Winding number $=0$ imply homotopic to a point?

Question

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 $U$?

Answer 1

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 $1$.

Answer 2

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 $1$.

Answer 3

I am gonna give an indirect proof using the fundamental group.

Let $p:=\gamma(0)\in\mathbb{C}\setminus 0$ and consider $[\gamma]\in\pi_1(\mathbb{C}\setminus 0,p)\cong\mathbb{Z}$, where the isomorphism is obtained by composition of $\pi_1(\mathbb{C}\setminus \{0\},p)\cong\pi_1 (S^1,p/|p|)\cong \mathbb{Z}$. It is easy to see that the image of $[\gamma]$ via this isomorphism is $\tilde{\theta}(1)$, where $\theta: \mathbb{C}\setminus 0\to S^1$ is the angle function and $\tilde{\theta}:\mathbb{C}\setminus 0\to \mathbb{R}$ its universal lifting. Since by definition $n(\gamma,0)=\tilde{\theta}(1)$, we can write $[\gamma]=n(\gamma,0)$.

Therefore, if $n(\gamma,0)=0$ then $[\gamma]$ is trivial in $\pi_1(\mathbb{C}\setminus \{0\},p)$, which means that $\gamma$ is homotopic to the constant path.