Proof that $\{ e \} \cdot \{ F \} = \{ F \} \cdot \{ e \} = \{ F \}$ In Mendelson's proof that $\{ e \} \cdot \{ F \} = \{ F \} \cdot \{ e \} = \{ F \}$, on pp. 145-147 of Introduction to Topology (where $\{ e \}$ is the constant path and $\{ F \}$ is an equivalence class of closed paths), he first defines $\{ F \} \cdot \{ e \}$ on the unit interval, and then defines a function $H(t, s)$ which projects $\{ F \} \cdot \{ e \}$ onto a unit interval where the length of the segment assigned to $\{ e \}$ is reduced to zero. That is, at $H(t, 0)$, the lengths of the two segments are as in $\{ F \} \cdot \{ e \}$, and at $H(t, .5)$, the length of $\{ e \}$ is half what it is at $H(t, 0)$, and at $H(t, 1)$, the length of $\{ e \}$ is $0$.
The piece of logic I'm looking to double check is why this only works with the constant path $\{ e \}$.
As I understand it, if I have a product $\{ F \} \cdot \{ G \}$ (where $\{ G \}$ is not $\{ e \}$), I can't reduce the length of $\{ G \}$ to $0$ on the unit interval because $\{ G \}$ has different values at different points on the interval. Therefore, I can't define a function that would transform $\{ F \} \cdot \{ G \}$ into just $\{ F \}$. There would always have to be some arbitrary non-zero length mapped to $\{ G \}$.
However, it would also seem that I can reduce the length of $\{ G \}$ on the unit interval to a length as close to zero as I like.
Is this correct?
Apologies for any errors in phrasing, I'm trying to restate the proof in my own words to better grasp it.
 A: Intuitively, the reason why it works for $\{e\}$ is that the constant path at the basepoint does not carry any information about $X$ and can be "crowded out" from the composed path equivalence class losslessly.
Formally, we define
$$F \cdot G : [0,1] \to X, F(s) = \begin{cases}
F(2s) & s \le 1/2 \\
G(2s-1) & s \ge 1/2
\end{cases}$$
Now try to define a homotopy from $F \cdot G$ to $F$ as in the case of $G = e$. This would have the form
$$H : [0,1] \times [0,1] \to X, H(s,t) = \begin{cases}
F(2s/(1+t)) & s \le (1+t)/2 \\
G((2s-1-t)/(1-t)) & s \ge (1+t)/2 , t < 1
\end{cases}$$
Note that for $t=1$ we get $H(s,1) = F(s)$. The idea is to stretch the subinterval carrying $F$ linearly from its initial size $[0,1/2]$ to $[0,(1+t)/2]$ which results in $[0,1]$ for $t = 1$. Doing so, the subinterval carrying $G$ shrinks linearly from its initial size $[1/2,1]$ to $[(1+t)/2,1]$ which results in the degenerate interval $[1,1]$ for $t = 1$.
In case $G = e$ we have $G((2s-1-t)/(1-t)) \equiv x_0$ = basepoint of $X$ so that we get a continuous $H$.
If $G \ne e$, then $H$ is not continuous. This is intuitively clear: The loop $G$ is present for all $t < 1$ (though it lives on a smaller and smaller subinterval of $[0,1]$), thus it cannot continuously vanish at $t = 1$.
To see this formally, consider a neighborhood $U$ of $x_0$ not containing the complete image $G([0,1])$. If $H$ were continuous, then there would be a neighborhood $V$ of $(1,1)$ in $[0,1] \times [0,1]$ such that $H(V) \subset U$. We find $s_0, t_0 \in [0,1)$ such that $(s_0,1] \times (t_0,1] \subset V$. Now let $t \in (t_0,1)$ such that $(1+t)/2 > s_0$. Then $G([0,1]) \subset H([0,1] \times \{t\}) \subset U$ which is impossible. 
