Lemma 37.9.1. Let $S$ be a scheme. Let $X \subset X'$ and $Y \subset Y'$ be two first order thickenings over $S$. Let $(a, a'), (b, b') : (X \subset X') \to (Y \subset Y')$ be two morphisms of thickenings over $S$. Assume that

$a = b$, and

the two maps $a^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ (Morphisms, Lemma 29.31.3) 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 $\mathcal{O}_ S$-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 $\mathcal{O}_ S$-algebra maps $(a')^\sharp $ and $(b')^\sharp $ is an $\mathcal{O}_ S$-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 $\mathcal{O}_ S$-derivation from $\mathcal{O}_ Y$ into $a_*\mathcal{C}_{X/X'}$. Some details omitted.
$\square$

## Comments (0)