Lemma 81.12.5. Let (R \to R', f) be a glueing pair, see above. The functor (81.12.0.1) is fully faithful on the full subcategory of algebraic spaces Y/X which are (a) glueable for (R \to R', f) and (b) have affine diagonal Y \to Y \times _ X Y.
Proof. Let Y, Z be two algebraic spaces over X which are both glueable for (R \to R', f) and assume the diagonal of Z is affine. Let a : U \times _ X Y \to U \times _ X Z over U and b : X' \times _ X Y \to X' \times _ X Z over X' be two morphisms of algebraic spaces which induce the same morphism c : U' \times _ X Y \to U' \times _ X Z over U'. We want to construct a morphism f : Y \to Z over X which produces the morphisms a, b on base change to U, X'. By the faithfulness of Lemma 81.12.4, it suffices to construct the morphism f étale locally on Y (details omitted). Thus we may and do assume Y is affine.
Let y \in |Y| be a point. If y maps into the open U \subset X, then U \times _ X Y is an open of Y on which the morphism f is defined (we can just take a). Thus we may assume y maps into the closed subset V(f) of X. Since R/fR = R'/fR' there is a unique point y' \in |X' \times _ X Y| mapping to y. Denote z' = b(y') \in |X' \times _ X Z| and z \in |Z| the images of y'. Choose an étale neighbourhood (W, w) \to (Z, z) with W affine. Observe that
(U \times _ X W) \times _{U \times _ X Z, a} (U \times _ X Y),\quad (U' \times _ X W) \times _{U' \times _ X Z, c} (U' \times _ X Y),
and
(X' \times _ X W) \times _{X' \times _ X Z, b} (X' \times _ X Y)
form an object of \textit{Spaces}(U \leftarrow U' \to X') with affine parts (this is where we use that Z has affine diagonal). Hence by Lemma 81.12.2 there exists a unique affine scheme V glueable for (R \to R', f) such that
(U \times _ X V, U' \times _ X V, X' \times _ X V)
is the triple displayed above. By fully faithfulness for the affine case (Lemma 81.12.2) we get a unique morphisms V \to W and V \to Y agreeing with the first and second projection morphisms over U and X' in the construction above. By Lemma 81.12.3 the morphism V \to Y is étale. To finish the proof, it suffices to show that there is a point v \in |V| mapping to y (because then f is defined on an étale neighbourhood of y, namely V). There is a unique point w' \in |X' \times _ X W| mapping to w. By uniqueness w' is mapped to z' under the map |X' \times _ X W| \to |X' \times _ X Z|. Then we consider the cartesian diagram
\xymatrix{ X' \times _ X V \ar[r] \ar[d] & X' \times _ X W \ar[d] \\ X' \times _ X Y \ar[r] & X' \times _ X Z }
to see that there is a point v' \in |X' \times _ X V| mapping to y' and w', see Properties of Spaces, Lemma 66.4.3. Of course the image v of v' in |V| maps to y and the proof is complete. \square
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