i s} )|} \in \mathbb{S}^1$ (i.e, $2 \pi \cdot n(c) = \tilde{\alpha}(1) - \tilde{\alpha}(0)$, where $\tilde{\alpha}: [0, 1] \to \mathbb{R}$ is any lifting of $\alpha$).
Now, $\pi: \mathbb{R} \to \mathbb{S}^1, t \mapsto e^{it}$, is a covering map, and by the path lifting lemma there exists a lift $\tilde{\alpha}: [0, 1] \to \mathbb{R}$ such that $e^{i \tilde{\alpha}(s)} = \alpha(s)$ for every $s \in [0, 1]$, or equivalenty, such that $f(e^{2 \pi i s}) = |f(e^{2 \pi i s})|e^{i \tilde{\alpha}(s)}$ for every $s \in [0, 1]$. Every $z \in \mathbb{S}^1$ can be expressed as $z = e^{2 \pi i s}$ for some $s \in [0, 1]$, but I don't know how to get such a $g$ from $\tilde{\alpha}$ and how to relate this to the condition that the winding number is zero. The other direction (proving that if such a $g$ exists, then the winding number must be zero) is still unclear to me as well. Can anyone help me with this? I'd be grateful. Thanks in advance!
 A: Your result is equivalent to the following special case:

Let $\phi: \mathbb{S}^1 \to \mathbb{S}^1$ be a continuous map. Then there exists a continuous $g: \mathbb{S}^1 \to \mathbb{R}$ such that $\phi(z) = e^{ig(z)}$ for every $z \in \mathbb{S}^1$ if and only if the path $c : [0, 1] \to \mathbb{S}^1$, defined by $c(s) = \phi(e^{2\pi i s})$, satisfies $n(c) = 0$.

The general case follows from the special case by considering $\phi(z) = f(z)/\lvert f(z) \rvert$.
The map $q : [0,1] \to \mathbb{S}^1, q(s) = e^{2 \pi i s}$, is a closed  map, hence a quotient map. We have $c = \phi \circ q$.

*

*Let $n(c) = 0$. Take any lift $\tilde c : [0,1] \to \mathbb R$. Then $\tilde c (1) = \tilde c (0)$. Since $q$ is a quotient map, we get a unique continuous $g : \mathbb{S}^1 \to \mathbb R$ such that $g \circ q = \tilde c $. Therefore
$$\pi \circ g \circ q = \pi \circ \tilde c = c = \phi \circ q .$$
Since $q$ is surjective, we conclude $\pi \circ g = \phi$ which means $e^{ig(z)} = \phi(z)$ for all $z \in \mathbb{S}^1 $.


*Let $g : \mathbb{S}^1 \to \mathbb R$ be a map such that $\pi \circ g = \phi$. Then
$$c = \phi \circ q = \pi \circ g \circ q .$$
This means that $\tilde c = g \circ q$ is a lift of $c$. But clearly  $\tilde c (1) = \tilde c (0)$. Hence $n(c) = 0$.
