Lemma 37.9.8. Let $S$ be a scheme. Let $X \subset X'$ be a first order thickening over $S$. Let $Y$ be a scheme over $S$. Let $a', b' : X' \to Y$ be two morphisms over $S$ with $a = a'|_ X = b'|_ X$. This gives rise to a commutative diagram

$\xymatrix{ X \ar[r] \ar[d]_ a & X' \ar[d]^{(b', a')} \\ Y \ar[r]^-{\Delta _{Y/S}} & Y \times _ S Y }$

Since the horizontal arrows are immersions with conormal sheaves $\mathcal{C}_{X/X'}$ and $\Omega _{Y/S}$, by Morphisms, Lemma 29.31.3, we obtain a map $\theta : a^*\Omega _{Y/S} \to \mathcal{C}_{X/X'}$. Then this $\theta$ and the derivation $D$ of Lemma 37.9.1 are related by Equation (37.9.1.1).

Proof. Omitted. Hint: The equality may be checked on affine opens where it comes from the following computation. If $f$ is a local section of $\mathcal{O}_ Y$, then $1 \otimes f - f \otimes 1$ is a local section of $\mathcal{C}_{Y/(Y \times _ S Y)}$ corresponding to $\text{d}_{Y/S}(f)$. It is mapped to the local section $(a')^\sharp (f) - (b')^\sharp (f) = D(f)$ of $\mathcal{C}_{X/X'}$. In other words, $\theta (\text{d}_{Y/S}(f)) = D(f)$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).