Lemma 29.32.16. Let $i : Z \to X$ be an immersion of schemes over $S$, and assume $i$ (locally) has a left inverse. Then the canonical sequence

$0 \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/S} \to \Omega _{Z/S} \to 0$

of Lemma 29.32.15 is (locally) split exact. In particular, if $s : S \to X$ is a section of the structure morphism $X \to S$ then the map $\mathcal{C}_{S/X} \to s^*\Omega _{X/S}$ induced by $\text{d}_{X/S}$ is an isomorphism.

Proof. Follows from Algebra, Lemma 10.131.10. Clarification: if $g : X \to Z$ is a left inverse of $i$, then $i^*c_ g$ is a right inverse of the map $i^*\Omega _{X/S} \to \Omega _{Z/S}$. Also, if $s$ is a section, then it is an immersion $s : Z = S \to X$ over $S$ (see Schemes, Lemma 26.21.11) and in that case $\Omega _{Z/S} = 0$. $\square$

Comment #8581 by on

After "if $g : X \to Z$ is a left inverse of $i$, then $i^*c_g$ is a right inverse of the map $i^*\Omega_{X/S} \to \Omega_{Z/S}$," one could say "by Modules, Lemma 17.28.13."

Comment #8584 by on

I don't know if this might be only me, but I got confused by the current phrasing of the proof to see how one obtains a globally splitting s.e.s. from a global retraction of $i$ over $S$. After posting my confusion here, R. van Dobben de Bruyn explained in a comment what I was missing.

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