$\xymatrix{ Z \ar[r]_ i \ar[rd]_ j & X \ar[d]^ f \\ & Y }$

be a commutative diagram of schemes where $i$ and $j$ are formally unramified and $f$ is formally smooth. Then the canonical exact sequence

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

of Lemma 37.7.11 is exact and locally split.

Proof. Denote $Z \to Z'$ the universal first order thickening of $Z$ over $X$. Denote $Z \to Z''$ the universal first order thickening of $Z$ over $Y$. By Lemma 37.7.10 here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram

$\xymatrix{ Z \ar[r]_{i'} \ar[rd]_{j'} & Z' \ar[r]_ a \ar[d]^ k & X \ar[d]^ f \\ & Z'' \ar[r]^ b & Y }$

In the proof of Lemma 37.7.11 we identified the sequence above with the sequence

$\mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'} \to (i')^*\Omega _{Z'/Z''} \to 0$

Let $U'' \subset Z''$ be an affine open. Denote $U \subset Z$ and $U' \subset Z'$ the corresponding affine open subschemes. As $f$ is formally smooth there exists a morphism $h : U'' \to X$ which agrees with $i$ on $U$ and such that $f \circ h$ equals $b|_{U''}$. Since $Z'$ is the universal first order thickening we obtain a unique morphism $g : U'' \to Z'$ such that $g = a \circ h$. The universal property of $Z''$ implies that $k \circ g$ is the inclusion map $U'' \to Z''$. Hence $g$ is a left inverse to $k$. Picture

$\xymatrix{ U \ar[d] \ar[r] & Z' \ar[d]^ k \\ U'' \ar[r] \ar[ru]^ g & Z'' }$

Thus $g$ induces a map $\mathcal{C}_{Z/Z'}|_ U \to \mathcal{C}_{Z/Z''}|_ U$ which is a left inverse to the map $\mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'}$ over $U$. $\square$

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