The Stacks project

Proof. Write $X = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a$ and $S = \mathop{\mathrm{lim}}\nolimits _{b \in B} S_ b$ as in Proposition 32.5.4, i.e., with $X_ a$ and $S_ b$ of finite type over $\mathbf{Z}$ and with affine transition morphisms.

Fix $b \in B$. By Proposition 32.6.1 applied to $S_ b$ and $X = \mathop{\mathrm{lim}}\nolimits X_ a$ over $\mathbf{Z}$ we find there exists an $a \in A$ and a morphism $f_{a, b} : X_ a \to S_ b$ making the diagram

commute. Let $I$ be the set of triples $(a, b, f_{a, b})$ we obtain in this manner.

Let $(a, b, f_{a, b})$ and $(a', b', f_{a', b'})$ be in $I$. Let $b'' \leq \min (b, b')$. By Proposition 32.6.1 again, there exists an $a'' \geq \max (a, a')$ such that the compositions $X_{a''} \to X_ a \to S_ b \to S_{b''}$ and $X_{a''} \to X_{a'} \to S_{b'} \to S_{b''}$ are equal. We endow $I$ with the preorder

where $h_{a, a'} : X_ a \to X_{a'}$ and $g_{b, b'} : S_ b \to S_{b'}$ are the transition morphisms. The remarks above show that $I$ is directed and that the maps $I \to A$, $(a, b, f_{a, b}) \mapsto a$ and $I \to B$, $(a, b, f_{a, b})$ are cofinal. If for $i = (a, b, f_{a, b})$ we set $X_ i = X_ a$, $S_ i = S_ b$, and $f_ i = f_{a, b}$, then we get an inverse system of morphisms over $I$ and we have

by Categories, Lemma 4.17.4 (recall that limits over $I$ are really limits over the opposite category associated to $I$ and hence cofinal turns into initial). This finishes the proof. $\square$