Exercise 3.7 in Greenberg Harper

Question

Exercise 3.7 in Greenberg-Harper's Algebraic Topology is showing that X is simply connected if and only if two paths that have the same endpoints are homotopic. I have a solution to this particular problem, but I was confused on the hint given. They give a hint about transforming the unit square. Elsewhere in the book, they use the unit square to pictorially describe the specific homotopy. What is the homotopy that the square is describing? Any help would be appreciated. Thanks. I have attached the following picture.

Answer 1

Hint in so many words:

$\tau^{-1}\cdot \sigma$ can be taken to be a map from $S^1\rightarrow X$. Thus a contractionen $C:I\times I \rightarrow X$ to the one of the end points (lets say) $\sigma(0)=\tau(0)=a$ exists.

In this contraction $\tau^{-1}\cdot\sigma$ lives on one edge of $I\times I$ (i.e. $C(0, t)$) and is contracted to $a=C(1,*)$ on the other edge. However since the endpoint $b=\sigma(1)=\tau(1)$ is where the paths are connected the other endpoint $a$ lives on the other edges of $I\times I$ under $C$: $a=C(*,0)=C(*,1)$.

The picture now defines a Homotopy $H$ that maps the left half of the lower side of the square continuouly to the right half of the lower side of the square. Thus $C\circ H$ defines the required Homotopy.