Free category

Remarks

Concatenation is associative and the empty paths are units for it, so $F(G)$ is a category. It has the universal property of a free object: for every category $\mathcal{C}$, each graph morphism from $G$ to the underlying graph of $\mathcal{C}$ extends to exactly one functor $F(G) \to \mathcal{C}$.

Example: let $G$ have one vertex and one loop $a$. The paths are $a^n = (a, \dots, a)$ with $n \ge 0$ letters, and $a^m \circ a^n = a^{m + n}$, so $F(G)$ is the one-object category of the monoid $(\N, +)$, the free monoid on one generator.

Non-example: the category with one object and morphisms $\Z/2\Z$ is generated by one loop $a$, but it satisfies $a \circ a = \id$. In the free category on one loop, $a \circ a$ has length $2$ and the identity has length $0$, so they differ. So the category of $\Z/2\Z$ is not free on that graph.