Lemma 76.17.1. Let S be a scheme. Let B be an algebraic space over S. Let X \subset X' and Y \subset Y' be two first order thickenings of algebraic spaces over B. Let (a, a'), (b, b') : (X \subset X') \to (Y \subset Y') be two morphisms of thickenings over B. Assume that
a = b, and
the two maps a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'} (Lemma 76.5.3) are equal.
Then the map (a')^\sharp - (b')^\sharp factors as
\mathcal{O}_{Y'} \to \mathcal{O}_ Y \xrightarrow {D} a_*\mathcal{C}_{X/X'} \to a_*\mathcal{O}_{X'}
where D is an \mathcal{O}_ B-derivation.
Proof.
Instead of working on Y we work on X. The advantage is that the pullback functor a^{-1} is exact. Using (1) and (2) we obtain a commutative diagram with exact rows
\xymatrix{ 0 \ar[r] & \mathcal{C}_{X/X'} \ar[r] & \mathcal{O}_{X'} \ar[r] & \mathcal{O}_ X \ar[r] & 0 \\ 0 \ar[r] & a^{-1}\mathcal{C}_{Y/Y'} \ar[r] \ar[u] & a^{-1}\mathcal{O}_{Y'} \ar[r] \ar@<1ex>[u]^{(a')^\sharp } \ar@<-1ex>[u]_{(b')^\sharp } & a^{-1}\mathcal{O}_ Y \ar[r] \ar[u] & 0 }
Now it is a general fact that in such a situation the difference of the \mathcal{O}_ B-algebra maps (a')^\sharp and (b')^\sharp is an \mathcal{O}_ B-derivation from a^{-1}\mathcal{O}_ Y to \mathcal{C}_{X/X'}. By adjointness of the functors a^{-1} and a_* this is the same thing as an \mathcal{O}_ B-derivation from \mathcal{O}_ Y into a_*\mathcal{C}_{X/X'}. Some details omitted.
\square
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