How to prove the square lemma? Let $F:I \times I \to X$ be a continuous map, and let $f,h$ and $k,g$ be paths in $X$ defined by 
$$f(s) = F(s,0)$$
$$g(s) = F(1,s)$$
$$h(s) = F(0,s)$$
$$k(s) = F(s,1)$$
Then $f \cdot g$ is homotopic to $h \cdot k$
I tried like $F(s,t)F(1−t,s)$, but it clearly will not work(not even well defined). Then I thought about finding a way to sort of relableling $I×I$ to $I$, letting the homotopy starting from the lower left of the square. But I cannot find an explicit way to write down this.
 A: You are looking at the composition of $F$ with four different paths which I will denote by $\gamma_f(s) = (s,0)$ and so on.
There is an obvious homotopy $H$ of $\gamma_f*\gamma_g$ with $\gamma_h*\gamma_k$ given by
$$H(\tau,s) = \cases{\tau\gamma_h(2s) + (1-\tau)\gamma_f(2s)&if $0\le s\le\tfrac{1}{2}$,\\\tau\gamma_k(2s-1) + (1-\tau)\gamma_g(2s-1)&if $\tfrac{1}{2}\le s\le1$.}$$
Now consider the composition $F\circ H$.
A: Here is another "constructive" proof.
Let G(t,s)= 
\begin{array}{ll}
F\left(2t,\dfrac{ts}{1-s}\right) & \mbox{if~~~~$0\leq t\leq\dfrac{1}{2}$,~~~~$0\leq s<1$};\\
\\
F\left(0,2t\right) & \mbox{if~~~~$0\leq t\leq\dfrac{1}{2}$,~~~~$s=1$};\\
\\ 
F\left(2t-1,\dfrac{t-1+s(2-t)}{s}\right) & \mbox{if~~~~$\dfrac{1}{2}\leq t\leq 1$,~~~~$0<s\leq 1$};\\
\\
F(1,2t-1) & \mbox{if~~~~$\dfrac{1}{2}\leq t\leq 1$,~~~~$s=0$}.\end{array} 
Then
G(t,0)= 
\begin{array}{ll}
F\left(2t,0\right) & \mbox{if~~~~$0\leq t\leq\dfrac{1}{2}$};\\
\\
F(1,2t-1) & \mbox{if~~~~$\dfrac{1}{2}\leq t\leq 1$}.\end{array}
G(t,1)= 
\begin{array}{ll}
F\left(0,2t\right) & \mbox{if~~~~$0\leq t\leq\dfrac{1}{2}$};\\
\\
F(2t-1,1) & \mbox{if~~~~$\dfrac{1}{2}\leq t\leq 1$}.\end{array} 
So $G(t,0)=f\ast g$ and that $G(t,1)=h\ast k$. It should be clear that $G$ is continuous, hence $f\ast g\sim h\ast k$.
