Proposition 11

Let X be a simply connected topological space and G a discrete group operating continuously on the right in X. If the subgroup of G generated by the union of the stabilizers of the points of X is equal to G, the space $X/G$ is simply connected.

Let $(E, p)$ be a covering of $X/G$. We have to prove that, for every point $x$ of E, there exists a continuous section $s$ of $p$ such that $s\circ p(x) =x($I, p. 70, cor. 2 of prop. 1). Let $q: X\rightarrow X/G$ denote the canonical mapping and choose a point $y$ of X such that $q(y) =p(x)$. Since the space X is simply connected, there exists a continuous mapping $f: X\rightarrow E$ such that $f(y) =x$ and $p\circ f=q($I, p. 125, prop. 3). Let $z$ be a point of X and $g$ an element of the subgroup of G stabilizing $z$. The mappings $t\mapsto f(t)$ and $t\mapsto f(t\cdot g)$ of X into E are continuous liftings to E of the mapping $q$ which coincide for $t=z$. Since the space X is connected, they are equal (I, p. 34, cor. 1 of prop. 11). Since the union of the stabilizer subgroups of the points of X generates G, one has $f(t\cdot g) =f(t)$ for every $t\in X$ and every $g\in G$. Passing to the quotient, one deduces from $f$ a continuous mapping $s: X/G\rightarrow E$ which is a continuous section of $p$ such that $s(p(x)) =x$.