I would appreciate help understanding the motivation for a line in the proof of Lemma 2.3 on page 114.
The Lemma states: Let $\gamma$ be a path in $X$ from $a$ to $b$. Let $\rho$ be any map from $[0,1]$ to $[0,1]$ such that $\rho(0) =0$ and $\rho(1) = 1$. Then $[\gamma \circ \rho] = [\gamma]$.
Proof: Note that $\gamma \circ \rho$ is a path from $a$ to $b$. A homotopy of $\gamma$ to $\gamma \circ \rho$ is given by:
$$\gamma_{t}(s) = \gamma((1 - t)s + t\rho(s))$$ with $0\leq s,t\leq 1$
(And this is where my question comes)
This homotophy is the composition of $\gamma$ and the homotopy $\alpha_{t} = (1-t)s + t\rho(s)$
$\gamma_{t}$ is well-defined and moves continuously with $t$. $\square$
of $\rho$, regarded as a path in $[0,1]$ from $0$ to $1$ and the identity path $\alpha_{0}(s)= s$ from $0$ to $1$.
I am wondering why he needed to make this elaboration regarding the input of $\gamma_{t}$. I would presume that having found the homotopy $\gamma_{t}$ above that he could claim the assertion $[\gamma \circ \rho]= [\rho]$
It seems to me that the latter homotopy, $\alpha_{t}$ shows that $s$ is homotopic to $\rho(s)$, but is this necessary given $\rho$ as stated in the lemma itself. Or perhaps, since homotopies are continuous, it is to claim $\gamma_{t}$ moves continuously with $t$
Thanks very much.
You are correct that the elaboration is unnecessary to the proof. But it is common in mathematical writing to try to give additional insights, especially when there is something general to learn from those insights that might be useful elsewhere. In this case, the additional insight is that if you precompose any function $f:X\to Y$ by any homotopy of the identity map on $X$ then you get a homotopy of $f$.