Lemma 97.11.2. Let $S$ be a scheme. Let $X \to Z \to B$ and $B' \to B$ be morphisms of algebraic spaces over $S$. Set $Z' = B' \times _ B Z$ and $X' = B' \times _ B X$. Then
in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})$.
Lemma 97.11.2. Let $S$ be a scheme. Let $X \to Z \to B$ and $B' \to B$ be morphisms of algebraic spaces over $S$. Set $Z' = B' \times _ B Z$ and $X' = B' \times _ B X$. Then
in $\mathop{\mathit{Sh}}\nolimits ((\mathit{Sch}/S)_{fppf})$.
Proof. The equality as functors follows immediately from the definitions. The equality as sheaves follows from this because both sides are sheaves according to Lemma 97.11.1 and the fact that a fibre product of sheaves is the same as the corresponding fibre product of pre-sheaves (i.e., functors). $\square$
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