The Stacks project

Proof. Let $f, g : Y \to Z$ be two morphisms of algebraic spaces over $X$ with $Y$ and $Z$ glueable for $(R \to R', f)$ such that $f$ and $g$ are mapped to the same morphism in the category $\textit{Spaces}(U \leftarrow U' \to X')$. We have to show the equalizer $E \to Y$ of $f$ and $g$ is an isomorphism. Working étale locally on $Y$ we may assume $Y$ is an affine scheme. Then $E$ is a scheme and the morphism $E \to Y$ is a monomorphism and locally quasi-finite, see Morphisms of Spaces, Lemma 67.4.1. Moreover, the base change of $E \to Y$ to $U$ and to $X'$ is an isomorphism. As $Y$ is the disjoint union of the affine open $V = U \times _ X Y$ and the affine closed $V(f) \times _ X Y$, we conclude $E$ is the disjoint union of their isomorphic inverse images. It follows in particular that $E$ is quasi-compact. By Zariski's main theorem (More on Morphisms, Lemma 37.43.3) we conclude that $E$ is quasi-affine. Set $B = \Gamma (E, \mathcal{O}_ E)$ and $A = \Gamma (Y, \mathcal{O}_ Y)$ so that we have an $R$-algebra homomorphism $A \to B$. Since $E \to Y$ becomes an isomorphism after base change to $U$ and $X'$ we obtain ring maps $B \to A_ f$ and $B \to A \otimes _ R R'$ agreeing as maps into $A \otimes _ R R'_ f$. Since $A$ is glueable for $(R \to R', f)$ we get a ring map $B \to A$ which is left inverse to the map $A \to B$. The corresponding morphism $Y = \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(B)$ maps into the open subscheme $E \subset \mathop{\mathrm{Spec}}(B)$ pointwise because this is true after base change to $U$ and $X'$. Hence we get a morphism $Y \to E$ over $Y$. Since $E \to Y$ is a monomorhism we conclude $Y \to E$ is an isomorphism as desired. $\square$