I just started to study higher homotopy groups in the book Introduction to topology by V.A. Vassiliev. In such book, the author defines the product of two homotopy classes in $\pi_{n}(X,x_0)$ as follows:
Let $a_0$ be a fixed point in the $n$-sphere $S^{n}$ and let $f,g:S^n \longrightarrow X$ be two maps such that $f(a_0)=g(a_0)=x_0$. Let $p:\mathbb{R}^{n+1} \longrightarrow \mathbb{R}^{n}$, $i:\mathbb{R}^{n} \longrightarrow \mathbb{R}^{n}$ be the maps defined by $p(x_0,x_1,\ldots,x_n)=(x_1,\ldots,x_n)$ and $i(x_1,\ldots,x_n)=(-x_1,x_2,\ldots,x_n)$, that is, $p$ is a projection and $i$ is an involution; and now let's define $fg:S^n \longrightarrow X$ by $(fg)(y)=(f \circ p)(y)$ if $y_0 \geq 0$ and $(fg)(y)=(g \circ i \circ p)(y)$ if $y_0 \leq 0$. Thus, the product of two homotopy classes $[f],[g]$ is defined as $[f] \cdot[g]:=[fg]$. It is worth mentioning that we are also considering $f$ and $g$ as maps $(B^n,S^{n-1}) \longrightarrow (X,x_0)$, where $B^n$ is the unit closed ball in $\mathbb{R}^{n}$ centered at the origin.
Now, the exercise proposed by the book is to show that the set of homotopy classes $\pi_{n}(X,x_0)$ endowed with this product is indeed a group. It's more or less clear that the identity element should be the homotopy class of the constant map $e(y)=x_0$, in order to prove this one may assume that $a_0=(-1,0,\ldots,0)$ (by considering a rotation of the sphere if it's necessary) and then given any $f:(S^n,a_0) \longrightarrow (X,x_0)$ we can consider the map $H:B^{n} \times [0,1] \longrightarrow X$ defined by $H(y,t)=f(y/(1-t/2))$ if $|y| \leq 1-t/2$ and $H(y,t)=x_0$ if $|y| > 1-t/2$; hence, $H$ induces an homotopy $S:S^n \times [0,1] \longrightarrow X$ between $f$ and $fe$. However, with this definition I don't see clearly who the inverse of an arbitrary homotopy class $[f]$ is. That is, I don't understand how to find $[f]^{-1}$.
Any help?
In advance thank you very much.