Corollary 2

Let B and X be topological spaces and let $p: X\rightarrow B$ be a continuous mapping. Let $(A_i)_{i\in I}$ be a family of subsets of B which is an open covering of B, or else a locally finite closed covering of B. If for every $i\in$ I, the mapping $p_{A_i}:\overset{-1}{p}(A_i)\rightarrow A_i$ is strict ( resp. universally strict), the mapping $p$ is strict ( resp. universally strict).

For every $i\in I$, put $Y_i=\overset{-1}{p}(A_i)$ and $p_i=p_{A_i}$. Let A be the sum space of the family $(A_i)_{i\in I}$, Y the sum space of the family $(Y_i)_{i\in I}$; let us denote by $f: A\rightarrow B$ (resp. $g: Y\rightarrow X,q: Y\rightarrow A$) the mapping deduced from the family of canonical injections $A_i\rightarrow B$ (resp. from the canonical injections $Y_i\rightarrow X$, from the mappings $p_i$). The square

Y $^g$ X

$qp$

A $^f$ B

is a cartesian square (examples, I, p. 10 and p. 14). By the corollary, I, p. 20, the mapping $f$ is universally strict. By proposition 11, it is therefore enough to prove that the mapping $q$ is strict (resp. universally strict) if the mappings $p_i$, for $i\in I$, are so. We are thus reduced to proving the corollary when the sets $A_i$, $i\in I$, constitute a partition of the space B into open and closed subsets, which we shall henceforth suppose.

Suppose that each of the mappings $p_i,i\in I$, is strict. If U is an open subset of X and saturated for the equivalence relation defined by $p$, the set $p_i(X_i\cap U)$ is open in $A_i$ and $p(U) =$ $\bigcup_{i\in I}p_i(X_i\cap U)$ is open in B. The mapping $p$ is therefore strict.

Suppose now that each of the mappings $p_i,i\in I$, is universally strict, and now prove that the mapping $p$ is universally strict. Let C be a topological space and $h: C\rightarrow B$ a continuous mapping. It is a question of proving that the mapping pr$_1: X\times_BC\rightarrow C$ is strict. The space C is identified with the sum space of the family of the $C_i=\overset{-1}{h}(A_i)$ and the space $X\times_BC$ is identified with the sum space of the family of the $X\times_BC_i= X_i\times_{A_i}C_i$. Since $p_i$ is universally strict, the mapping pr$_1: X_i\times_{A_i}C_i\rightarrow C_i$ is strict. By what precedes, the mapping pr$_1: X\times_BC\rightarrow C$ is strict. This proves that the mapping $p$ is universally strict.