I am trying to understand the fundamental group of a topological space X and I have two questions. (First I write the definitions to be sure I understood them well.) A path is a continuous map $\alpha:[0,r]\rightarrow X$. It is normalized if r=1. Otherwise, it can be normalized by $\alpha(rt):=\alpha_1(t)$ so that $t$ takes values in $[0,1]$.One can define the composition of paths as follows: \begin{equation} (\alpha\cdot\beta)(t)= \begin{cases} \beta(t), & 0\leq t\leq r_1 \\ \alpha(t-r_1), & r_1\leq t\leq r_1+r_2 \end{cases} \end{equation} and $\alpha(r_1)=\beta(0)$.
Q.1 Could someone prove the associativity for say $\alpha, \beta,$ and $\gamma$ and tell me the pictorial interpretation of associativity?
(my attempt to answer due to Eric Wofsey's and Hurkyl's comments):
$(\alpha\cdot\beta)\cdot\gamma(t)$ means firstly applying $\gamma$ and then applying the composition $(\alpha\cdot\beta)$.Hence:
\begin{equation}
(\alpha\cdot\beta)\cdot\gamma(t)=
\begin{cases}
\gamma(t), & 0\leq t\leq r_3 \\
\beta(t-r_3), & r_3\leq t\leq r_1+r_3\\
\alpha(t-(r_1+r_3)), & r_1+r_3\leq t\leq r_1+r_2+r_3
\end{cases}
\end{equation}
and $\gamma(r_3)=\beta(0)$ and $\beta(r_1)=\alpha(0)$.
Therefore both are equivalent.
$\alpha\cdot(\beta\cdot\gamma)(t)$ means firstly applying the composition $\beta\cdot\gamma$ and then applying $(\alpha)$ to it.Hence:
\begin{equation}
\alpha\cdot(\beta\cdot\gamma)(t)=
\begin{cases}
\gamma(t), & 0\leq t\leq r_3 \\
\beta(t-r_3), & r_3\leq t\leq r_1+r_3\\
\alpha(t-(r_1+r_3)), & r_1+r_3\leq t\leq r_1+r_2+r_3
\end{cases}
\end{equation}
and $\gamma(r_3)=\beta(0)$ and $\beta(r_1)=\alpha(0)$.
Therefore, $(\alpha\cdot\beta)\cdot\gamma(t)=\alpha\cdot(\beta\cdot\gamma)(t)$.
Q.2 Also, I read in the book Knots and Surfaces by Gilbert (here) that "the composition of normalized paths is not associative." Why is that?