Proposition 3

Let

${E'}^{f'}$ E

${B'}^f$ B

be a cartesian square. Suppose that in a neighbourhood of every point of B, the mapping $f$ has a continuous section. Then, if $(E', p')$ is a locally trivial fibre $B'$-space ( resp. a covering), the B-space $(E, p)$ is a locally trivial fibre space ( resp. a covering).

It is a question of proving that every point $a$ of B possesses a neighbourhood U such that the U-space induced by $(E, p)$ over U is a locally trivial fibre space (resp. a covering). Take for U a neighbourhood of $a$ over which there exists a continuous section $s$ of $f$. Let $i: U\rightarrow B$ and $j:\overset{-1}{p}(U)\rightarrow E$ denote the canonical injections. Since square (1) is cartesian, there exists a unique continuous mapping $s':\overset{-1}{p}(U)\rightarrow E'$ such that $f'\circ s'=j$ and $p'\circ s'=s\circ p_U$. The square

U $^sB'$

is thus commutative and its composite with square (1) is the cartesian square

$\overset{-1}{p}(U)^j$ E

$p_Up$

U $^iB$.

By proposition 7 of I, p. 15, square (2) is cartesian. Corollary 2 permits one to conclude.