I want to find a example of closed paths freely homotopic but not homotopic (I do not have many tools, like fundamental group, then has to be the simplest way possible).
I thought at the following:
$U=\mathbb{R}^2\smallsetminus\{(-1,0),(1,0)\}$;
$\lambda:[0,2\pi]\longrightarrow U, \lambda(t)=(1-cos(t),sin(t))$;
$\mu:[0,2\pi]\longrightarrow U, \mu(t)=(cos(t)-1,sin(t))$.
Clearly they are not homotopic. For free homotopy I thought of exchanging the two paths, but I couldn't write it.
Anyone have any idea?
Thanks.
A free homotopy between two closed paths $\lambda, \mu:[a,b]\longrightarrow X\subset \mathbb{R}^n$ is a continuos function $H:[a,b]\times[0,1]\longrightarrow X$, such that $H(s,0)=\lambda(s)$, $H(s,1)=\mu(s)$ and $H(a,t)=H(b,t), \forall s\in[a,b], \forall t \in[0,1]$.
Best Answer
Hint. It is not difficult to show that free homotopy classes of closed paths correspond to conjugacy classes in the fundamental group (assume the space is path connected for this)
Therefore you should look for an example with nonabelian fundamental group.