Lemma 91.28.2. Let $S$ be a scheme. Let $X \to B$ and $Y \to B$ be morphisms of algebraic spaces over $S$. The object $E$ in (91.28.0.1) satisfies $H^ i(E) = 0$ for $i = 0, -1$ and for a geometric point $(\overline{x}, \overline{y}) : \mathop{\mathrm{Spec}}(k) \to X \times _ B Y$ we have

$H^{-2}(E)_{(\overline{x}, \overline{y})} = \text{Tor}_1^ R(A, B) \otimes _{A \otimes _ R B} C$

where $R = \mathcal{O}_{B, \overline{b}}$, $A = \mathcal{O}_{X, \overline{x}}$, $B = \mathcal{O}_{Y, \overline{y}}$, and $C = \mathcal{O}_{X \times _ B Y, (\overline{x}, \overline{y})}$.

Proof. The formation of the cotangent complex commutes with taking stalks and pullbacks, see Lemmas 91.18.9 and 91.18.3. Note that $C$ is a henselization of $A \otimes _ R B$. $L_{C/R} = L_{A \otimes _ R B/R} \otimes _{A \otimes _ R B} C$ by the results of Section 91.8. Thus the stalk of $E$ at our geometric point is the cone of the map $L_{A/R} \otimes C \to L_{A \otimes _ R B/R} \otimes C$. Therefore the results of the lemma follow from the case of rings, i.e., Lemma 91.15.2. $\square$

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