Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 11F (Topics In Analysis)

Let $u(t)=\gamma (t)/|\gamma (t)|$, a continuous map into the unit circle. By uniform continuity, partition $[0,1]$ so finely that on each interval beginning at $t_j$, $|u(t)/u(t_j)-1|<1$. This ratio lies in the right half-plane and has a continuous complex argument with value zero at $t_j$. Starting from ${\theta }_0$, add these local arguments successively, matching endpoint values. This constructs a continuous real lift with $u(t)=e^{i\theta (t)}$.

Any two such lifts with the same initial value differ by a continuous function taking values in $2\pi \mathbb{Z}$, and hence coincide. Since the curve closes, $e^{i(\theta (1)-\theta (0))}=1$. Define its winding number byChanging the chosen initial argument adds a constant multiple of $2\pi$ to the whole lift, so the winding number is independent of that choice. No differentiability of the curve is required.

For each integer $m$, take ${\gamma }_m(t)=e^{2\pi imt}$. It is a continuous closed curve avoiding zero and has the continuous argument $2\pi mt$. ConsequentlyPositive and negative integers correspond to opposite senses of traversal, and the constant curve gives zero.

Choose continuous argument lifts ${\theta }_1,{\theta }_2$ for the two curves. Their sum is a lift for their pointwise product, becauseSubtract its initial value from its final value to obtainIn particular, winding numbers subtract under pointwise division of curves avoiding zero.

The continuous closed curve $q(t)={\gamma }_2(t)/{\gamma }_1(t)$ satisfies $|q(t)-1|<1$. It lies in the right half-plane, which admits a single continuous complex argument in $(-\pi /2,\pi /2)$. Since $q$ closes, this argument has equal endpoint values, giving $w(q)=0$. Additivity under multiplication then yieldsThe strict inequality ensures that the quotient stays in a region avoiding zero and admitting a single argument branch.

The appropriate homotopy invariance of winding number is: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for each $s$, then all loops $t\mapsto H(s,t)$ have the same winding number. The basepoint may move; every intermediate curve must close and avoid zero.

Compactness gives $m={\min }_{s,t}|H(s,t)|>0$. Uniform continuity gives $\delta >0$ such that $|s-s^{\prime }|<\delta$ implies $|H(s,t)-H(s^{\prime },t)|<m$ for every $t$. Part (d) then makes the winding numbers equal for nearby $s,s^{\prime }$. Subdivide $[0,1]$ into steps smaller than $\delta$ and chain these equalities. ThereforeThis proves the theorem directly for continuous loops, without requiring an integral formula or smooth homotopy.