How to prove this ? I have tried to use $F$ but it doesn't fulfill the conditions. Thank you.

Question

Let $X$ be a topological space, $I=[0,1]$ be a subspace of usual space $\mathbb{R}$ and $\dot{I}=\{0,1\}$. Let $F:I \times I \rightarrow X$ be continuous. Let $\alpha, \beta, \gamma, \delta$ be paths in $X$. If $F(t,0)=\alpha(t), F(t,1)=\beta(t), F(0,t)=\gamma(t), F(1,t)=\delta(t)$, then prove that $\alpha \simeq \gamma \star \beta \star \delta^{-1}$ rel $\dot{I}$. Note : if $f$ and $g$ two paths in $X$ with $f(0)=g(0)$ and $f(1)=g(1)$, $f \simeq g$ rel $\dot{I}$ if and only if there exists a continuous map $F:I \times I \rightarrow X$ such that $F(t,0)=f(t), F(t,1)=g(t), F(0,t)=f(0), F(1,t)=g(0)$.

Answer 1

Try to verify the starting and ending point of $\alpha$ and $\gamma*\beta*\delta^{-1}$.

Note: clarify the variables, they should be $\alpha(s),\beta(s)$ and $\gamma(t),\delta(t)$, where $s$ is the horizontal axis and $t$ is the vertical axis. (to help visualize)

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$F(0,0)=\alpha(0)=\gamma(0)\implies$ the starting point is the same.

$F(0,1)=\gamma(1)=\beta(0)\implies$ the starting point of $\beta$ is identified with the endpoint of $\gamma$.

$F(1,1)=\beta(1)=\delta(1)\implies$ the endpoint of $\beta$ needs to be identified with the endpoint of $\delta$ so now it should be $\delta^{-1}$ to correct the direction.

$F(1,0)=\alpha(1)=\delta(0)\implies$ the endpoint is the same as we reverse the direction of $\delta$.

Also, $\alpha\simeq\beta$, $\gamma\simeq\delta$ by the assumption.

Now, we define $\phi=\gamma*\beta*\delta^{-1}$

$\phi(s)$ have the same starting and ending point as $\alpha$, and $\alpha,\phi\in X$

Define $\phi(s)$ by, $$ \phi(s)= \begin{cases} \gamma(3s) &\text{for } s\in[0,1/3]\\ \beta(3s-1) &\text{for } s\in[1/3,2/3]\\ \delta(3-3s)=\delta^{-1}(3s-2) &\text{for } s\in[2/3,1] \end{cases} $$

Define $H(s,t)$ by, $$ H(s,t)= \begin{cases} F(s-st,3st)=\gamma &\text{for } s\in[0,1/3],t\in[0,1] \\ F((1+2t)(s-\frac{1}{3}t),t)=\beta &\text{for } s\in[1/3,2/3],t\in[0,1]\\ F(t-(st-s),t-3(s-2/3)t)=\delta^{-1} &\text{for } s\in[2/3,1],t\in[0,1] \end{cases}$$

$H(s,0)=F(s,0)=\alpha(s)$, $H(s,1)=\gamma*\beta*\delta^{-1}$ and each branch is continuous and $H$ is well-defined and hence is a homotopy between $\alpha$ and $\phi$.