Lemma 106.7.4. Let $\mathcal{X}$ be an algebraic stack over a scheme $S$ whose structure morphism $\mathcal{X} \to S$ is locally of finite presentation. Let $(A' \to A, x)$ be a deformation situation. If there exists a faithfully flat finitely presented $A'$-algebra $B'$ and an object $y'$ of $\mathcal{X}$ over $B'$ lifting $x|_{B' \otimes _{A'} A}$, then there exists an object $x'$ over $A'$ lifting $x$.
Proof. Let $I = \mathop{\mathrm{Ker}}(A' \to A)$. Set $B'_1 = B' \otimes _{A'} B'$ and $B'_2 = B' \otimes _{A'} B' \otimes _{A'} B'$. Let $J = IB'$, $J_1 = IB'_1$, $J_2 = IB'_2$ and $B = B'/J$, $B_1 = B'_1/J_1$, $B_2 = B'_2/J_2$. Set $y = x|_ B$, $y_1 = x|_{B_1}$, $y_2 = x|_{B_2}$. Let $F$ be the fppf sheaf of Lemma 106.7.2 (which applies, see footnote in the proof of Lemma 106.7.3). Thus we have an equalizer diagram
\[ \xymatrix{ F(A') \ar[r] & F(B') \ar@<1ex>[r] \ar@<-1ex>[r] & F(B'_1) } \]
On the other hand, we have $F(B') = \text{Lift}(y, B')$, $F(B'_1) = \text{Lift}(y_1, B'_1)$, and $F(B'_2) = \text{Lift}(y_2, B'_2)$ in the terminology from Artin's Axioms, Section 98.21. These sets are nonempty and are (canonically) principal homogeneous spaces for $T_ y(J)$, $T_{y_1}(J_1)$, $T_{y_2}(J_2)$, see Artin's Axioms, Lemma 98.21.2. Thus the difference of the two images of $y'$ in $F(B'_1)$ is an element
\[ \delta _1 \in T_{y_1}(J_1) = T_ x(I) \otimes _ A B_1 \]
The equality in the displayed equation holds by Lemma 106.7.3 applied to $A' \to B'_1$ and $x$ and $y_1$, the flatness of $A' \to B'_1$ which also guarantees that $J_1 = I \otimes _{A'} B'_1$. We have similar equalities for $B'$ and $B'_2$. A computation (omitted) shows that $\delta _1$ gives a $1$-cocycle in the Čech complex
\[ T_ x(I) \otimes _ A B \to T_ x(I) \otimes _ A B_1 \to T_ x(I) \otimes _ A B_2 \to \ldots \]
By Descent, Lemma 35.9.2 this complex is acyclic in positive degrees and has $H^0 = T_ x(I)$. Thus we may choose an element in $T_ x(I) \otimes _ A B = T_ y(J)$ whose boundary is $\delta _1$. Replacing $y'$ by the result of this element acting on it, we find a new choice $y'$ with $\delta _1 = 0$. Thus $y'$ maps to the same element under the two maps $F(B') \to F(B'_1)$ and we obtain an element o $F(A')$ by the sheaf condition. $\square$
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