Algebraic stacks satisfy the (strong) Rim-Schlessinger condition
Lemma 98.4.1. Let $S$ be a scheme. Let be a pushout in the category of schemes over $S$ where $X \to X'$ is a thickening and $X \to Y$ is affine, see More on Morphisms, Lemma 37.14.3. Let $\mathcal{Z}$ be an algebraic stack over $S$. Then the functor of fibre categories is an equivalence of categories.
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.
$\square$
\[ \xymatrix{ X \ar[r] \ar[d] & X' \ar[d] \\ Y \ar[r] & Y' } \]
\[ \mathcal{Z}_{Y'} \longrightarrow \mathcal{Z}_ Y \times _{\mathcal{Z}_ X} \mathcal{Z}_{X'} \]
\[ (\mathit{Sch}/Y)_{fppf} \times _{y, \mathcal{Z}} \mathcal{W} \quad \text{and}\quad (\mathit{Sch}/X')_{fppf} \times _{x', \mathcal{Z}} \mathcal{W} \]
\[ (\mathit{Sch}/Y_1') \to (\mathit{Sch}/V') \to (\mathit{Sch}/W) = \mathcal{W} \to \mathcal{Z} \]
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (3)
Comment #3821 by slogan_bot on
Comment #6883 by Cedric Luger on
Comment #6980 by Johan on