The following lecture proof the existence of the inverse for the fundamental group starting from minuite 15:
https://www.youtube.com/watch?v=VVxPwbUCep0&list=PLCgncMh0TrClLOU991FDEQK5LF_Aqm4FS&index=5
But "Introduction to knot theory book" of Richard H. Crowell, prove it as follows:
But I want to transform the homotopy given in the book to what given on youtube, because what is given in the book is not clear for me, could anyone help me in doing so please?
First of all I recommend to read my answer to Proving the existence of identity of the fundamental group of homotopy theory $ \pi ( X,p)$ where you can see the difference between the approach by Crowell and Fox and the "standard" approach to the fundamental group. Both approaches are equivalent (i.e. result in the same fundamental group) but are technically different.
For your question observe that identity paths in the sense of Crowell and Fox have stopping time $0$. In the "standard" approach all paths have stopping time $1$, thus on the level of individual paths you do not have identity elements. However, if you work on the level of equivalence classes of paths, then you easily see that the equivalence class of the constant path at a point $p$ is an identity.
Crowell and Fox prove that $a \cdot a^{-1} \simeq e_1$ and $a^{-1} \cdot a \simeq e_2$. To this end they construct a fixed-endpoint family of paths $h_s$ between $e_1$ and $a \cdot a^{-1}$ (the other case is treated similarly). The stopping $\lVert h_s \rVert$ time of $h_s$ must vary from $0$ for $s= 0$ and $2\lVert a \rVert$ for $s = 1$. This is most easily done by setting $\lVert h_s \rVert = 2s \lVert a \rVert$. The definition of $h_s$ is now fairly obvious (see Figure 12 in the book): For $t \in [0,s \lVert a \rVert]$ go "ascending" with $a$ from $a(0)$ to $a(s \lVert a \rVert)$, for $t \in [s \lVert a \rVert, 2s \lVert a \rVert]$ go "ascending" with $a^{-1}$ from $a^{-1}(1- s \lVert a \rVert)$ to $a^{-1}(1)$, the latter being the same as going "descending" with $a$ from $a(s \lVert a \rVert) = a^{-1}(1- s \lVert a \rVert)$ to $a(0) = a^{-1}(1)$.