2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Do you know how to do it for $C_1=C_2$, i.e. loops in the circle? If not, read about the fundamental group of the circle in relation to the group of covering transformations of the universal cover. Once you are done reading, see if the same proof works for the annulus (or use the fact that one deformation retracts to the other). $\endgroup$ – Moishe Kohan Sep 19 '14 at 10:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.