Lemma 4

For every path $c= (c_1, c_2)$ in $X\times_YX$, we have $\beta ([c_1]) =\beta ([c_2])$.

Let $c$ be such a path.

Let $x\in X$, and let $K_x$ be its stabilizer in G. By TG, III, p. 32, prop. 8, there exists an open neighbourhood $U_x$ of $x$ in X such that $K_x\cdot U_x=$ $U_x,g\cdot U_x\cap U_x$ = $\emptyset$ for every $g\in$ G - $K_x$, and such that the mapping $f$ induces a homeomorphism of $U_x/K_x$ onto an open neighbourhood $V_x$ of $f(x)$ in Y. As X is unlaceable and the restriction of the mapping $f$ to $U_x$ is open and closed, we may further suppose that $U_x$ is connected and that the image of the canonical homomorphism $\pi_1(U_x, x)\rightarrow \pi_1(X, x)$ is reduced to the identity element. The open sets $(V_x)_{x\in X}$ thus constructed form an open covering of Y. By lemma 4 of III, p. 272, applied to the compact space $\mathbf{I}$ and to the open sets $(f\circ c_1)^{-1}(V_x)$ for $x\in X$, there exists an integer $n\geqslant 1$ such that for every $i\in \{1, . . . , n\}$, there exists a point $x_i$ in X such that $c_1([\frac{i-1}{n},\frac{i}{n}])$

is contained in $\overset{-1}{f}(V_{x_i})$. As $f\circ c_1=f\circ c_2,c_2([\frac{i-1}{n},\frac{i}{n}])$ is also

contained in $\overset{-1}{f}(V_{x_i})$.

For $j= 1$ or 2 and for $i\in \{1, . . . , n\}$, let $c_{j,i}$ denote the path in X defined by $t\mapsto c_j(\frac{i+t-1}{n})$; we have $c_j=c_{j,1}* \cdots *c_{j,n}($III, p. 291, remark 1). Hence, to prove that $\beta ([c_1]) =\beta ([c_2])$, it is enough to prove that $\beta ([c_{1,i}]) =\beta ([c_{2,i}])$ for every integer $i\in \{1, . . . , n\}$. Replacing the pair of paths $(c_1, c_2)$ by the pair $(c_{1,i}, c_{2,i})$ if necessary, we thus suppose that there exists $x\in X$ such that $(f\circ c_1)([0,1])\subset V_x$.

The inverse image of $V_x$ by $f$ is the disjoint union of the connected components $g\cdot U_x$, where $g$ runs through a system of representatives in G of $G/K_x$. For $i= 1$ or 2, let $g_i$ be an element of G such that the point $x_i=$ $g_i\cdot c_i(0)$ belongs to $U_x$. The image of the path $g_i\cdot c_i$ is then contained in $U_x$. By definition of the groupoid $\varpi (X)/G$, we have $\beta ([g_i\cdot c_i]) =\beta ([c_i])$, which makes it possible to suppose that the images of the paths $c_1$ and $c_2$ are contained in $U_x$ and that $g_1=g_2=e$.

For $s= 0$ or 1, let $d_s$ be a path in $U_x$ joining $x$ to $c_1(s)$ and let $g_s$ be an element of $K_x$ such that $g_s\cdot c_2(s) =c_1(s)$. The paths $d_0*c_1*\overline{d_1}$ and $d_0*(g_0\cdot c_2)*(g_0g_1^{-1}\cdot \overline{d_1})$ are loops at $x$ in $U_x$. They are therefore strictly homotopic in X to the constant loop at $x$, because the image of the canonical homomorphism $\pi_1(U_x, x)\rightarrow \pi_1(X, x)$ reduces to the identity element. Their classes have in particular the same image under the groupoid morphism $\beta$, whence

Since $\beta ([g\cdot c]) =\beta ([c])$ for every element $g\in G$ and every path $c$ in X, it follows that $\beta ([c_1]) =\beta ([c_2])$, as was to be proved.

By the lemma, the two groupoid morphisms $\beta \circ \varpi$(pr$_1$) and $\beta \circ \varpi$(pr$_2$) of $\varpi (X\times_YX)$ into Coeg($f$) are equal. By the universal property of the coequaliser, there therefore exists a unique groupoid morphism $\alpha ':$ Coeg($f$)$\rightarrow \varpi (X)/G$ such that $\beta =\alpha '\circ \gamma$. The morphism $\alpha '\circ \alpha$ is the unique groupoid morphism $\varphi$ of $\varpi (X)/G$ into itself such that $\varphi \circ \beta =\beta$; hence $\alpha '\circ \alpha$ = Id$_{\varpi(X)/G}$. Analogously, $\alpha \circ \alpha '=$ Id$_{Coeg(f)}$. Consequently, $\alpha$ is an isomorphism.