Corollary 1

In a graph, every class of paths contains a unique path without retracing.

Let G be a graph. Let us denote by $j: G\rightarrow \varpi_G$ the canonical quiver morphism.

The existence, for every path $c$ of G, of an equivalent path without retracing is immediate by induction on the length of $c$ (cf. II, p. 157).

Let A be an orientation of G and let $G'$ be the groupoid associated with the free group F(A) constructed on A (II, p. 163, Example 1). Let $\psi$ be the morphism of quivers from G into $G'$ such that, for every $f\in A,\psi (f)$ is the element $f$ of F(A) and $\psi (\overline{f})$ is the element $f^{-1}$ of F(A). Let $\psi '$ be the unique morphism of groupoids from $\varpi_G$ into $G'$ such that $\psi '\circ j=\psi$.

Let $c,c'$ be paths without backtracking equivalent modulo $\mathscr{R}$. They have the same source and the same end. To prove that they are equal, it is enough to prove that the sequences $(f_1, . . . , f_n)$ and $(g_1, . . . , g_m)$ of their edges are equal. The image under $\psi '$ of the common class of $c$ and $c'$ is equal to $\psi (f_1). . . \psi (f_n)$ and to $\psi (g_1). . . \psi (g_m)$. Now the terms of the sequences $(\psi (f_1), . . . , \psi (f_n))$ and $(\psi (g_1), . . . , \psi (g_m))$ belong to the subset $A\cup A^{-1}$ of F(A) and two consecutive elements of these sequences are not inverses of one another. By A, I, p. 84, Prop. 7, these two sequences are equal. It follows that the sequences $(f_1, . . . , f_n)$ and $(g_1, . . . , g_m)$ are equal, and therefore that $c=c'$, whence uniqueness.