Understanding why $f*\overline{f}$ is homotopic to $e_{x_0}$

Question

I'm trying to understand why $f*\overline{f}$ is homotopic to $e_{x_0}$, where $\overline{f}$ is the inverse of $f$.

Munkres' "Topology" gives an explanation which I am not sure I understand, and links like these (pg. 3) only end up confusing me further. Where did $f_1$ and $f_0$ come from? They have not been defined earlier!

Edit: I am looking for an explanation of this. My attempts at understanding this from other online links has been largely futile.

Any help would be greatly appreciated.

Answer 1

In the linked PDF on page 3

$f_t$ is the path that equals $f$ on the $[0,1-t]$ and stationary on $[1-t,1]$, while $g_t$ is the inverse path of $f_t$.

So $f_t$ is like $f$, only that it stops at time $1-t$ and stays at the point $f(1-t)$ during the remaining time $[1-t,1]$. The reverse path $g_t$ has starting point $f(1-t)$, but it is fixed at that point during $[0,t]$ and only starts moving at time $t$, so $g(s)=f(1-s)$. In particular, $$f_0=f \text{ and } f_1=f(1-1)=f(0)=x_0$$ A homotopy $F$ is given by $$F(s,t)=(f_t\cdot g_t)(s)= \begin{cases}f(2s), &\text{ if }2s\le 1-t\\ f(1-t), &\text{ if }1-t\le2s\le1+t\\ f(2-2s), &\text{ if }2s-1\ge t \end{cases}$$

Answer 2

This picture could give you some intuition.

You have to imagine the back and the forth paths are superposing. The animation makes a gap just for visualization. The idea is to make a back and forth trip on the path $f \ast \bar f$, but going nearest and nearest until you just don't move at all.