Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 3 / 2 / b / i / Solution

There is one constant path and $r$ loop generators; every other path is a word in those loops. Distinct words are distinct basis paths, so there are no relations among the generators. Thereforethe free associative algebra on $r$ generators, with each loop sent to its corresponding generator. For $r=1$ this is $k[x_1]$; for $r\ge 2$ it is noncommutative. The quiver with one loop is the $r=1$ case of the quiver with r loops.