Lemma 106.7.2. Let $\mathcal{X}$ be an algebraic stack over a base scheme $S$. Assume $\mathcal{I}_\mathcal {X} \to \mathcal{X}$ is locally of finite presentation. Let $(A' \to A, x)$ be a deformation situation. Then the functor

\[ F : B' \longmapsto \{ \text{lifts of }x|_{B' \otimes _{A'} A}\text{ to } B'\} /\text{isomorphisms} \]

is a sheaf on the site $(\textit{Aff}/\mathop{\mathrm{Spec}}(A'))_{fppf}$ of Topologies, Definition 34.7.8.

**Proof.**
Let $\{ T'_ i \to T'\} _{i = 1, \ldots n}$ be a standard fppf covering of affine schemes over $A'$. Write $T' = \mathop{\mathrm{Spec}}(B')$. As usual denote

\[ T'_{i_0 \ldots i_ p} = T'_{i_0} \times _{T'} \ldots \times _{T'} T'_{i_ p} = \mathop{\mathrm{Spec}}(B'_{i_0 \ldots i_ p}) \]

where the ring is a suitable tensor product. Set $B = B' \otimes _{A'} A$ and $B_{i_0 \ldots i_ p} = B'_{i_0 \ldots i_ p} \otimes _{A'} A$. Denote $y = x|_ B$ and $y_{i_0 \ldots i_ p} = x|_{B_{i_0 \ldots i_ p}}$. Let $\gamma _ i \in F(B'_ i)$ such that $\gamma _{i_0}$ and $\gamma _{i_1}$ map to the same element of $F(B'_{i_0i_1})$. We have to find a unique $\gamma \in F(B')$ mapping to $\gamma _ i$ in $F(B'_ i)$.

Choose an actual object $y'_ i$ of $\textit{Lift}(y_ i, B'_ i)$ in the isomorphism class $\gamma _ i$. Choose isomorphisms $\varphi _{i_0i_1} : y'_{i_0}|_{B'_{i_0i_1}} \to y'_{i_1}|_{B'_{i_0i_1}}$ in the category $\textit{Lift}(y_{i_0i_1}, B'_{i_0i_1})$. If the maps $\varphi _{i_0i_1}$ satisfy the cocycle condition, then we obtain our object $\gamma $ because $\mathcal{X}$ is a stack in the fppf topology. The cocycle condition is that the composition

\[ y'_{i_0}|_{B'_{i_0i_1i_2}} \xrightarrow {\varphi _{i_0i_1}|_{B'_{i_0i_1i_2}}} y'_{i_1}|_{B'_{i_0i_1i_2}} \xrightarrow {\varphi _{i_1i_2}|_{B'_{i_0i_1i_2}}} y'_{i_2}|_{B'_{i_0i_1i_2}} \xrightarrow {\varphi _{i_2i_0}|_{B'_{i_0i_1i_2}}} y'_{i_0}|_{B'_{i_0i_1i_2}} \]

is the identity. If not, then these maps give elements

\[ \delta _{i_0i_1i_2} \in \text{Inf}_{y_{i_0i_1i_2}}(J_{i_0i_1i_2}) = \text{Inf}_ y(J) \otimes _ B B_{i_0i_1i_2} \]

Here $J = \mathop{\mathrm{Ker}}(B' \to B)$ and $J_{i_0 \ldots i_ p} = \mathop{\mathrm{Ker}}(B'_{i_0 \ldots i_ p} \to B_{i_0 \ldots i_ p})$. The equality in the displayed equation holds by Lemma 106.7.1 applied to $B' \to B'_{i_0 \ldots i_ p}$ and $y$ and $y_{i_0 \ldots i_ p}$, the flatness of the maps $B' \to B'_{i_0 \ldots i_ p}$ which also guarantees that $J_{i_0 \ldots i_ p} = J \otimes _{B'} B'_{i_0 \ldots i_ p}$. A computation (omitted) shows that $\delta _{i_0i_1i_2}$ gives a $2$-cocycle in the Čech complex

\[ \prod \text{Inf}_ y(J) \otimes _ B B_{i_0} \to \prod \text{Inf}_ y(J) \otimes _ B B_{i_0i_1} \to \prod \text{Inf}_ y(J) \otimes _ B B_{i_0i_1i_2} \to \ldots \]

By Descent, Lemma 35.9.2 this complex is acyclic in positive degrees and has $H^0 = \text{Inf}_ y(J)$. Since $\text{Inf}_{y_{i_0i_1}}(J_{i_0i_1})$ acts on morphisms (Artin's Axioms, Remark 98.21.4) this means we can modify our choice of $\varphi _{i_0i_1}$ to get to the case where $\delta _{i_0i_1i_2} = 0$.

Uniqueness. We still have to show there is at most one $\gamma $ restricting to $\gamma _ i$ for all $i$. Suppose we have objects $y', z'$ of $\textit{Lift}(y, B')$ and isomorphisms $\psi _ i : y'|_{B'_ i} \to z'|_{B'_ i}$ in $\textit{Lift}(y_ i, B'_ i)$. Then we can consider

\[ \psi _{i_1}^{-1} \circ \psi _{i_0} \in \text{Inf}_{y_{i_0i_1}}(J_{i_0i_1}) = \text{Inf}_ y(J) \otimes _ B B_{i_0i_1} \]

Arguing as before, the obstruction to finding an isomorphism between $y'$ and $z'$ over $B'$ is an element in the $H^1$ of the Čech complex displayed above which is zero.
$\square$

## Comments (0)