If $F,G: I \rightarrow Z$ are closed path at $z \in Z$, where Z is a topological space. Why is $F \cdot G: I \rightarrow Z$ a closed path?
This does not appear to make sense to me considering my idea of a closed path. Which is unfortunately undefined in my textbook. Introduction to Topology by Bert Mendelson.
My understanding is that a closed path at $z$ is simply a path passing through $z$ with endpoints defined. However, the definition leads me to think that $z$ must be an endpoint. As the authors defined the path $F \cdot G$ as
$$(F \cdot G)(t) = F(2t) \;\;\;\; 0 \leq t \leq \frac{1}{2}$$ $$(F \cdot G)(t) = G(2t-1) \;\;\;\; \frac{1}{2} \leq t \leq 1$$
However, that means that $F(1) = z$ and $G(0) = z$, and unless both $F$ and $G$ are the path $F(t) = z$ and $G(t) = z$ then it's not true that we can simply swap $F$ and $G$ here. So I'm quite confused by what this definition is supposed to mean.
The only possible meaning of "closed path" I can think of that makes sense in this context is that of "closed curve," what in topology we most commonly call a "loop". It is just to say that $F(0) = F(1) = G(0) = G(1) = z$, and with this definition the formula given for the product makes sense and defines a closed path at $z$, and in fact is the same formula you'll find for the concatenation of loops in other texts (e.g. Hatcher's Algebraic Topology).