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a) Let $H: [0,K] \times [0,1] \rightarrow \mathbb{R}^2 $ be a homotopy of closed curves, so $H$ is continuous and for every $\sigma \in [0,1]$ it holds that $ c_{\sigma}:[0,K] \rightarrow \mathbb{R}^2$, $c_{\sigma}(t) := c(t,\sigma) $ is a $C^0$-closed curve.

Consider $ p \in \mathbb{R}^2$ with $p \notin H([0,K] \times [0,1]) $.

Show: $$ W_{c_0}(p) = W_{c_1}(p) $$ (winding number around $p$)

b) Consider now a $C^0$-closed curve $s : [0,K] \rightarrow \mathbb{R}^2$. Let $p,q \in \mathbb{R}^2$ be in the same connected component of $\mathbb{R}^2$\ $s([0,K])$

Show: $$ W_s(p) = W_s(q) $$

Attempt:

So for a) :First I tried to use the definition of the winding number around a point. $W_{c_o}(p)$:= $\frac{1}{2\pi}$ $( \Theta(K) - \Theta(0))$, where $\Theta$ is the angle function of $\frac{ c_o - p}{||c_o -p||}$, but I failed. So I've started to consider $H$. We know that $H$ is continuous and $[0,K] \times [0,1] $ is compact. So $H$ is uniformly continuous, which means that for $\varepsilon = 1$ there is a $\delta > 0$ such that $| H_{s1}(t) - H_{s2}(t) | < 1$ $\forall t \in [0,K]$ and $|s_1 - s_2 | < \delta $, where $s_1, s_2 \in [0,1]$. This implies ("hopefully") that $| \frac{H_{s1}(t)}{H_{s2}(t)} - 1 | < \frac{1}{|H_{s2}(t)|}$. Like you see I'm really lost in this exercise. I really need help here.

b) For this part I have only an little idea. So I think that we somehow can use a) for this. We could try to get the claim of a) to solve b), but to be honest I don't know how the " transform" b) to a).

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    $\begingroup$ For (a), do you know that the winding number is continuous? Because then you have a map from a connected set in the integers, which are disconnected... $\endgroup$ – Steve D May 16 '18 at 19:54
  • $\begingroup$ We never mentioned that the winding number is continuous in my lecture. I'm sorry. $\endgroup$ – RukiaKuchiki May 17 '18 at 13:25
  • $\begingroup$ You might want to work it out yourself then. This is (imho) the cleanest way to show both (a) [using connectedness of $[0,K]\times[0,1]$] and (b) [using connectedness of the component]. $\endgroup$ – Steve D May 17 '18 at 16:20

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