Product of any 2 p-based loops is again a p-based loop.

Question

Knowing that the definition of product of 2 paths is as given here:

The definition of the product of 2 paths.

How can I show that: Product of any 2 p-based loops is again a p-based loop.

The book said that it is clear, but can anyone tell me a justification for this please?

EDIT:

A path in a topological space $X$ is then a continuous mapping $$a:[0, ||a||] \rightarrow X.$$ where the number $a$ is the stopping time and it is assumed that $||a|| \geq 0.$

A path whose initial and terminal points coincide is called a loop.

Answer 1

It's indeed trivial:
If $a$ and $b$ are $p$-loops, that means $$p=a(0)=a(\|a\|)=b(0)=b(\|b\|)$$ Thus, by the definition of product of paths, we will have $(ab)(0)=a(0)=p$ and $(ab)(\|ab\|)=(ab)(\|a\|+\|b\|)=b(\|b\|) =p$.