The Stacks project

Definition 98.23.5. Let $S$ be a locally Noetherian base. Let $\mathcal{X}$ be a category fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. Assume that $\mathcal{X}$ satisfies (RS*). A naive obstruction theory is given by the following data

1. for every $S$-algebra $A$ such that $\mathop{\mathrm{Spec}}(A) \to S$ maps into an affine open $\mathop{\mathrm{Spec}}(\Lambda ) \subset S$ and every object $x$ of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(A)$ we are given an object $E_ x \in D^-(A)$ and a map $\xi _ x : E \to \mathop{N\! L}\nolimits _{A/\Lambda }$,

2. given $(x, A)$ as in (1) there are transformations of functors

   $$\text{Inf}_ x( - ) \to \mathop{\mathrm{Ext}}\nolimits ^{-1}_ A(E_ x, -) \quad \text{and}\quad T_ x(-) \to \mathop{\mathrm{Ext}}\nolimits ^0_ A(E_ x, -)$$

3. for $(x, A)$ as in (1) and a ring map $A \to B$ setting $y = x|_{\mathop{\mathrm{Spec}}(B)}$ there is a functoriality map $E_ x \to E_ y$ in $D(A)$.

These data are subject to the following conditions

1. in the situation of (3) the diagram

   $$\xymatrix{ E_ y \ar[r]_{\xi _ y} & \mathop{N\! L}\nolimits _{B/\Lambda } \\ E_ x \ar[u] \ar[r]^{\xi _ x} & \mathop{N\! L}\nolimits _{A/\Lambda } \ar[u] }$$

   is commutative in $D(A)$,

2. given $(x, A)$ as in (1) and $A \to B \to C$ setting $y = x|_{\mathop{\mathrm{Spec}}(B)}$ and $z = x|_{\mathop{\mathrm{Spec}}(C)}$ the composition of the functoriality maps $E_ x \to E_ y$ and $E_ y \to E_ z$ is the functoriality map $E_ x \to E_ z$,

3. the maps of (2) are isomorphisms compatible with the functoriality maps and the maps of Remark 98.21.3,

4. the composition $E_ x \to \mathop{N\! L}\nolimits _{A/\Lambda } \to \Omega _{A/\Lambda }$ corresponds to the canonical element of $T_ x(\Omega _{A/\Lambda }) = \mathop{\mathrm{Ext}}\nolimits ^0(E_ x, \Omega _{A/\Lambda })$, see Remark 98.21.8,

5. given a deformation situation $(x, A' \to A)$ with $I = \mathop{\mathrm{Ker}}(A' \to A)$ the composition $E_ x \to \mathop{N\! L}\nolimits _{A/\Lambda } \to \mathop{N\! L}\nolimits _{A/A'}$ is zero in

   $$\mathop{\mathrm{Hom}}\nolimits _ A(E_ x, \mathop{N\! L}\nolimits _{A/\Lambda }) = \mathop{\mathrm{Ext}}\nolimits ^0_ A(E_ x, \mathop{N\! L}\nolimits _{A/A'}) = \mathop{\mathrm{Ext}}\nolimits ^1_ A(E_ x, I)$$

   if and only if $x$ lifts to $A'$.