Lemma 96.10.2. Let $S$ be a scheme. Let $Z \to B$, $X \to B$, and $B' \to B$ be morphisms of algebraic spaces over $S$. Set $Z' = B' \times _ B Z$ and $X' = B' \times _ B X$. Then

$\mathit{Mor}_{B'}(Z', X') = B' \times _ B \mathit{Mor}_ B(Z, X)$

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 96.10.1 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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