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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