How do I show that the winding number of this closed continuous path is zero? I have the following problem:

Show that if $\gamma:[a,b]\rightarrow U$ is a closed continuous path, $p\in \mathbb{R}^2$ and $U\subset \mathbb{R}^2\setminus \{p\}$ is an open subset on which there is a continuous angle function with vertex $p$, then $W(\gamma,p)=0$. Where $W$ denotes the winding number.


We have the following definition of a continuous angle function. A continuous angle function is a function $$\phi:Sect_{\Theta_1, \Theta_2} \rightarrow [0,2\pi]$$ where $Sect_{\Theta_1, \Theta_2}$ is the image of the following function $$S_{\Theta_1, \Theta_2}:Strip_{\Theta_1, \Theta_2}\rightarrow \mathbb{R}^2\setminus \{p\};\,\,(r,\Theta)\mapsto p+(r\cos(\Theta),r\sin(\Theta))$$ and $Strip_{\Theta_1, \Theta_2}=\{(r,\Theta)|r>0, \Theta_1<\Theta<\Theta_2\}$ with $\Theta_1,\Theta_2\in [0,2\pi]$ and $0<\Theta_2-\Theta_1<2\pi$

My Idea was the following:
Let us subdivide the intervall $$a=t_0<t_1<...<t_n=b$$Since on $U$ there is a continuous angle function we know that there exists a "general" sector $\mathfrak{O}$ with vertex $p$ such that $\gamma([t_i,t_{i+1}])\subset \mathfrak{O}$ for all $0\leq i\leq n-1$. Now on this sector one can chose an arbitrary angle function $$\phi:\mathfrak{O}\rightarrow \mathbb{R}$$and we can define $$w_i=\frac{\phi(\gamma(t_{i+1}))-\phi(\gamma(t_i))}{2\pi}$$ and by definition $W(\gamma,p)=\sum_{i=0}^{n-1}w_i=\sum_{i=0}^{n-1}\frac{\phi(\gamma(t_{i+1}))-\phi(\gamma(t_i))}{2\pi}=\frac{-\phi(\gamma(t_{0}))+\phi(\gamma(t_n))}{2\pi} \stackrel{\gamma(a)=\gamma(b)}{=}\frac{0}{2\pi}=0$
Does this works like this or did I made a thinking error?
Thanks for your help
 A: The definition of the winding number is given in your question How do i show that continuous maps between circles have the same degree iff they are homotopic?
You do not give a precise definition of a "continuous angle function", so let me suggest the following:

*

*A continuous angle function on an open subset $U \subset \mathbb R^2 \setminus \{p\}$ is a continuous function $\phi : U \to \mathbb R$ such that for all $x \in U$
$$x = p + \lVert x - p \rVert(\cos \phi(x),\sin \phi(x)). $$
In other words, the points of $U$ can be represented in polar coordinates centered at $p$ via a continuous angle function on $U$.
Let us agree to call any open $U \subset \mathbb R^2 \setminus \{p\}$ on which there exists a continuous angle function an angle region.

Note that we can always write $x = p + \lVert x - p \rVert(\cos \Theta,\sin \Theta)$ with some polar angle $\Theta$, but this angle is not uniquely determined by $x$. In fact, we have $(\cos \Theta,\sin \Theta) = (\cos \Theta',\sin \Theta')$ if and only if $\Theta' = \Theta+2k\pi$ for some $k \in \mathbb Z$. Of course you can make the polar angle unique by requiring for example $\Theta \in [0,2\pi)$, but this is not a good idea if we are interested to find a continuous angle function on an open $U$ containing a point of the form $x = p + t(1,0)$ with $t > 0$.
Anyawy, if $\phi$ is a continuous angle function on $U$, then also $\phi_k(x) = \phi(x) +2k\pi$ is one for each $k \in \mathbb Z$.
I guess that you have a proof that all sectors around $p$ are angle regions.
However, let me mention that your above definiton of a sector is inadequate. No sector contains the horizontal ray $R_p = \{p + t(1,0) \mid t > 0\}$ starting and $p$ and going to the right, i.e. the sectors do not cover $\mathbb R^2 \setminus \{p\}$. This defect comes from the requirement $\Theta_1,\Theta_2\in [0,2\pi]$ because $Strip_{\Theta_1, \Theta_2}=\{(r,\Theta)|r>0, \Theta_1<\Theta<\Theta_2\}$ does not contain a point with "polar angle" $0$ or $2 \pi$, i.e. no point of $R_p$. The solution of this problem is easy: Simply drop the requirement $\Theta_1,\Theta_2\in [0,2\pi]$. Then you have $Strip_{\Theta_1, \Theta_2} = Strip_{\Theta_1 + 2k\pi, \Theta_2+2k\pi}$ for all $k \in \mathbb Z$, i.e. strips are no longer determined by unique "boundary angles", but this does not matter.
Lemma 1. Let $\gamma : [a,b] \to \mathbb R^2 \setminus \{p\}$ be a continuous path. Assume that $\gamma([a,b])$ is contained in two angle regions $U_i  \subset \mathbb R^2 \setminus \{p\}$ with continuous angle functions $\phi_i$. Then $\phi_1(\gamma(b)) - \phi_1(\gamma(a)) = \phi_2(\gamma(b)) - \phi_2(\gamma(a))$.
That is, for each continuous path whose image is contained in some angle region, the argument increase $A(\gamma,p) = \phi(\gamma(b)) - \phi(\gamma(a))$ along $\gamma$ is well-defined: Just pick $U$ with continuous angle function $\phi$ such that $\gamma([a,b]) \subset U$.
Proof. Consider the continuous functions $\psi_i = \phi_i \gamma : [a,b] \to \mathbb R$. For each $t \in [a,b]$ we have $\psi_1(t) = \psi_2(t) + 2k_t\pi$ for some $k_t \in \mathbb Z$. Thus $d = \psi_1 - \psi_2$ takes values in the set $S = \{2k\pi \mid k \in \mathbb Z\}$. But the IVT shows that is impossible that $d(t_1) \ne d(t_2)$ for $t_1 \ne t_2$ because then $d$ would also have values not in $S$. Thus $d$ is constant and the lemma follows.
Lemma 2.  Let $\gamma : [a,b] \to \mathbb R^2 \setminus \{p\}$ be a continuous path such that $\gamma([a,b])$ is contained in an angle region. Let  $J_i = [t_{i-1},t_i]$ be a partitioning of $[a,b]$ so that in particular each $\gamma \mid_{J_i}$ has image in an angle region. Then $A(\gamma,p) = \sum_i A(\gamma \mid_{J_i}, p)$.
Proof. This is obvious since we can take the same $U$ and same $\phi$ for $\gamma$ and the $\gamma \mid_{J_i}$.
Lemma 3.  Let $\gamma : [a,b] \to \mathbb R^2 \setminus \{p\}$ be a continuous path. Let $J_i = [t_{i-1},t_i]$ and $J'_k = [t'_{k-1},t'_k]$ be two partitions of $[a,b]$ such that all $\gamma \mid_{J_i}$ and all $\gamma \mid_{J'_k}$ have images in angle regions $U_i$ and $U'_k$, respectively. Then $\sum_i A(\gamma \mid_{J_i},p) = \sum_k A(\gamma \mid_{J'_k},p)$.
Proof. There exists a partition $J''_l =  [t''_{l-1},t''_l]$ of $[a,b]$ such that each $t''_j$ is one of the points $t_i$ or $t'_j$.  Lemma 2 shows that $\sum_i A(\gamma \mid_{J_i},p) = \sum_l A(\gamma \mid_{J''_l},p)$ since each $J_i$ is partitioned in suitable $J''_l$. Simililarly $\sum_k A(\gamma \mid_{J_k},p) = \sum_l A(\gamma \mid_{J''_l},p)$ and the lemma follows.
Now for each continuous path $\gamma : [a,b] \to \mathbb R^2 \setminus \{p\}$ we can find a partition $J_i = [t_{i-1},t_i]$ of $[a,b]$ such that each $\gamma \mid_{J_i}$ is contained in an angle region (we may take sectors since they cover $\mathbb R^2 \setminus \{p\}$, but we are completely free in our choice). Then by Lemma 3 the angle increase along $\gamma$
$$A(\gamma,p) = \sum_i A(\gamma \mid_{J_i},p)$$
is well-defined. It does not depend on the choices of partition and angle regions.
The winding number of a closed  continuous path is then defined by
$$W(\gamma,p) = \frac{1}{2\pi}A(\gamma,p). $$
Working with sections as angle regions gives your orginal  definition. But we can use more general angle regions.
The proof of your claim $W(\gamma,p) = 0$ is now trivial. In fact, the single angle region $U$ (with continuous angle function $\phi$) suffices to get $2\pi W(\gamma,p) = A(\gamma,p) = \phi(\gamma(b)) - \phi(\gamma(b)) = 0$ (note that $\gamma(b) = \gamma (a)$).
Remark:
There are weird angle regions. Let $X = \{(0,0)\} \cup \{(e^t(\cos t, \sin t)\mid t \in \mathbb R \}$. This is an infinite spiral plus the origin. It is closed in $\mathbb R^2$, thus its complement $U$ is open. It is an angle region for the origin. All its continuous angle functions are surjections $\phi : U \to \mathbb R$.
