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The Stacks project

Lemma 70.10.1. Let f : X \to Y be a morphism of quasi-compact and quasi-separated algebraic spaces over \mathbf{Z}. Then there exists a direct set I and an inverse system (f_ i : X_ i \to Y_ i) of morphisms algebraic spaces over I, such that the transition morphisms X_ i \to X_{i'} and Y_ i \to Y_{i'} are affine, such that X_ i and Y_ i are quasi-separated and of finite type over \mathbf{Z}, and such that (X \to Y) = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i).

Proof. Write X = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a and Y = \mathop{\mathrm{lim}}\nolimits _{b \in B} Y_ b as in Proposition 70.8.1, i.e., with X_ a and Y_ b quasi-separated and of finite type over \mathbf{Z} and with affine transition morphisms.

Fix b \in B. By Lemma 70.4.5 applied to Y_ 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 Y_ b making the diagram

\xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \\ X_ a \ar[r] & Y_ b }

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 Lemma 70.4.5 again, there exists an a'' \geq \max (a, a') such that the compositions X_{a''} \to X_ a \to Y_ b \to Y_{b''} and X_{a''} \to X_{a'} \to Y_{b'} \to Y_{b''} are equal. We endow I with the preorder

(a, b, f_{a, b}) \geq (a', b', f_{a', b'}) \Leftrightarrow a \geq a',\ b \geq b',\text{ and } g_{b, b'} \circ f_{a, b} = f_{a', b'} \circ h_{a, a'}

where h_{a, a'} : X_ a \to X_{a'} and g_{b, b'} : Y_ b \to Y_{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, Y_ i = Y_ b, and f_ i = f_{a, b}, then we get an inverse system of morphisms over I and we have

\mathop{\mathrm{lim}}\nolimits _{i \in I} X_ i = \mathop{\mathrm{lim}}\nolimits _{a \in A} X_ a = X \quad \text{and}\quad \mathop{\mathrm{lim}}\nolimits _{i \in I} S_ i = \mathop{\mathrm{lim}}\nolimits _{b \in B} Y_ b = Y

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


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