Proposition 9

Let X and Y be topological spaces and let $f$ be a continuous mapping of X into Y. Suppose that the space X is locally arcwise connected, that the space Y is simply connected, and that the mapping $f$ is strict and surjective. Then the groupoid $\varpi (Y)$ is generated by the image of $\varpi (X)$ under $\varpi (f)$.

We may suppose that the spaces X and Y are non-empty. Let G denote the subgroupoid of $\varpi (Y)$ generated by the image of $\varpi (X)$ under $\varpi (f)$. Since the mapping $f$ is surjective, the set of vertices of G is equal to Y. By prop. 8, Coeg($f$) is a transitive groupoid. Since $\varpi '(f)$ induces the identity on the vertices, the image of Coeg($f$) under $\varpi '(f)$ is a transitive subgroupoid of $\varpi (Y)$. The image of $\varpi (X)$ under $\gamma$ generates Coeg($f$) (II, p. 200, corollary); since one has $\varpi (f) =$ $\varpi '(f)\circ \gamma$, the groupoid G is transitive.

Let $y_0$ be a point of Y and let H denote the subgroup $G_{y_0}$ of $\pi_1(Y, y_0)$. By Th. 1 of IV, p. 342, there exists a connected covering $(T, p)$ of Y and a point $t_0$ of the fibre $T_{y_0}$ of which H is the stabilizer.

If $x$ is a point of X, the set Fl$_{y_0,f(x)}(G)$ is not empty, since G is transitive. For $u\in$ Fl$_{y_0,f(x)}(G)$, the point $t_0\cdot u$ is a point of the fibre $T_{f(x)}$, independent of $u$ since the group H, the isotropy group of G at $y_0$, fixes $t_0$. Let us denote this point by $\sigma (x)$; let $\sigma : X\rightarrow T$ denote the resulting mapping and $s: X\rightarrow X\times_YT$ the mapping $x\mapsto (x, \sigma (x))$. By construction, the mapping $s$ is compatible with the canonical operations of $\varpi (X)$ in X and $X\times_YT$. It then follows from lemma 1 of III, p. 312 that $s$ is continuous. The mapping $\sigma$ is therefore continuous; moreover it is compatible with the equivalence relation defined by $f$, for $\sigma (x)$ depends only on $f(x)$. Since $f$ is strict and surjective, there exists a unique continuous mapping $\overline{\sigma}: Y\rightarrow T$ such that $\overline{\sigma}\circ f=\sigma$, so that the covering T admits a section. Since T is connected, the mapping $p: T\rightarrow Y$ is a homeomorphism (I, p. 31, cor. 4 of prop. 6), and one has $H =\pi_1(Y, y_0)$, whence $G_{y_0}=\pi_1(Y, y_0)$. Since G is transitive, it follows that $G =\varpi (Y)$.