Problem in understanding the solution of the problem [Hatcher 1.1.6]

Question

I have completely understood the solution except the following fact(Please see 5th line from last) ,

How the fact that $S^1\times I $ deformation retracts onto $S^1\times \{0\}$ implies that $\beta*\bar{\alpha}$ is path homotopic to $e_1*\bar{\gamma}$. I am not able to write an explicit homotopy.

Answer 1

First lets prove a lemma:

Let $H:X\times I\to X$ be any homotopy with identity for $t=0$. Assume that $\lambda:I\to X$ is a path. Then $\lambda(t)$ is homotopic to $H(\lambda(t), 1)$.

Proof. Indeed, define $H':I\times I\to X$ by $H'(t, s)=H(\lambda(t), s)$. By the assumption $H(\cdot, 0)$ is the identity thus $H'(t, 0)=\lambda(t)$. On the other hand obviously $H'(t, 1)=H(\lambda(t), 1)$ which completes the proof. $\Box$

Now apply that lemma to $G$. You obtain that $\beta(t)$ is homotopic to

$$G(\beta(t), 1)=G(1, t, 1)=(1,0)=e_1(t)$$

On the other hand $\alpha(t)$ is homotopic to $G(\alpha(t), 1)$. And what is that?

$$G(\alpha(t), 1)=G\big(f(\beta(t)), 1\big)=G(f(1, t), 1)=G(e^{2\pi i t}, t, 1)=(e^{2\pi i t}, 0)=\gamma(t)$$