Let $C_1$ and $C_2$ be simple, closed curves in $\mathbb{R}^2$ such that $C_1$ lies in the region bounded by $C_2$, and the origin $O$ lies in the region bounded by $C_1$. Define an annulus $A$ as the region bounded by and including $C_1$ and $C_2$.
Let $\gamma_0,\gamma_1:[0,1]\rightarrow A$ be paths such that $\gamma_0(0)=\gamma_1(0)$ is a point on $C_1$ and $\gamma_0(1)=\gamma_1(1)$ is a point on $C_2$.
Define winding number $W(\gamma,O)$ about the origin as $$W(\gamma,O)=\frac{\theta(1)-\theta(0)}{2\pi}$$ where $\theta:[0,1] \rightarrow \mathbb{R}$ is the continuous angular coordinate of $\gamma$.
Proposition: $\gamma_0$ and $\gamma_1$ are homotopic $\iff$ $W(\gamma,O)=W(\gamma,O)$.
I have managed to show for the part $\implies$ using the lifting lemma. However, for the converse part, it gets difficult to find the homotopy, especially $C_1$ and $C_2$ can be wobbly curves. I asked people around and they said that the proposition is true, but none gives me the rigorous proof. I even looked in some books but the settings are different, ie for closed paths, circular annulus, or a punctured plane.
I would like to get some ideas, references or proper proof for that from anyone here. Thank you so much in advance.
Edited:
Possible idea: Since $A$ is a doubly connected region, by theorem in conformal theory, there exists a bijective analytic conformal map $\phi: A \rightarrow A^r$ where $A^r$ is a circular annulus (provided $C_1$ and $C_2$ are analytic curves). Then, obtain a homotopy between $A^r$ and $S^1$. Then, find the path homotopy between the images of the transformed curves. We can return to original $A$ and obtain the path homotopy.
Do you think the idea is possibly correct?