Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 12F (Topics in Analysis)

For a continuous loop in $\mathbb{C}\setminus \{0\}$, choose a continuous argument lift $\theta :[0,1]\to \mathbb{R}$ with $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift is constructed by subdividing the interval into pieces whose images lie in an argument chart and matching the arguments at their endpoints. The winding number of a continuous closed path isChanging the initial lift adds one constant multiple of $2\pi$ and does not change this integer.

A loop homotopy is a continuous $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ with $H(0,t)={\gamma }_0(t)$, $H(1,t)={\gamma }_1(t)$ and $H(s,0)=H(s,1)$ for every $s$. Compactness gives a positive lower bound $r$ for $|H|$ and uniform continuity. Partition the $s$ interval finely enough that neighboring loops $\eta ,\delta$ satisfy $|\eta (t)-\delta (t)|<r$. Their quotient lies in $|z-1|<1$, and has a single-valued continuous argument with equal endpoint values. Its winding number is zero. Arguments of a product add, so $w(\eta )=w(\delta )+w(\eta /\delta )=w(\delta )$. Applying this at every step provesThis proves homotopy invariance of winding number directly, including continuous curves without differentiability.

The principal argument is continuous and single-valued on $\mathbb{C}\setminus (-\infty ,0]$. The given loop stays in this slit plane, and its initial and final values have the same principal argument. The definition therefore gives $\boxed{w(\gamma ;0)=0}$.

Here the target $C$ of $g$ is the unit circle. The boundary loop $g(e^{2\pi it})$ extends through the disk, so the radial homotopy $g(s{e}^{2\pi it})$ contracts it to a constant loop. Thus its winding number is zero. On the boundary put $h(z)=g(z)/z^n$, also circle-valued. Adding argument lifts gives $w(h)=0-n=-n$. If $z^n+g(z)$ never vanished, $h$ would avoid $-1$, which is the only negative real point of the unit circle. Part (iii) would give $w(h)=0$, contradicting $n\ne 0$. Therefore there is a boundary point with $z^n+g(z)=0$. The circle-valued hypothesis is essential; it is not a claim about arbitrary complex-valued $g$.