Theorem 3

Let X be a delaceable topological space and let G be a discrete group operating properly in X; let $f: X\rightarrow X/G$ be the canonical surjection. The canonical morphism of groupoids $\varpi ''(f):\varpi (X)/G\rightarrow \varpi (X/G)$ introduced above is an isomorphism.

Let Coeg($f$) denote the coequalizer of the two morphisms of groupoids from $\varpi (X\times_YX)$ to $\varpi (X)$ induced by the projections pr$_1$ and pr$_2$; let $\gamma :\varpi (X)\rightarrow$ Coeg($f$) be the canonical morphism of groupoids. Since the image of the mapping $(m$, pr$_2): G\times X\rightarrow X\times X$ is the subspace $X\times_YX$ of $X\times X$, the two morphisms of groupoids $\gamma \circ \varpi (m)$ and $\gamma \circ \varpi$(pr$_2$), from $|G|\times \varpi (X)$ to Coeg($f$), are equal. By the universal property of $\varpi (X)/G$, there exists a unique morphism of groupoids $\alpha :\varpi (X)/G\rightarrow$ Coeg($f$) such that $\gamma =\alpha \circ \beta$.

Let also $\varpi '(f)$ denote the unique morphism of groupoids from Coeg($f$) to $\varpi (Y)$ such that $\varpi (f) =\varpi '(f)\circ \gamma$. By IV, p. 399, Example 2, the hypotheses of Theorem 1 of IV, p. 398 are satisfied and the morphism $\varpi '(f)$ is an isomorphism. Since

we have $\varpi ''(f) =\varpi '(f)\circ \alpha$. Thus, to prove Theorem 3, it is enough to prove that the morphism $\alpha$ is an isomorphism.