Proposition 4

Let

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

$p'p$

${B'}^f$ B

be a cartesian square. If the mapping $p$ is separated, the same is true of $p'$. Conversely, if $p'$ is separated and if $f$ is universally strict and surjective, then $p$ is separated.

Consider the square

$q'q$

${B'}^f$ B

where the mapping $\varphi$ is induced by the mapping $f'\times f': X'\times X'\rightarrow X\times X$. Recall (I, p. 13, example 1) that this square is cartesian and that one has $\overset{-1}{\varphi}(\Delta_X) = \Delta_{X'}$.

If the mapping $p$ is separated, the set $\Delta_X$ is closed in $X\times_BX$; consequently, $\Delta_{X'}$ is a closed subset of $X'\times_{B'}X'$, which proves that the mapping $p'$ is separated.

If the mapping $f$ is surjective, the mapping $\varphi$ is surjective (I, p. 10, cor.) ; if $f$ is universally strict, $\varphi$ is strict (I, p. 20, def. 6). Suppose the mapping $p'$ separated. Then $\Delta_{X'}$ is closed in $X'\times_{B'}X'$. Since $\Delta_{X'}=\overset{-1}{\varphi}(\Delta_X)$ and since $\varphi$ is surjective and strict, $\Delta_X$ is closed in $X\times_BX$, which proves that the mapping $p$ is separated.