Proposition 3

Let

${X'}^{f'}$ X

$p'p$

${B'}^f$ B

be a cartesian square. For every continuous section $s$ of $p'$, the mapping $f'\circ s$ is a continuous lifting of $f$ to X. The mapping $s\mapsto f'\circ s$ is a bijection of $\mathscr{C}_{B'}(B'; X')$ onto $\mathscr{C}_B(B'; X)$.

If $s: B'\rightarrow X'$ is a continuous section of $p'$, we have $p\circ f'\circ s=$ $f\circ p'\circ s=f$, which proves that $f'\circ s$ is a continuous lifting of $f$ to X, i.e. a B-morphism from $B'$ into X. Conversely, let $g: B'\rightarrow X$ be a B-morphism. We have $f\circ$ Id$_{B'}=p\circ g$; there therefore exists, by definition of a cartesian square, a unique continuous mapping $s: B'\rightarrow X'$ such that $p'\circ s=$ Id$_{B'}$ and $f'\circ s=g$, whence the proposition.