Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 1 / 20I (Algebraic Topology)

Use path concatenation in the order of traversal and let $\bar{u}(t)=u(1-t)$. A loop $\alpha$ based at $x_0$ determines the loop $\bar{u}\ast \alpha \ast u$ based at $x_1$. Thus the change of basepoint isomorphism isA based homotopy of $\alpha$ can be concatenated with the fixed outer paths to give a based homotopy of the resulting loops, so this map is well-defined on the fundamental group. Different parenthesizations and affine reparametrizations of path concatenation are based homotopic.

For loops $\alpha ,\beta$ based at $x_0$, the product of their transported loops is represented byThe middle $u\ast \bar{u}$ contracts relative to its endpoints, leaving $\bar{u}\ast (\alpha \ast \beta )\ast u$. Hence $u_{\#}$ is a group homomorphism. The reverse path induces the inverse: transporting back produces $u\ast \bar{u}\ast \alpha \ast u\ast \bar{u}$, whose outer retraced paths contract. Similarly the other composite contracts at $x_1$. ThereforeThis proves the asserted group isomorphism without needing any global path-connectedness hypothesis beyond the supplied path between the two points.

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.

View the torus as the topological group $S^1\times S^1$ under coordinatewise multiplication, and let $x_0$ be the common basepoint. From the given free homotopy constructThis is continuous, has $K(s,1)=x_0$, and at $s=0,1$ agrees with $\alpha ,\beta$ because $H(0,1)=H(1,1)=x_0$. Thus it exhibits the requested based homotopy explicitly.

Algebraically ${\pi }_1(S^1\times S^1)\cong {\mathbb{Z}}^2$ is Abelian, so conjugate elements are equal. Free and based homotopy classes of loops with a specified common basepoint therefore coincide here.

Let $a,b$ be the two oriented circle generators of the fundamental group of a bouquet of circles, ${\pi }_1(S^1\vee S^1)\cong F(a,b)$. TakeTheir loops are freely homotopic: slide the basepoint of the $a$ loop along the path $b$, thereby introducing the route $b^{-1}$ to the original loop and $b$ back. Equivalently the change-of-basepoint construction along a loop realizes conjugation, and the retraced connecting portions can be contracted with a moving marked point.

They are not based homotopic, since their reduced words in the free group are distinct: $a$ has length one and $b^{-1}ab$ has length three with no cancellation. This illustrates that free homotopy remembers conjugacy classes, whereas based homotopy remembers actual elements of the fundamental group.