Lemma 81.12.6. Let (R \to R', f) be a glueing pair, see above. Any object (V, V', Y') of \textit{Spaces}(U \leftarrow U' \to X') with V, V', Y' quasi-affine is isomorphic to the image under the functor (81.12.0.1) of a separated algebraic space Y over X.
Proof. Choose n', T' \to Y' and n_1, T_1 \to V as in Properties, Lemma 28.18.6. Picture
Observe that T_1 \times _ V V' and V' \times _{Y'} T' are affine (namely the morphisms V' \to V and V' \to Y' are affine as base changes of the affine morphisms U' \to U and U' \to X'). By construction we see that
In other words, the affine schemes \mathbf{A}^{n'}_{T_1} and \mathbf{A}^{n_1}_{T'} are part of a triple making an affine object of \textit{Spaces}(U \leftarrow U' \to X'). By Lemma 81.12.2 there exists a morphism of affine schemes T \to X and isomorphisms U \times _ X T \cong \mathbf{A}^{n'}_{T_1} and X' \times _ X T \cong \mathbf{A}^{n_1}_{T'} compatible with the isomorphisms displayed above. These isomorphisms produce morphisms
satisfying the property of Properties, Lemma 28.18.6 with n = n' + n_1 and moreover define a morphism from the triple (U \times _ X T, U' \times _ X T, X' \times _ X T) to our triple (V, V', Y') in the category \textit{Spaces}(U \leftarrow U' \to X').
By Lemma 81.12.2 there is an affine scheme W whose image in \textit{Spaces}(U \leftarrow U' \to X') is isomorphic to the triple
By fully faithfulness of this construction, we obtain two maps p_0, p_1 : W \to T whose base changes to U, U', X' are the projection morphisms. By Lemma 81.12.3 the morphisms p_0, p_1 are flat and of finite presentation and the morphism (p_0, p_1) : W \to T \times _ X T is a closed immersion. In fact, W \to T \times _ X T is an equivalence relation: by the lemmas used above we may check symmetry, reflexivity, and transitivity after base change to U and X', where these are obvious (details omitted). Thus the quotient sheaf
is an algebraic space for example by Bootstrap, Theorem 80.10.1. Since it is clear that Y/X is sent to the triple (V, V', Y'). The base change of the diagonal \Delta : Y \to Y \times _ X Y by the quasi-compact surjective flat morphism T \times _ X T \to Y \times _ X Y is the closed immersion W \to T \times _ X T. Thus \Delta is a closed immersion by Descent on Spaces, Lemma 74.11.17. Thus the algebraic space Y is separated and the proof is complete. \square
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