Proposition 1

Let G, $G'$ be groupoids, let $\varphi$ be a morphism from G into $G'$, and let A be a subset of the set of vertices of G which meets every orbit of G. In order that $\varphi$ be a homotopism, it is necessary and sufficient that the following conditions be satisfied:

(i) the mapping Orb($\varphi$ ) from Orb(G) into Orb(G$'$), deduced from $\varphi$ by passing to orbits, is bijective;

(ii) for every $a\in A$, the homomorphism $\varphi_a: G_a\rightarrow G'_{\varphi(a)}$ is bijective.

Suppose first that $\varphi$ is a homotopism and let $\psi$ be an inverse of $\varphi$ up to homotopy. Then,

Orb($\psi$ )$\circ$ Orb($\varphi$ ) $=$ Orb($\psi \circ \varphi$ ) $=$ Orb(Id$_G$) $=$ Id$_{Orb(G)}$,

for two homotopic groupoid morphisms induce the same mapping by passing to orbits. Analogously, Orb($\varphi$ )$\circ$ Orb($\psi$ ) $=$ Id$_{Orb(G')}$. The mapping Orb($\varphi$ ) is therefore bijective, whence assertion (i). We also have, for every vertex $a$ of G,

since $\psi \circ \varphi$ is homotopic to Id$_G$, the homomorphism $(\psi \circ \varphi )_a$ is bijective (cf. p. 180), so that $\varphi_a$ is injective and $\psi_{\varphi(a)}$ is surjective. On interchanging the roles of $\varphi$ and $\psi$, we see that $\varphi_a$ is also surjective, whence condition (ii).

Suppose now that conditions (i) and (ii) are satisfied, and let us prove that $\varphi$ is a homotopism.

Let us first treat the case where each orbit of $G'$ is reduced to a point.

For each vertex $b$ of $G'$, choose a vertex $u(b)$ of G belonging to A and whose image under $\varphi$ is $b$. This is possible because the mapping Orb($\varphi$ ) is surjective and A meets each orbit of G. Let $f$ be an arrow of $G'$; we have $o(f) =t(f)$ by hypothesis; put $b=o(f)$ and $a=u(b)$. By condition (ii), there exists a unique arrow $v(f)\in G_a$ whose image under $\varphi$ is $f$. The pair $\psi = (u, v)$ is a groupoid morphism from $G'$ into G. Let us prove that $\psi$ is inverse to $\varphi$ up to homotopy. We already have $\varphi \circ \psi =$ Id$_{G'}$ by construction of $\psi$.

Let $x$ be a vertex of G. Put $a=\psi (\varphi (x))$; this is an element of A such that $\varphi (a) =\varphi (x)$. Since Orb($\varphi$ ) is injective, $a$ belongs to the orbit of $x$ in G and there exists an arrow $f$ in G joining $a$ to $x$. The arrow $h(x) =\psi (\varphi (f))^{-1}f$ then joins $a$ to $x$ and one has $\varphi (h(x)) =e_{\varphi(a)}$.

Let us show that the mapping $h:$ Som(G) $\rightarrow$ Fl(G) resulting is a homotopy joining $\psi \circ \varphi$ to Id$_G$. Condition (i) of definition 1 is satisfied, by construction. Let $f$ be an arrow of G$,x$ its origin and $y$ its term. One has $\varphi (x) =\varphi (y)$ ; put $a=\psi (\varphi (x)) =\psi (\varphi (y))$. The arrows $h(x)f h(y)^{-1}$ and $\psi \circ \varphi (f)$ belong to $G_a$ and both have image $\varphi (f)$ in $G'_{\varphi(a)}$. Since the mapping $\varphi_a$ is injective, one has $h(x)f h(y)^{-1}=\psi \circ \varphi (f)$. Condition (ii) of definition 1 is thus satisfied.

Let us now prove proposition 1 in the general case. Let X be a maximal oriented forest of $G'($II, p. 157, prop. 1). Let $G''$ be the groupoid deduced from $G'$ by contraction of the arrows of X and let $\varphi ': G'\rightarrow G''$ be the canonical morphism. The morphism $\varphi '$ satisfies the conditions of prop. 1 (II, p. 170, remark 1 and p. 178, cor. 2), and hence the same is true of the morphism $\varphi '\circ \varphi$.

Since the orbits of $G''$ are reduced to points, it follows from the special case already proved that $\varphi '$ and $\varphi '\circ \varphi$ are homotopisms, so that $\varphi$ is likewise one. This completes the proof of the proposition.