# Second fundamental exact sequence of sheaves of differentials. Sufficient condition to get a splitting s.e.s

$$\def\spec{\operatorname{Spec}}$$I am trying to understand the proof of Lemma 0474 of the Stacks Project. I'll give some context to its statement before discussing its proof: In commutative algebra, if $$k,A$$ are rings and $$A\to B$$ is a surjective ring homomorphism with kernel $$I\subset A$$, then one has an exact sequence of $$B$$-modules $$\label{2nd_fund}\tag{1} I/I^2\to\Omega_{A/k}\otimes_AB\to\Omega_{B/k}\to 0.$$ Moreover, the morphism $$I/I^2\to\Omega_{A/k}\otimes_AB$$ has a $$B$$-linear retraction (hence the sequence becomes a splitting s.e.s.) if and only if the $$k$$-algebra map $$A/I^2\to B$$ has a section that is a $$k$$-algebra homomorphism. All of this can be read in Matsumura's Commutative Algebra, Ch. 10, (26.I), Theorem 58 (Matsumura calls \eqref{2nd_fund} the “second fundamental exact sequence”). A sufficient condition for $$A/I^2\to B$$ to have a section that is a $$k$$-algebra homomorphism is that $$A\to B$$ has a section that is a $$k$$-algebra homomorphism (this is the sufficient condition stated in Lemma 02HP).

If $$i:Z\to X$$ is now an immersion of schemes (or more generally, of ringed spaces; by definition a topological immersion onto a locally closed subset and such that $$f^{-1}\mathcal{O}_Y\to\mathcal{O}_X$$ is onto) over $$S$$, then one has the exact sequence $$\label{eq}\tag{2} \mathcal{C}_{Z/X}\to i^*\Omega_{X/S}\to\Omega_{Z/S}\to 0.$$ See the Stacks Project, Lemma 01UZ. Here, $$\mathcal{C}_{Z/X}$$ is the conormal sheaf of $$Z$$ in $$X$$.

Now, Lemma 0474 gives a sufficient condition for \eqref{eq} to become a splitting s.e.s., namely, that $$i$$ has a retraction over $$S$$. However, I cannot understand the proof, which is “it follows from Algebra, Lemma 02HP.” To prove that \eqref{eq} becomes a splitting s.e.s., I assume that one has to somehow construct an $$\mathcal{O}_Z$$-linear retraction of $$\mathcal{C}_{Z/X}\to i^*\Omega_{X/S}$$. But how can this be done? I understand how one can construct a local retraction by means of Algebra, Lemma 02HP. Namely, if we use:

1. that the sheaf of relative differentials is compatible with restrictions, Lemma 01US,

2. $$\Omega_{\spec A/\spec k}\cong\widetilde{\Omega_{A/k}}$$, for any rings $$A,k$$ (Lemma 01UT), and

3. Lemma 01I9, point (1),

then in \eqref{eq} we can restrict the schemes $$Z,X,S$$ to affines and \eqref{eq} becomes \eqref{2nd_fund} tildified over $$Z$$.

Nevertheless:

I don't know how one can construct a global retraction using solely the algebraic result.

• One way to argue is that Tag 02HP tells you that the right exact sequence is actually exact on the left as well, so then splitting it can also be done on the right (see Tag 010G). This is what is written in the proof of Tag 0474. Does this answer your question? Jul 20, 2023 at 14:38
• @R.vanDobbendeBruyn Okay, yes. Such a long question I wrote for such a short answer haha. I guess I missed the implicit syllogism because of the relatively vague phrasing of proof of Lemma 0474. Thanks! ^^ Jul 20, 2023 at 15:12

Lemma 1. Consider Lemma 08TF. If $$\mathcal{O}_2\to\mathcal{O}_2'$$ has a section that is a homomorphism of $$\mathcal{O}_1$$-algebras, then the map $$\mathcal{I}/\mathcal{I}^2\to \Omega_{\mathcal{O}_2/\mathcal{O}_1}\otimes_{\mathcal{O}_2}\mathcal{O}_2'$$ has a $$\mathcal{O}_2'$$-linear retraction.
Then, one would proceed in the following way: Assume $$i:Z\to X$$ is an immersion of ringed spaces. By shrinking $$X$$ if necessary, we can assume $$i$$ is a closed immersion. Let $$\mathcal{J}\subset\mathcal{O}_X$$ be the ideal sheaf associated with $$Z$$. On Lemma 08TF, set $$\mathcal{O}_2\to\mathcal{O}_2'$$ to be $$i^{-1}(\mathcal{O}_X\to\mathcal{O}_X/\mathcal{J})$$, $$\mathcal{I}=i^{-1}\mathcal{J}$$ and $$\mathcal{O}_1=f^{-1}\mathcal{O}_S$$, where $$f:Z\to S$$ is the structure morphism, so that we get an exact sequence $$i^{-1}(\mathcal{J}/\mathcal{J}^2)\to\Omega_{i^{-1}\mathcal{O}_X/f^{-1}\mathcal{O}_S}\otimes_{i^{-1}\mathcal{O}_X}\mathcal{O}_Z\to\Omega_{Z/S}\to 0.$$ The mid term equals $$i^*\Omega_{X/S}$$ (use Lemma 08RR), whereas the left term is $$\mathcal{C}_{Z/X}$$. Now suppose $$i$$ has a retraction over $$S$$. This implies that $$i^{-1}(\mathcal{O}_X\to\mathcal{O}_X/\mathcal{J})$$ has a $$f^{-1}\mathcal{O}_S$$-linear section. Hence, we can apply Lemma 1.