The Stacks project

Proof. Consider the set $I$ of pairs $(U, V)$ where $U \subset X$ and $V \subset S$ are affine open and $U \to S$ factors through $V$. For $i \in I$ denote $(U_ i, V_ i)$ the corresponding pair. Set $F_ i = \mathit{Mor}_{V_ i}(Z_{V_ i}, U_ i)$. It is immediate that $F_ i$ is a subfunctor of $\mathit{Mor}_ S(Z, X)$. Then we claim that conditions (2)(a), (2)(b), (2)(c) of Schemes, Lemma 26.15.4 which proves the lemma.

Condition (2)(a) follows from Lemma 37.68.2.

To check condition (2)(b) consider $T/S$ and $b \in \mathit{Mor}_ S(Z, X)$. Thinking of $b$ as a morphism $T \times _ S Z \to X$ we find an open $b^{-1}(U_ i) \subset T \times _ S Z$. Clearly, $b \in F_ i(T)$ if and only if $b^{-1}(U_ i) = T \times _ S Z$. Since the projection $p : T \times _ S Z \to T$ is finite hence closed, the set $U_{i, b} \subset T$ of points $t \in T$ with $p^{-1}(\{ t\} ) \subset b^{-1}(U_ i)$ is open. Then $f : T' \to T$ factors through $U_{i, b}$ if and only if $b \circ f \in F_ i(T')$ and we are done checking (2)(b).

Finally, we check condition (2)(c) and this is where our condition on $X \to S$ is used. Namely, consider $T/S$ and $b \in \mathit{Mor}_ S(Z, X)$. It suffices to prove that every $t \in T$ is contained in one of the opens $U_{i, b}$ defined in the previous paragraph. This is equivalent to the condition that $b(p^{-1}(\{ t\} )) \subset U_ i$ for some $i$ where $p : T \times _ S Z \to T$ is the projection and $b : T \times _ S Z \to X$ is the given morphism. Since $p$ is finite, the set $b(p^{-1}(\{ t\} )) \subset X$ is finite and contained in the fibre of $X \to S$ over the image $s$ of $t$ in $S$. Thus our condition on $X \to S$ exactly shows a suitable pair exists. $\square$