How to define this homotopy? I was trying to prove that $Fe \sim F$. Here $\sim$ denote homotopy and $e$ the constant path and $F$ a path. 
My idea is: The $Fe$ means that for the first half of the time it is $F$ and for the rest it is $e$. $F$ means it is $F$ for all the time. Therefore the homotopy makes the $e$ smaller until it is only one point. Denote the homotopy to be defined by $H(x,t)$. The $x$ is the time in the paths and the $t$ is the time for the homotopy. It means at $t=0$, $H$ is $F$ for $0 \le x \le ?$ and $H$ is $e$ for $? \le x \le 1$. The $?$ is an expression of $t$. At $t=0$ the $?$ is equal to $1/2$ and at $t=1$ it is equal to $1$. So I determined that $?$ should probably equal to ${t+1 \over 2}$. Then 
$$ \begin{array}{ccc}
H(x,t) = & F(!) & 0 \le x \le {t+1 \over 2}\\
     & e(!!) & {t+1 \over 2} \le x \le 1  
\end{array}$$
here $!$ and $!!$ are expressions of $x$. Since $!$ must be in $[0,1]$ it follows that $! = {x \over {t+1 \over 2}} = {2x \over t + 1}$. 
My problem is this: $!!$ should be ${x - {t+1 \over 2} \over 1 - {t+1 \over 2}}$ but this expression is $\infty $ for $t=1$. 
Can you please help me correct my mistake? 
 A: There's a better approach to this problem which you can use for other situations like this. The beauty of this is that you essentially just draw the intuitive picture of what's happening and then just read the information off.
Let $\alpha : I \to X$ be a path with $\alpha(0) = x$ and $\alpha(1)$ = y. Following Hatcher's notation, we have that 
$$
\alpha * \operatorname{const}_x(s) =
\begin{cases}
\alpha(2s) & \text{if } 0 \leq s \leq 1/2, \\
\operatorname{const}_x = x & \text{if } 1/2 \leq s \leq 1.
\end{cases}
$$
and 
$$
\operatorname{const}_x * \alpha (s) =
\begin{cases}
\operatorname{const}_x = x & \text{if } 0 \leq s \leq 1/2, \\
\alpha(2s - 1) & \text{if } 1/2 \leq s \leq 1.
\end{cases}
$$
We want to exhibit a homotopy $H : I \times I \to X$ such that $H_0(s) = \alpha * \operatorname{const}_x(s)$ and $H_1(s) = \alpha(s)$. Pictorially, we have,

Each of these can be described as a piecewise linear function simply be reading off the equations of the lines partitioning the square. Explicitly, for the first square we have,
$$
\varphi : I \to I, \qquad s \mapsto
\begin{cases}
2t & \text{if } 0 \leq t \leq 1/2, \\
1 & \text{if } 1/2 \leq t \leq 1,
\end{cases}
$$
and for the second square we have,
$$
\psi : I \to I, \qquad t \mapsto
\begin{cases}
1 & \text{if } 0 \leq t \leq 1/2, \\
-2t + 2 & \text{if } 1/2 \leq t \leq 1,
\end{cases}
$$
The key now is that these two are homotopic to the $\operatorname{id}_I$ via
$$
\varphi_s(t) = (1 - s) \varphi(t) + st \qquad \text{ and } \qquad \psi_s(t) = (1 - s) \psi(t) + st.
$$
The desired homotopies are then $\alpha \big( \varphi_s(t)\big)$ and $\alpha \big( \psi_s(t)\big)$.
