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Suppose $\gamma$ is a smooth closed curve in $U=\mathbb{C} - \{0\}$. Suppose the winding number of $\gamma$ around 0 is 0. Is $\gamma$ homotopic to a point in $U$?

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The curve $\gamma\colon[a,b]\to\mathbb C-\{0\}$ can be written as $\gamma(t)=r(t)e^{i\theta(t)}$ for $a\le t\le b$ with $r,\theta$ continuos and $r>0$. Wlog. $\gamma(a)=\gamma(b)=1$, i.e. $r(a)=r(b)=1$, $\theta(a)=0$ and (because the winding number is zero) $\theta(b)=0$. We can define $\sqrt[n]{}$ on $\gamma([a,b])$ simply by letting $\sqrt[n]{\gamma(t)}=\sqrt[n]{r(t)}\cdot e^{\frac ini\theta(t)}$. If $n>\frac\pi2\max|\theta|$, we see that $\Re(\sqrt[n]{\gamma(t)})>0$. This allows one to easily contract $\sqrt[n]{\gamma(t)}$ to $1$ within the right half plane, e.g. by letting. $$H(\tau,t)=(1-\tau)\left(\sqrt[n]{\gamma(t)}-1\right)+1.$$ As a consequence, $$(H(\tau,t))^n=\left((1-\tau)\left(\sqrt[n]{\gamma(t)}-1\right)+1\right)^n$$ is a homotopy that retracts $\gamma$ to $1$.

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  • $\begingroup$ I read somewhere that the winding number about zero is an isomorphism from the fundamental group on $U$ to $\mathbb{Z}$. This would also imply that winding number zero means homotopic to a point. $\endgroup$ – Digital Gal Oct 25 '12 at 23:43
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We can write $\gamma\colon [0,1] \to U$ as $\gamma(t) = r(t) e^{i\theta(t)}$, where $r, \theta$ are continuous and $r>0$. Wlog. $\gamma(0)=\gamma(1)=1$, thus $r(0)=r(1)=1$ and we can assume that $\theta(0)=\theta(1) =0$, since the winding number of $\gamma$ around $0$ is $0$. Let $r_s(t) = (1-s)r(t) +s$ and $\theta_s(t) = (1-s)\theta(t)$. Then $H_s(t) = r_s(t)e^{i\theta_s(t)}$ is the desired homotopy of $\gamma$ to $1$.

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