Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 20I / c / Solution

Let $H(s,z)$ be the free homotopy from $\alpha$ to $\beta$ and put $v(s)=H(s,1)$. Because both endpoint loops are based at $x_0$, this is itself a loop based at $x_0$.

Cut the circle parameter at 1 and regard $H$ as a map on a square. Traversing its boundary gives the concatenation $\alpha \ast v\ast \bar{\beta }\ast \bar{v}$. The boundary is null-homotopic since the map extends over the square. Thus, in the fundamental group,Hence the corresponding elements are conjugate group elements. The moving basepoint path is precisely what distinguishes free homotopy from based homotopy.