Proposition 1

Every morphism of principal fibre spaces is an isomorphism.

Let $f: B\times G\rightarrow B\times G$ be a morphism of the trivial principal fibre space $(B\times G$, pr$_1)$ into itself and let $\varphi : B\times G\rightarrow G$ be the continuous mapping pr$_2\circ f$, so that $f(b, g) = (b, \varphi (b, g))$. For every $b\in B$ and all $g,h\in G$, one has $\varphi (b, gh) =\varphi (b, g)\cdot h$, whence $\varphi (b, g) =\varphi (b, e)\cdot g$, where $e$ denotes the identity element of G. The morphism $f$ is therefore an isomorphism whose inverse morphism $f^{-1}$ is defined by $f^{-1}(b, g) =$ $(b, \varphi (b, e)^{-1}g)$ for $(b, g)\in B\times G$.

Now let E and $E'$ be principal fibre spaces and let $f: E\rightarrow E'$ be a morphism of principal fibre spaces. In view of property (FP), for every point $b\in B$, there exists an open neighbourhood U of $b$ such that the principal fibre spaces $E_U$ and $E'_U$ are trivializable. By passing to subspaces, $f$ induces a morphism $f_U: E_U\rightarrow E'_U$ of principal fibre spaces; by what precedes, this morphism $f_U$ is an isomorphism. In particular, $f_b: E_b\rightarrow E'_b$ is a bijection. There then exists a unique mapping $g: E'\rightarrow E$ which induces the bijection $f_b^{-1}$ by passing to subspaces. By Proposition 4 of I, p. 19, the mapping $g$ is continuous, hence $f$ is an isomorphism of principal fibre spaces.