Let $\gamma:[0,1] \rightarrow \mathbb{C}$ be a continuous map never taking the value 0 and satisfying $\gamma(0)=\gamma(1)$. Define the degree (or winding number) $w(\gamma ; 0)$ of $\gamma$ about 0 . Prove the following:

(i) $w(1 / \gamma ; 0)=w\left(\gamma^{-} ; 0\right)$, where $\gamma^{-}(t)=\gamma(1-t)$.

(ii) If $\sigma:[0,1] \rightarrow \mathbb{C}$ is continuous, $\sigma(0)=\sigma(1)$ and $|\sigma(t)|<|\gamma(t)|$ for each $0 \leqslant t \leqslant 1$, then $w(\gamma+\sigma ; 0)=w(\gamma ; 0)$.

(iii) If $\gamma_{m}:[0,1] \rightarrow \mathbb{C}, m=1,2, \ldots$, are continuous maps with $\gamma_{m}(0)=\gamma_{m}(1)$, which converge to $\gamma$ uniformly on $[0,1]$ as $m \rightarrow \infty$, then $w\left(\gamma_{m} ; 0\right)=w(\gamma ; 0)$ for sufficiently large $m$.

Let $\alpha:[0,1] \rightarrow \mathbb{C} \backslash\{0\}$ be a continuous map such that $\alpha(0)=\alpha(1)$ and $\left|\alpha(t)-e^{2 \pi i t}\right| \leqslant 1$ for each $t \in[0,1]$. Deduce from the results of (ii) and (iii) that $w(\alpha ; 0)=1$.

[You may not use homotopy invariance of the winding number without proof.]