Paths on a surface 
Inversion
When we say that a real number \(a\) (non-zero) has an inverse for the product, we are saying that there is a number \(b\) that when multiplied with \(a\) we obtain the identity element (\(e=1\)), that is, such that \(a.b=b.a=1\). Furthermore, for the usual product of real numbers, we also know that \(b=\frac{1}{a}\).
For the product of paths (of homotopy classes), we verify invertibility in an analogous way: given a path, show the existence of a path such that when multiplied with it (on the left or right) we obtain a path homotopic to a constant.
Formaly, given \(f\) from \([0,1]\) to \(S\), there is \(f^{*}\) from \([0,1]\) to \(S\) such that \(f.f^{*}\) and \(f^{*}.f\) are homotopic to \(e\) and to \(e'\), respectively. Here \(f^{*}\) is the path that covers \(f\) in the opposite direction, that is, \(f^{*}\) from \([0,1]\) to \(S\) is given by \(f^{*}=f(1-t)\).
Considering the product on paths covered in one hour, the products above will differ on the time and velocity which the paths are covered.
In the case of \(f.f^{*}\):
Corresponds to covering \(f\) in the first half and hour and \(f^{*}\) in the last one. This means, covering \(f\) in the first half an hour, with twice the velocity, and in the last half an hour reverse in the opposite direction towards the initial point, also with twice the velocity. This product is homotopic to the constant path \(e\), equal to the initial point of \(f\) and, formaly, \(g_{1}=f.f^{*}\) from \([0,1]\) to \(S\) can be defined by: \[g_{1}(t)=\begin{cases} \begin{array}{lcc} f(2t) & \mbox{if} & 0\leq t\leq0.5\\ f(2-2t) & \mbox{if} & 0.5<t\leq1 \end{array}\end{cases}\]
In the case of \(f^{*}.f\):
Corresponds to covering \(f^{*}\) in the first half and hour and \(f\) in the last one. This means, starting at the end point of \(f\) we cover the path in reverse direction for the first half an hour (with twice the speed) and in the last half an hour we cover \(f\) with the original direction (also with twice the velocity), back to the end point of \(f\). This product is homotopic to the constant path \(e'\) (equal to the end point of \(f\)) and, formaly, \(g_{2}=f^{*}.f\) from \([0,1]\) to \(S\) can be defined by: \[g_{2}(t)=\begin{cases} \begin{array}{lcc} f(1-2t) & \mbox{if} & 0\leq t\leq0.5\\ f(2t-1) & \mbox{if} & 0.5<t\leq1 \end{array}\end{cases}\]
Therefore, the product paths are different from \(e\) or from \(e'\)! However, it can be shown that they are homotopic.
Formaly, an homotopy \(H\) from \([0,1]\times[0,1]\) to \(S\) (between \(g_{1}\) and the constant path \(f(0)\)) can be given by \[H(t,x)=\begin{cases} \begin{array}{lcc} f((1-x)2t) & \mbox{if} & 0\leq t\leq0.5\\ f((1-x)(-2t+2)) & \mbox{if} & 0.5<t\leq1 \end{array}\end{cases}\]
