Proposition 3

Let G be a groupoid, let H be a distinguished subgroupoid of G and let $p: G\rightarrow G/H$ be the canonical morphism. Let $\varphi : G\rightarrow G'$ be a groupoid morphism such that $H\subset$ Ker($\varphi$ ). There exists a unique groupoid morphism $\overline{\varphi}: G/H\rightarrow G'$ such that $\overline{\varphi}\circ p=\varphi$.

One says that $\overline{\varphi}$ is the groupoid morphism deduced from $\varphi$ by passing to the quotient.

The uniqueness of such a morphism is obvious, since the mappings Som($p$) and Fl($p$) are surjective.

Let $a$ and $b$ be vertices of G. If they are in the same orbit of H, there exists an arrow $f$ joining $a$ to $b$ in H, and one has $\varphi (f) =e_{\varphi(a)}$. In particular, $\varphi (a) =\varphi (b)$. Consequently, the mapping Som($\varphi$ ) defines, by passing to the quotient, a mapping $\overline{\varphi}_1:$ Orb(H) $\rightarrow$ Som(G$'$). Let $f$ and $g$ be arrows in G. If $f$ and $g$ are equivalent modulo $\mathscr{R}$, there exist arrows $x$ and $y$ in H such that $f$ = $xgy$. Consequently, $\varphi (f) =\varphi (x)\varphi (g)\varphi (y) =\varphi (g)$ since $\varphi (x)$ and $\varphi (y)$ are identity elements. The mapping Fl($\varphi$ ) therefore defines, by passing to the quotient, a mapping $\overline{\varphi}_2:$ Fl(G)$/\mathscr{R}\rightarrow$ Fl(G$'$).

Let $f$ and $g$ be composable arrows of G; let us denote by $u$ and $v$ their classes in Fl(G$/H$). One has $\overline{\varphi}_2(uv) =\varphi (f g) =\varphi (f)\varphi (g) =$ $\overline{\varphi}_2(u)\overline{\varphi}_2(v)$.

The pair $\overline{\varphi}= (\overline{\varphi}_1, \varphi_2)$ is a groupoid morphism from $G/H$ into $G'$ and one has $\overline{\varphi}\circ p=\varphi$.