Proposition 1

For square (1) to be Cartesian, it is necessary and sufficient that it be commutative and that, for every B-space Y and every pair of B-morphisms $u: Y\rightarrow B',v: Y\rightarrow X$, there exist a unique B-morphism $w: Y\rightarrow X'$ such that $p'\circ w=u$ and $f'\circ w=v$.

Suppose that square (1) is Cartesian. Let Y be a B-space and let $u: Y\rightarrow B',v: Y\rightarrow X$ be B-morphisms. The mappings $f\circ u$ and $p\circ v$ are both equal to the projection of the B-space Y; the unique continuous mapping $w$ such that $p'\circ w=u$ and $f'\circ w=v$ is then a B-morphism. This proves the necessity of the condition.

Conversely, suppose this condition satisfied. Let Y be a topological space and let $u: Y\rightarrow B',v: Y\rightarrow X$ be continuous mappings such that $f\circ u=p\circ v$. When Y is endowed with the structure of a B-space defined by $f\circ u,u$ and $v$ are B-morphisms. Every continuous mapping $w: Y\rightarrow X'$ such that $p'\circ w=u$ and $f'\circ w=v$ being a B-morphism, there exists one and only one.