Lemma

Given composable arrows $u$ and $v$ of $G/H$, one can choose representatives $f$ and $g$ of $u$ and $v$ in Fl(G) which are composable. The class modulo $\mathscr{R}$ of the product $f g$ does not depend on the choice of $f$ and $g$.

Let $f$ and $g$ be arbitrary representatives of $u$ and $v$ in Fl(G). The end of $f$ and the origin of $g$ belong to one and the same orbit of H, hence can be joined by an arrow $x$ of H (II, p. 162, prop. 1). The arrows $f x$ and $g$ are then representatives of $u$ and $v$ in F which are composable in G. This proves the first assertion.

Let now $f,g$ on the one hand, and $f',g'$ on the other hand, be representatives of $u$ and $v$ which are composable in G. By hypothesis, there exist arrows $x,y,z,t$ in H such that one has $f'=xf y$ and $g'=zgt$. Then one has $yz\in H_{t(f)}$ and $f'g'=xf yzgt=xf(yz)f^{-1}f gt$. Since H is a normal subgroupoid of G, the arrow $f(yz)f^{-1}$ is an arrow of H. The same is therefore true of the arrow $xf(yz)f^{-1}$, which proves that $f'g'$ and $f g$ are equivalent modulo $\mathscr{R}$.

By virtue of this lemma, there exists a unique law of composition $m$ in the quiver $G/H$ such that, for every pair $(u, v)$ of composable arrows, $m(u, v)$ is the class modulo $\mathscr{R}$ of the product $f g$, for every pair $(f, g)$ of composable arrows of G such that $u$ is the class of $f$ and $v$ that of $g$. Endowed with this law of composition, $G/H$ is a groupoid. Let $p_1:$ Som(G) $\rightarrow$ Som(G$/H$) and $p_2:$ Fl(G) $\rightarrow$ Fl(G$/H$) be the canonical surjections. The pair $p= (p_1, p_2)$ is a morphism of groupoids of G into $G/H$ whose kernel is the normal subgroupoid H.