The Stacks project

Proof. Recall that $\text{Inf}_ x(M)$ is the set of automorphisms of the trivial deformation of $x$ to $A[M]$ which induce the identity automorphism of $x$ over $A$. The trivial deformation is the pullback of $x$ to $\mathop{\mathrm{Spec}}(A[M])$ via $\mathop{\mathrm{Spec}}(A[M]) \to \mathop{\mathrm{Spec}}(A)$. Let $G \to \mathop{\mathrm{Spec}}(A)$ be the automorphism group algebraic space of $x$ (this exists because $\mathcal{X}$ is an algebraic space). Let $e : \mathop{\mathrm{Spec}}(A) \to G$ be the neutral element. The discussion in More on Morphisms of Spaces, Section 76.17 gives

By the same token

Since $G \to \mathop{\mathrm{Spec}}(A)$ is locally of finite presentation by assumption, we see that $\Omega _{G/A}$ is locally of finite presentation, see More on Morphisms of Spaces, Lemma 76.7.15. Hence $e^*\Omega _{G/A}$ is a finitely presented $A$-module. Moreover, $\Omega _{G_ B/B}$ is the pullback of $\Omega _{G/A}$ by More on Morphisms of Spaces, Lemma 76.7.12. Therefore $e_ B^*\Omega _{G_ B/B} = e^*\Omega _{G/A} \otimes _ A B$. we conclude by More on Algebra, Lemma 15.65.4. $\square$

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

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

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

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

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

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. $\square$

Proof. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. We first reduce the lemma to the case where $x$ lifts to $U$. Recall that $T_ x(M)$ is the set of isomorphism classes of lifts of $x$ to $A[M]$. Therefore Lemma 106.7.2¹ says that the rule

is a sheaf on the small étale site of $\mathop{\mathrm{Spec}}(A)$; the tensor product is needed to make $A[M] \to A_1[M \otimes _ A A_1]$ a flat ring map. We may choose a faithfully flat étale ring map $A \to A_1$ such that $x|_{A_1}$ lifts to a morphism $u_1 : \mathop{\mathrm{Spec}}(A_1) \to U$, see for example Sheaves on Stacks, Lemma 96.19.10. Write $A_2 = A_1 \otimes _ A A_1$ and set $B_1 = B \otimes _ A A_1$ and $B_2 = B \otimes _ A A_2$. Consider the diagram

The rows are exact by the sheaf condition. We have $M \otimes _ A B_ i = (M \otimes _ A A_ i) \otimes _{A_ i} B_ i$. Thus if we prove the result for the middle and right vertical arrow, then the result follows. This reduces us to the case discussed in the next paragraph.

Assume that $x$ is the image of a morphism $u : \mathop{\mathrm{Spec}}(A) \to U$. Observe that $T_ u(M) \to T_ x(M)$ is surjective since $U \to \mathcal{X}$ is smooth and representable by algebraic spaces, see Criteria for Representability, Lemma 97.6.3 (see discussion preceding it for explanation) and More on Morphisms of Spaces, Lemma 76.19.6. Set $R = U \times _\mathcal {X} U$. Recall that we obtain a groupoid $(U, R, s, t, c, e, i)$ in algebraic spaces with $\mathcal{X} = [U/R]$. By Artin's Axioms, Lemma 98.21.6 we have an exact sequence

where the zero on the right was shown above. A similar sequence holds for the base change to $B$. Thus the result we want follows if we can prove the result of the lemma for $T_ u(M)$ and $T_{e \circ u}(M)$. This reduces us to the case discussed in the next paragraph.

Assume that $\mathcal{X} = X$ is an algebraic space locally of finite presentation over $S$. Then we have

by the discussion in More on Morphisms of Spaces, Section 76.17. By the same token

Since $X \to S$ is locally of finite presentation, we see that $\Omega _{X/S}$ is locally of finite presentation, see More on Morphisms of Spaces, Lemma 76.7.15. Hence $x^*\Omega _{X/S}$ is a finitely presented $A$-module. Clearly, we have $y^*\Omega _{X/S} = x^*\Omega _{X/S} \otimes _ A B$. we conclude by More on Algebra, Lemma 15.65.4. $\square$

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

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

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

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. $\square$