Proposition 5

Let G be a graph; let us denote by $j$ the canonical quiver morphism of G into $\varpi_G$. Let $\varphi$ be a quiver morphism of G into a groupoid $G'$ such that $\varphi (\overline{f}) =\varphi (f)^{-1}$ for every arrow $f$ of G. Then there exists a unique groupoid morphism $\varphi '$ of $\varpi_G$ into $G'$ such that $\varphi '\circ j=\varphi$.

One defines a mapping $u:$ Ch(G) $\rightarrow$ Fl(G$'$) by setting, for every path $c= (a_0, f_1, a_1, . . . , f_n, a_n)$ in G$,u(c) =e_{a_0}\varphi (f_1). . . \varphi (f_n)$. For every pair $(c, c')$ of composable paths in G, one has $u(cc') =$ $u(c)u(c')$. The mapping $u$ is compatible with the equivalence relation $\mathscr{R}$ defined above, whence, by passing to the quotient, a mapping $u':$ Ch(G)$/\mathscr{R}\rightarrow$ Fl(G$'$). Let us then set $\varphi '=$ (Som($\varphi$ )$, u'$). It is a groupoid morphism of $\varpi_G$ into $G'$ such that $\varphi '\circ j=\varphi$.

Let $\psi$ be a groupoid morphism of $\varpi_G$ into $G'$ such that $\psi \circ j=\varphi$. The equalizer of $\varphi '$ and $\psi$ is a subgroupoid of $\varpi_G$ which contains $j(G)$. Hence it is equal to $\varpi_G$, because $\varpi_G$ is generated by $j(G)$, and one has $\psi =\varphi '$.