The Stacks project

Proof. Let $y'$ be an object of left hand side. The sheaf $\mathit{Isom}(y', y')$ on the category of schemes over $Y'$ is representable by an algebraic space $I$ over $Y'$, see Algebraic Stacks, Lemma 94.10.11. We conclude that the functor of the lemma is fully faithful as $Y'$ is the pushout in the category of algebraic spaces as well as the category of schemes, see Pushouts of Spaces, Lemma 81.6.1.

Let $(y, x', f)$ be an object of the right hand side. Here $f : y|_ X \to x'|_ X$ is an isomorphism. To finish the proof we have to construct an object $y'$ of $\mathcal{Z}_{Y'}$ whose restrictions to $Y$ and $X'$ agree with $y$ and $x'$ in a manner compatible with $f$. In fact, it suffices to construct $y'$ fppf locally on $Y'$, see Stacks, Lemma 8.4.8. Choose a representable algebraic stack $\mathcal{W}$ and a surjective smooth morphism $\mathcal{W} \to \mathcal{Z}$. Then

are algebraic stacks representable by algebraic spaces $V$ and $U'$ smooth over $Y$ and $X'$. The isomorphism $f$ induces an isomorphism $\varphi : V \times _ Y X \to U' \times _{X'} X$ over $X$. By Pushouts of Spaces, Lemmas 81.6.2 and 81.6.7 we see that the pushout $V' = V \amalg _{V \times _ Y X} U'$ is an algebraic space smooth over $Y'$ whose base change to $Y$ and $X'$ recovers $V$ and $U'$ in a manner compatible with $\varphi$.

Let $W$ be the algebraic space representing $\mathcal{W}$. The projections $V \to W$ and $U' \to W$ agree as morphisms over $V \times _ Y X \cong U' \times _{X'} X$ hence the universal property of the pushout determines a morphism of algebraic spaces $V' \to W$. Choose a scheme $Y_1'$ and a surjective étale morphism $Y_1' \to V'$. Set $Y_1 = Y \times _{Y'} Y_1'$, $X_1' = X' \times _{Y'} Y_1'$, $X_1 = X \times _{Y'} Y_1'$. The composition

corresponds by the $2$-Yoneda lemma to an object $y_1'$ of $\mathcal{Z}$ over $Y_1'$ whose restriction to $Y_1$ and $X_1'$ agrees with $y|_{Y_1}$ and $x'|_{X_1'}$ in a manner compatible with $f|_{X_1}$. Thus we have constructed our desired object smooth locally over $Y'$ and we win. $\square$