Theorem 1

Make the following hypotheses:

(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.