(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.

By definition of the inverse image groupoid (II, p. 166, example 4), A is the set of vertices of the groupoid $f^*G$ and one has Fl$_{a,b}(f^*G) =$ Fl$_{f(a),f(b)}(G)$ for every pair $(a, b)$ of elements of A. Moreover, one has Som($\varphi$ ) $=f$ and Fl($\varphi$ ) induces the identical mapping of Fl$_{a,b}(f^*G)$ into Fl$_{f(a),f(b)}(G)$. Consequently, the mapping Orb($\varphi$ ) is bijective and the homomorphism $\varphi_a: (f^*G)_a\rightarrow G_{f(a)}$ is an isomorphism, for every $a\in A$. The hypotheses of proposition 1 are therefore satisfied.

It follows in fact from remark 1 of II, p. 170 and corollary 2 of II, p. 178 that the hypotheses of proposition 1 are satisfied.

This subquiver being the image of the quiver $G_1$ under the quiver morphism $\pi \circ \theta_1$, the proposition follows at once from the construction of Coh($\varphi , \psi$ ).

The triple (Coh($\varphi , \psi$ )$, \alpha , h$) has the following universal property: if $G'$ is a groupoid, $\alpha '$ a morphism of groupoids from G into $G'$ and $h':$ Som(H) $\rightarrow$ Fl(G$'$) a homotopy joining $\alpha '\circ \varphi$ to $\alpha '\circ \psi$, there exists a unique morphism of groupoids $\eta :$ Coh($\varphi , \psi$ )$\rightarrow G'$ such that

(3) $\alpha '=\eta \circ \alpha$ and $h'=$ Fl($\eta$ )$\circ h$.

In view of the definition of the groupoid deduced by contraction of arrows, the contraction of the loops (1) implies that $\alpha$ is a morphism of groupoids, and the contraction of the loops (2) that $h$ is a homotopy joining $\alpha \circ \varphi$ to $\alpha \circ \psi$.

Let $G',\alpha ',h'$ be as in the statement. Let $\eta_1$ be the morphism of quivers from $G_1$ into $G'$ such that Som($\eta_1$) is equal to Som($\alpha '$) and such that Fl($\eta_1$) coincides with Fl($\alpha '$) on Fl(G) and with $h'$ on Som(H). There exists a unique morphism of groupoids $\eta_2:$ Grp(G$_1$)$\rightarrow G'$ such that $\eta_1=\eta_2\circ \theta_1$. Since $\alpha '$ is a morphism of groupoids and $h'$ is a homotopy joining $\alpha '\circ \varphi$ to $\alpha '\circ \psi ,\eta_2$ defines by passing to the quotient a morphism of groupoids $\eta$ from Coh($\varphi , \psi$ ) into $G'($II, p. 170, proposition 3). This morphism satisfies relations (3), and it is the only one (II, p. 185, proposition 2).

The morphism $\alpha$ is the composite of the morphisms $\alpha_1: G\rightarrow G_1$, $\theta_1: G_1\rightarrow$ Grp(G$_1$) and $\pi :$ Grp(G$_1$)$\rightarrow$ Coh($\varphi , \psi$ ). The mapping $\theta_1$ induces a bijection of the set of connected components of the quiver $G_1$ onto the set of orbits of Grp(G$_1$), and the mapping Orb($\pi$ )$:$ Orb(Grp(G$_1$))$\rightarrow$ Orb(Coh($\varphi , \psi$ )) is bijective (II, p. 170, remark 1). It is therefore enough to prove that the mapping of Orb(G) into $\pi_0(G_1)$ deduced from $\alpha_1$ is surjective and that its fibres are the connected components of the framework of the pair $(\varphi , \psi )$. The surjectivity results from the fact that the mapping Som($\alpha_1$) is the identity mapping. The equivalence relation in Som(G) defined by “$a$ and $b$ are in the same connected component of $G_1$” is generated by the relations “there exists an arrow of G joining $a$ to $b$” and “there exists a vertex $h$ of H such that $\varphi (h) =a$ and $\psi (h) =b$”. This equivalence relation is compatible with the mapping of Som(G) into Orb(G), and the relation deduced from it in Orb(G) is generated by the relation “there exists an orbit $\eta$ of H such that Orb($\varphi$ )$(\eta ) =\alpha$ and Orb($\psi$ )$(\eta ) =\beta$”. It is therefore the relation “$\alpha$ and $\beta$ are in the same connected component of the framework of the pair $(\alpha , \beta )$”.

(i) The mapping $k$ is a homotopy joining $\eta \circ \alpha$ to $\eta '\circ \alpha$;

(ii) For every vertex $a$ of H, one has

By definition, in order that $k$ be a homotopy joining $\eta$ to $\eta '$, it is necessary and sufficient that the following two conditions be fulfilled (recall that Som(G) = Som(Coh($\varphi , \psi$ ))) :

a) For every vertex $x$ of Coh($\varphi , \psi$ )$,k(x)$ joins $\eta (x)$ to $\eta '(x)$ ;

b) For every pair $(x, y)$ of vertices of Coh($\varphi , \psi$ ) and every arrow $f\in$ Fl$_{x,y}$(Coh($\varphi , \psi$ )), one has $\eta (f)k(y) =k(x)\eta '(f)$.

By Prop. 2, it is enough to verify condition b) when $f$ belongs to the image of Fl($\alpha$ ) or to that of $h$, so that b) is equivalent to the conjunction of the two conditions c) and d) below:

c) For $x\in$ Som(G), $y\in$ Som(G) and $g\in$ Fl$_{x,y}(G)$, one has $\eta (\alpha (g))k(y) =k(x)\eta '(\alpha (g))$;

d) For every $a\in$ Som(H), one has $\eta (h(a))k(\psi (a)) =k(\varphi (a))\eta '(h(a))$. Condition (i) is equivalent to the conjunction of a) and c), and condition (ii) is condition d), whence the corollary.

(i) the morphism of groupoids $v$ is a homotopism;

(ii) the mapping Orb($u$)$:$ Orb(H) $\rightarrow$ Orb(H$'$), deduced from $u$ by passing to orbits, is bijective;

(iii) there exists in each orbit of H a point $a$ such that the homomorphism $u_a: H_a\rightarrow H'_{u(a)}$ is surjective.

Then, the groupoid morphism $w:$ Coh($\varphi , \psi$ )$\rightarrow$ Coh($\varphi ', \psi '$) is a homotopism.

Let $G''$ be the groupoid deduced from $G'$ by contraction of the arrows of a maximal oriented forest. The canonical morphism $v': G'\rightarrow G''$ is a homotopism (II, p. 184, corollary 2 of prop. 1). The two diagrams

H $^{\varphi}$ G ${H'}^{\varphi'}G'$

$u\psi v'\circ v$ et Id$_{H'}\psi 'v'$

give rise to groupoid morphisms $w'_1:$ Coh($\varphi , \psi$ )$\rightarrow$ Coh($v'\circ \varphi ', v'\circ \psi '$) and $w'_2:$ Coh($\varphi ', \psi '$)$\rightarrow$ Coh($v'\circ \varphi ', v'\circ \psi '$) ; we have $w'_1=w'_2\circ w$. It is therefore enough to prove that $w'_1$ and $w'_2$ are homotopisms. Since the mappings Som($v'\circ v$)$:$ Som(G) $\rightarrow$ Som(G$''$) and Som($v'$)$:$ Som(G$'$)$\rightarrow$ Som(G$''$) are surjective, it is consequently enough to prove the theorem under the additional assumption that the mapping Som($v$) is surjective, an assumption that we shall make throughout the remainder of the proof.

Let us prove successively the following assertions:

– The mapping Orb($w$) is bijective;

– For every vertex $a$ of G, the homomorphism $w_a$ is surjective;

– For every vertex $a$ of G, the homomorphism $w_a$ is injective.

a) By assumption, the mapping Orb($u$) is bijective; the same is true of the mapping Orb($v$) by II, p. 182, prop. 1, since $v$ is a homotopism. The morphism of quivers from the framework of the pair $(\varphi , \psi )$ to that of the pair $(\varphi ', \psi ')$ defined by the mappings Orb($u$) and Orb($v$) is therefore an isomorphism. In particular, the mapping deduced from it by passing to connected components is bijective. Prop. 4 of II, p. 185 then implies that the mapping Orb($w$) is bijective.

(b) Let $f'$ be an arrow of $G'$, and let us denote by $a'$ its origin and by $b'$ its term. Since the mapping Som($v$) is surjective, there exist vertices $a$ and $b$ in G such that $a'=v(a)$ and $b'=v(b)$. Since the morphism $v$ is a homotopism, there exist an arrow $f$ of G joining $a$ to $b$ and an element $g\in G_a$ such that $v(g) =f'v(f)^{-1}($II, p. 182, prop. 1), whence $f'=v(gf)$. This proves that the mapping Fl($v$) is surjective.

Let us then prove that the mapping Fl($w$) is surjective. Its image contains that of Fl($\alpha '$), for one has $\alpha '\circ v=w\circ \alpha$ and the mapping Fl($v$) is surjective. Let $b$ be a vertex of $H'$; let $a$ be a vertex of H such that $b$ and $u(a)$ are in the same orbit of $H'$, and let $f$ be an arrow of $H'$ joining $u(a)$ to $b$. Then,

since $h'$ is a homotopy joining $\alpha '\circ \varphi '$ to $\alpha '\circ \psi '$. The arrow $h'(u(a)) =$ $w(h(a))$ belong to the image of Fl($w$), and so do the two arrows $\alpha '(\psi '(f))$ and $\alpha '(\varphi '(f))$ by what precedes. It follows that the arrow $h'(b)$ belong to the image of Fl($w$), which proves that the image of Fl($w$) contains that of $h'$. By prop. 2 of II, p. 185, the mapping Fl($w$) is surjective.

Let $g'\in$ Coh($\varphi ', \psi '$)$_{u(a)}$. Let $g$ be an arrow of Coh($\varphi , \psi$ ) such that $w(g) =g'$. Let us denote by $x$ and $y$ the origin and the end of $g$; one has $u(x) =$ $u(y) =u(a)$. Let $g_1$ (resp. $g_2$) be an arrow of G joining $a$ to $x$ (resp. $a$ to $y$) whose image under $v$ is $e_{u(a)}$. Then, $\alpha (g_1)g\alpha (g_2)^{-1}$ is an element of Coh($\varphi , \psi$ )$_a$ whose image under $w_a$ is $g'$. Hence, for every vertex $a$ of G, the homomorphism $w_a$ is surjective.

c) Let us prove that, for every vertex $a$ of G, the homomorphism $w_a$ is injective. By considering successively the diagrams

one is thus reduced to treating the two following cases: 1) One has $H'= H$ and $u=$ Id$_H; 2)$ One has $G'= G$ and $v=$ Id$_G$.

1) Suppose that one has $H'= H$ and $u=$ Id$_H$.

Consider a morphism of groupoids $v': G'\rightarrow G$ which is an inverse of $v$ up to homotopy and a homotopy $k:$ Som(G) $\rightarrow$ Fl(G) joining $v'\circ v$ to Id$_G$. By Remark 1 of II, p. 181, the mappings $\alpha \circ k\circ \varphi$ and $\alpha \circ k\circ \psi$ are homotopies joining respectively $\alpha \circ v'\circ \varphi '$ to $\alpha \circ \varphi$ and $\alpha \circ v'\circ \psi '$ to $\alpha \circ \psi$. Hence (cf. II, p. 180), the mapping

$h_1:$ Som(H) $\rightarrow$ Fl(Coh($\varphi , \psi$ )),

is a homotopy joining $\alpha \circ v'\circ \varphi '$ to $\alpha \circ v'\circ \psi '$. By the universal property of cohomotopers (II, p. 185, prop. 3), there exists a unique groupoid morphism $w':$ Coh($\varphi ', \psi '$)$\rightarrow$ Coh($\varphi , \psi$ ) such that $\alpha \circ v'=$ $w'\circ \alpha '$ and $h_1=$ Fl($w'$)$\circ h'$.

H $^{\varphi}_{\psi}$ G $^{\alpha}$ Coh($\varphi , \psi$ )

Id$_Hvw$

H $^{\varphi'}_{\psi'}{G'}^{\alpha'}$ Coh($\varphi ', \psi '$)

Id$_Hv'w'$

H $^{\varphi}_{\psi}$ G $^{\alpha}$ Coh($\varphi , \psi$ ) .

In particular one has

Since $k$ is a homotopy joining $v'\circ v$ to Id$_G,\alpha \circ k$ is a homotopy joining $w'\circ w\circ \alpha$ to $\alpha$. Since Fl($w'\circ w$)$\circ h=$ Fl($w'$)$\circ h'=h_1$, one has, for every vertex $x$ of H,

Fl($w'\circ w$)$\circ h(x)\cdot (\alpha \circ k\circ \psi )(x) =h_1(x)\cdot (\alpha \circ k\circ \psi )(x) = (\alpha \circ k\circ \varphi )(x)\cdot h(x)$,

by definition of $h_1$. By prop. 5 of II, p. 186, applied to the groupoid morphisms $w'\circ w$ and Id$_{Coh(\varphi ,\psi)}$, the mapping $\alpha \circ k$ is a homotopy joining $w'\circ w$ to Id$_G$. In particular, $w'\circ w$ is a homotopism.

For every vertex $a$ of G, the group homomorphism $(w'\circ w)_a$ is therefore bijective (II, p. 182, prop. 1). It follows that the homomorphism $w_a$ is injective, whence the result in case A).

2) Suppose that one has $G'= G$ and $v=$ Id$_G$.

Let $x$ be a vertex of $H'$. Since the mapping Orb($u$) is surjective, there exist a vertex $a$ of H and an arrow $f$ of $H'$ joining $u(a)$ to $x$. The arrows $\alpha (\varphi '(f))^{-1},h(a)$ and $\alpha (\psi '(f))$ join respectively $\varphi '(x)$ to $\varphi '(u(a)) =\varphi (a),\varphi (a)$ to $\psi (a)$ and $\psi '(u(a)) =\psi (a)$ to $\psi '(x)$, and therefore are composable in Coh($\varphi , \psi$ ). Put

Let us verify that the resulting arrow $h_2(x)$ does not depend on the chosen elements $a$ and $f$. Let $a'$ be a vertex of H and let $f'$ be an arrow of $H'$ joining $u(a')$ to $x$. Since the mapping Orb($u$) is injective, the vertices $a$ and $a'$ of H belong to the same orbit and there exists an arrow $c\in$ Fl(H) joining $a$ to $a'$. Then $u(c)f'f^{-1}$ is a loop at $u(a)$ in $H'$. By hypothesis (iii), there exists a vertex $b$ of H such that the homomorphism $u_b$ is surjective and an arrow $c'$ of H joining $b$ to $a$; then, Int($u(c')$)$(u(c)f'f^{-1})$ is a loop at $u(b)$ in H, and is therefore the image under $u_b$ of a loop $c''$ at $b$. Consequently, the arrow $g= (c'c)^{-1}c''c'$ of H joins the vertex $a'$ to the vertex $a$ and satisfies $f'=u(g)f$. We then have

since $h$ is a homotopy joining $\alpha \circ \varphi$ to $\alpha \circ \psi$. This proves the announced independence.

By construction, we have $h_2(u(x)) =h(x)$ for all $x\in$ Som(H). Thus we have defined a mapping $h_2$ of Som(H$'$) into Fl(Coh($\varphi , \psi$ )). Let $c$ be an arrow of $H'$, and let $x$ denote its origin and $y$ its end. Let $a$ be a vertex of H and $f$ an arrow of $H'$ joining $u(a)$ to $x$. Then $f c$ is an arrow of $H'$ joining $u(a)$ to $y$. By definition of $h_2$, we thus have

This proves that $h_2$ is a homotopy joining $\alpha \circ \varphi '$ to $\alpha \circ \psi '$.

By the universal property of cohomotopizers, there exists a unique morphism of groupoids $w':$ Coh($\varphi ', \psi '$)$\rightarrow$ Coh($\varphi , \psi$ ) such that $w'\circ \alpha '=\alpha$ and $h_2$ = Fl($w'$)$\circ h'$. We have $w'\circ w\circ \alpha =w'\circ \alpha '=\alpha$ and Fl($w'\circ w$)$\circ h=$ Fl($w'$)$\circ h'\circ$ Som($u$) $=h_2\circ u=h$ by definition of $h_2$. By the universal property of cohomotopizers, this implies that we have $w'\circ w=$ Id$_{Coh(\varphi ,\psi)}$. In particular, for every $a\in$ Som(G), the homomorphism $w_a$ is injective.

It then follows from Prop. 1 of II, p. 182 that the morphism $w$ is a homotopism, whence the theorem.