Proposition 3

Let E and B be topological spaces, $p: E\rightarrow B$ a continuous mapping and G a topological group operating on the right on E. The following conditions are equivalent:

(i) Endowed with the mapping $p$, E is a principal fibre space with base B and group G;

(ii) For every $x\in E$ and every $g\in G$, one has $p(x\cdot g) =p(x)$, the mapping $\theta : (x, g)\mapsto (x, x\cdot g)$ is a homeomorphism of $E\times G$ onto $E\times_BE$ and every point of B possesses a neighbourhood over which there exists a continuous section of the mapping $p$.

(i)$\Rightarrow$(ii) : Suppose that $(E, p)$ is a principal fibre space. The group G operates continuously and freely in E, with the fibres of $p$ as orbits. Hence the mapping $\theta$ is bijective and continuous. More precisely, if $E\times_BE$ is endowed with the operation of G defined by $(x, y)\cdot g= (x, y\cdot g)$, the mapping $\theta$ is an E-morphism of the trivial principal fibre space $E\times G$ into the principal fibre space $(E\times_BE\rightarrow E$, pr$_1)$ (example 5), hence an isomorphism (prop. 1). The other assertions of (ii) result immediately from the definition.

(ii)$\Rightarrow$(i) : Put $\varphi =$ pr$_2\circ \theta^{-1}$, so that, for every $(x, y)\in$ $E\times_BE$, one has

The mapping $\varphi : E\times_BE\rightarrow G$ is continuous and, for $(x, y)\in E\times_BE$, the element $\varphi (x, y)$ is the unique element $g$ of G such that $y=x\cdot g$. Let $b$ be a point of B; let U be a neighbourhood of $b$ and let $s$ be a continuous section of $p$ over U. For $(u, g)\in U\times G$, put $f(u, g) =s(u)\cdot$ $g$; the mapping $f$ is a U-morphism of $U\times G$ into $\overset{-1}{p}(U)$. For $y\in \overset{-1}{p}(U)$, put $f'(y) = (p(y), \varphi (s(p(y)), y))$. The mapping $f'$ is a U-morphism of $\overset{-1}{p}(U)$ into $U\times G$. One has

On the other hand, for $(u, g)\in U\times G$, one has

Hence $f$ is a U-isomorphism. It follows that $(E, p)$ is a principal fibre space with base B and group G.