Suppose that we are given two paths $f:I\to X$ and $f^{-1}(t):=f(1-t)$. Let $c_p$ be the constant map at $p\in X$. It is to be proven that $c_p\simeq f\circ f^{-1}$, where $\circ$ represents concatenation of paths.
The book gives the following diagram.
Then somehow concludes from it that the desired Homotopy is:
$H(s,t):=\begin{cases}f(2s); & 0\le s\le t/2 \\ f(t); & t/2\le s\le 1-\frac t2\\ f(2-2s); & 1-\frac t2\le s\le 1 \end{cases}$
I don't understand how this $H$ is obtained using the picture. I don't even understand why this $H$ is the desired homotopy: I know that $H$ is continuous by pasting lemma, and that $H(_,0)=f(0), H(_,1)= ?$. Can anyone please help me understand how to get homotopy using the picture?
I also want to add that I am sure that I will find the homotopy from various sources but it will be like- 'Define H like this and verify that this works'. But I'm not looking for that. I'm looking for a way to construct the homotopy using picture. Can anyone please help me understand that? Thanks.
The picture wants to say that for each $r \in [0,1]$ the desired homotopy $H$ is constant on each polygonal chain $C_r$ consisting of the three line segments $a_r$ going from $(\frac r 2,1)$ to $(\frac r 2,r)$, $b_r$ going from $(\frac r 2,r)$ to $(1-\frac r 2 ,r)$ and $c_r$ going from $(1-\frac r 2,r)$ to $(1-\frac r 2,1)$. These line segments have length $1- r$ and are three edges of a square.
We have $a_r = \{(\frac r 2,y) \mid r \le y \le 1 \}$, $b_r = \{ (s,r) \mid \frac r 2 \le s \le 1 -\frac r 2 \}$ and $c_r = \{(1- \frac r 2,y) \mid r \le y \le 1 \}$.
Thus on $b_t$ the homotopy has to take the value $(f \circ f^{-1})(\frac t 2) = f(2 \frac t 2) = f(t)$ which agrees with $(f \circ f^{-1})(1-\frac t 2) = f^{-1}(2(1- \frac t 2) -1) = f^{-1}(1-t) = f(t)$. Therefore $$H(s,t) = f(t) \text{ for } \frac t 2 \le s \le 1 - \frac t 2 .$$
On $a_r$ the homotopy has to take the value $(f \circ f^{-1})(\frac r 2) = f(r)$. The triangle $T_l$ with vertices $(0,0), (0,1), (\frac 1 2, 1)$ is the set of points $(s,t) \in [0,1]^2$ with $0 \le s \le \frac t 2$. Each $(s,t) \in T_l$ lies on a unique $a_r$; clearly we have $r = 2s$. Therefore $$H(s,t) = f(2s) \text{ for } 0 \le s \le \frac t 2 .$$
On $c_r$ the homotopy has to take the value $(f \circ f^{-1})(1- \frac r 2) = f(r)$. The triangle $T_r$ with vertices $(0,1), (1,1), (\frac 1 2, 1)$ is the set of points $(s,t) \in [0,1]^2$ with $1 - \frac t 2 \le s \le 1$. Each $(s,t) \in T_r$ lies on a unique $c_r$; clearly we have $r = 2(1-s)$. Therefore $$H(s,t) = f(2(1-s)) \text{ for } 1 -\frac t 2 \le s \le 1 .$$
By picture and formal definition we get for all $s \in [0,1]$ $$H(s,0) = f(0) = p ,$$ $$H(s,1) = (f \circ f^{-1})(s) .$$