Proposition 8

Let

${E'}^{f'}$ E

$p'p$

${B'}^f$ B

be a cartesian square. If the mapping $p$ is étale, the mapping $p'$ is étale. Conversely, if the mapping $p'$ is étale and the mapping $f$ is universally strict and surjective, the mapping $p$ is étale.

Suppose that the mapping $p$ is étale. It is in particular open and $p'$ is open (I, p. 17, Prop. 8). Consider then the cartesian square (I, p. 13, Example 1)

$q'q$

${B'}^fB$. One has $\Delta_{E'}=\overset{-1}{\varphi}(\Delta_E)$ (loc. cit.). Moreover, the diagonal $\Delta_E$ is open in $E\times_BE$ (Prop. 7), hence the diagonal $\Delta_{E'}$ is open in $E'\times_{B'}E'$. This proves that the mapping $p'$ is étale (loc. cit.).

Suppose now that the mapping $f$ is surjective and universally strict and that the mapping $p'$ is étale. Then $p'$ is open, hence $p$ is open (I, p. 21, Prop. 11, a)). On the other hand $\Delta_{E'}$ is an open subset of $E'\times_{B'}E'$. Since $\Delta_{E'}=\overset{-1}{\varphi}(\Delta_E)$ and the mapping $\varphi$ is surjective and strict (I, p. 20, Def. 6), $\Delta_E$ is open in $E\times_BE$. By Prop. 7, the mapping $p$ is étale.