Example 1

Let

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

$p'p$

${B'}^f$ B

be a cartesian square. Then Corollary 2 yields a cartesian square

${B'}^f$ B .

We have the relation $\overset{-1}{\varphi}(\Delta_X) = \Delta_{X'}$. In fact, by prop. 2 of I, p. 8, it is enough to consider the case where $X'= B'\times_BX$. Let then $((b, x),(b, x'))$ be an element of $X'\times_{B'}X'$ with $b\in B'$ and $x, x'\in X$. This element belongs to $\overset{-1}{\varphi}(\Delta_X)$ if and only if $x=x'$.