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In class we defined the winding number as follows: If $\gamma$ is a loop on $\mathbb{R}^2$ that does not pass through a point $p$, the winding number $W( \gamma, p)$ is an integer $n$ that $\gamma$ represents $n$ times the canonical generator in the fundamental group $\pi_1(\mathbb{R}^2\setminus \{p\})$. Essentially, it is thought of as the number of turns $\gamma$ makes about $p$.

I've been working on problems to better my understanding of this topic. I can't figure out this one, and was wondering if anyone could help me out?

Let $p$ and $q$ be distinct points on the plane and $X = \mathbb{R}^2 \setminus \{p, q\}$. If $\gamma$ is a loop in $X$ such that $W(\gamma, p) = W(\gamma, q) = 0$, does it follow that $\gamma$ represents the trivial element of the fundamental group $\pi_1(X)$?

Thank you so much!

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A lead-up: Do you know what $\pi_1(X)$ is? How would you prove it? –  Dylan Moreland Jun 2 '12 at 1:00
    
I know $X$ is homotopically equivalent to $S^1 \vee S^1$, 2 circles joined by a common point. We also proved in class that $\pi_1(X)$ is a free group with 2 generators. –  Maria Jun 2 '12 at 1:58

2 Answers 2

up vote 5 down vote accepted

Fix a basepoint $x_0\in X$. Let $\gamma_p$ be a loop at $x_0$ going around $p$ exactly once, and $q$ zero times. Similarly, let $\gamma_q$ be a loop at $x_0$ passing around $q$ exactly once and $p$ zero times. Then if for a path $\gamma$, $\gamma^{-1}$ means traverse $\gamma$ backwards, consider the loop $\gamma=\gamma_p\gamma_q\gamma_p^{-1}\gamma_q^{-1}$. Its equivalence class is nontrivial in $\pi_1(X)$ but it has $W(\gamma,p)=W(\gamma,q)=0$.

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+1 So easy and nice! –  DonAntonio Jun 2 '12 at 1:22

From the analytic viewpoint. you can show that any curve a punctured plane is homotopic to a piecewise linear curve (connect the dots + uniform continuity). Then, from complex analysis we have $$W(\gamma, P) = {1\over 2 \pi i}\int_{\gamma} {dz\over z - P}.$$ if $\gamma$ is a rectifiable curve. If $W(\gamma, P) = n$, then $\gamma$ is homotopic to a circular curve winding about the point $P$ for $n$ "orbits."

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How does this answer the question? –  Rahul Jun 2 '12 at 1:36
    
You are so wrong for making topology depend on complex analysis. –  akkkk Jun 3 '12 at 12:16
    
It's all mathematics. The connections between mathematical fields make mathematics interesting and cool. Siloization is bad for the field. –  ncmathsadist Jun 3 '12 at 12:29

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