hcju/stacksqa huggingface dataset:
The implications (1) $\Rightarrow$ (2) $\Rightarrow$ (3) are formal. Assume (3) a assume given an arbitrary diagram (\ref{equation-universal-first-order-thickening}). Choose a presentation $T' = U'/R'$, see Spaces, Definition \ref{spaces-definition-presentation}. We may assume that $U' = \coprod U'_i$ is a disjoint union of affines, so $R' = U' \times_{T'} U' = \coprod_{i, j} U'_i \times_T' U'_j$. For each pair $(i, j)$ choose an affine open covering $U'_i \times_T' U'_j = \bigcup_k R'_{ijk}$. Denote $U_i, R_{ijk}$ the fibre products with $T$ over $T'$. Then each $U_i \subset U'_i$ and $R_{ijk} \subset R'_{ijk}$ is a first order thickening of affine schemes. Denote $a_i : U_i \to Z$, resp.\ $a_{ijk} : R_{ijk} \to Z$ the composition of $a : T \to Z$ with the morphism $U_i \to T$, resp.\ $R_{ijk} \to T$. By (3) applied to $a_i : U_i \to Z$ we obtain unique morphisms $a'_i : U'_i \to Z'$. By (3) applied to $a_{ijk}$ we see that the two compositions $R'_{ijk} \to R'_i \to Z'$ and $R'_{ijk} \to R'_j \to Z'$ are equal. Hence $a' = \coprod a'_i : U' = \coprod U'_i \to Z'$ descends to the quotient sheaf $T' = U'/R'$ and we win.
This is formal. Namely, by Lemma \ref{lemma-check-universal-first-order-thickening} it suffices to consider solid commutative diagrams (\ref{equation-universal-first-order-thickening}) with $T'$ an affine scheme. The composition $T \to Z \to Y$ lifts uniquely to $T' \to Y$ as $Y \to X$ is assumed formally \'etale. Hence the fact that $Z \subset Z'$ is a universal first order thickening over $Y$ produces the desired morphism $a' : T' \to Z'$.
Proof of (1). By Lemma \ref{lemma-check-universal-first-order-thickening} it suffices to consider solid commutative diagrams (\ref{equation-universal-first-order-thickening}) with $T'$ an affine scheme. The composition $T \to Z \to Y$ lifts uniquely to $T' \to Y'$ as $Y'$ is the universal first order thickening. Then the fact that $Z' \to Y'$ is \'etale implies (see Lemma \ref{lemma-representable-property-formally-property}) that $T' \to Y'$ lifts to the desired morphism $a' : T' \to Z'$. \medskip\noindent Proof of (2). Let $T \subset T'$ be a first order thickening over $X$ and let $a : T \to Y$ be a morphism. Set $W = T \times_Y Z$ and denote $c : W \to Z$ the projection Let $W' \to T'$ be the unique \'etale morphism such that $W = T \times_{T'} W'$, see Theorem \ref{theorem-topological-invariance}. Note that $W' \to T'$ is surjective as $Z \to Y$ is surjective. By assumption we obtain a unique morphism $c' : W' \to Z'$ over $X$ restricting to $c$ on $W$. By uniqueness the two restrictions of $c'$ to $W' \times_{T'} W'$ are equal (as the two restrictions of $c$ to $W \times_T W$ are equal). Hence $c'$ descends to a unique morphism $a' : T' \to Y'$ and we win.
Choose any commutative diagram $$ \xymatrix{ V \ar[d] \ar[r] & U \ar[d] \\ Z \ar[r] & X } $$ where $V$ and $U$ are schemes and the vertical arrows are \'etale. Note that $V \to U$ is a formally unramified morphism of schemes, see Lemma \ref{lemma-formally-unramified}. Combining Lemma \ref{lemma-check-universal-first-order-thickening} and More on Morphisms, Lemma \ref{more-morphisms-lemma-universal-thickening} we see that a universal first order thickening $V \subset V'$ of $V$ over $U$ exists. By Lemma \ref{lemma-universal-thickening-over-formally-etale} part (1) $V'$ is a universal first order thickening of $V$ over $X$. \medskip\noindent Fix a scheme $U$ and a surjective \'etale morphism $U \to X$. The argument above shows that for any $V \to Z$ \'etale with $V$ a scheme such that $V \to Z \to X$ factors through $U$ a universal first order thickening $V \subset V'$ of $V$ over $X$ exists (but does not depend on the chosen factorization of $V \to X$ through $U$). Now we may choose $V$ such that $V \to Z$ is surjective \'etale (see Spaces, Lemma \ref{spaces-lemma-lift-morphism-presentations}). Then $R = V \times_Z V$ a scheme \'etale over $Z$ such that $R \to X$ factors through $U$ also. Hence we obtain universal first order thickenings $V \subset V'$ and $R \subset R'$ over $X$. As $V \subset V'$ is a universal first order thickening, the two projections $s, t : R \to V$ lift to morphisms $s', t': R' \to V'$. By Lemma \ref{lemma-etale-morphism-of-universal-thickenings} as $R'$ is the universal first order thickening of $R$ over $X$ these morphisms are \'etale. Then $(t', s') : R' \to V'$ is an \'etale equivalence relation and we can set $Z' = V'/R'$. Since $V' \to Z'$ is surjective \'etale and $v'$ is the universal first order thickening of $V$ over $X$ we conclude from Lemma \ref{lemma-universal-thickening-over-formally-etale} part (2) that $Z'$ is a universal first order thickening of $Z$ over $X$.
An immersion of algebraic spaces is by definition a representable morphism. Hence by Morphisms, Lemmas \ref{morphisms-lemma-open-immersion-unramified} and \ref{morphisms-lemma-closed-immersion-unramified} an immersion is unramified (via the abstract principle of Spaces, Lemma \ref{spaces-lemma-representable-transformations-property-implication}). Hence it is formally unramified by Lemma \ref{lemma-unramified-formally-unramified}. The other assertions follow by combining Lemmas \ref{lemma-first-order-infinitesimal-neighbourhood} and \ref{lemma-infinitesimal-neighbourhood-conormal} and the definitions.
The morphism $h$ is formally unramified by Lemma \ref{lemma-base-change-formally-unramified}. It is clear that $X \times_Y W'$ is a first order thickening. It is straightforward to check that it has the universal property because $W'$ has the universal property (by mapping properties of fibre products). See Lemma \ref{lemma-conormal-functorial-flat} for why this implies that the map of conormal sheaves is surjective.
Flatness is preserved under base change, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-flat}. Hence the first statement follows from the description of $W'$ in Lemma \ref{lemma-universal-thickening-fibre-product}. It is clear that $X \times_Y W'$ is a first order thickening. It is straightforward to check that it has the universal property because $W'$ has the universal property (by mapping properties of fibre products). See Lemma \ref{lemma-conormal-functorial-flat} for why this implies that the map of conormal sheaves is an isomorphism.
The map $c_{h'}$ is the map defined in Lemma \ref{lemma-functoriality-differentials}. If $i : Z \to Z'$ is the given closed immersion, then $i^*c_{h'}$ is a map $h^*\Omega_{X/S} \to \Omega_{Z'/S} \otimes \mathcal{O}_Z$. Checking that it is an isomorphism reduces to the case of schemes by \'etale localization, see Lemma \ref{lemma-universal-thickening-localize} and Lemma \ref{lemma-localize-differentials}. In this case the result is More on Morphisms, Lemma \ref{more-morphisms-lemma-differentials-universally-unramified}.
We know that there is a canonical exact sequence $$ \mathcal{C}_{Z/Z'} \to \Omega_{Z'/S} \otimes \mathcal{O}_Z \to \Omega_{Z/S} \to 0. $$ see Lemma \ref{lemma-differentials-relative-immersion}. Hence the result follows on applying Lemma \ref{lemma-differentials-universally-unramified}.
The map $h : Z' \to Y'$ in (1) comes from Lemma \ref{lemma-universal-thickening-functorial}. The assertion that $Y \times_{Y'} Z'$ is the universal first order thickening of $Z$ over $Y$ is clear from the universal properties of $Z'$ and $Y'$. By Lemma \ref{lemma-transitivity-conormal} we have an exact sequence $$ (i')^*\mathcal{C}_{Y \times_{Y'} Z'/Z'} \to \mathcal{C}_{Z/Z'} \to \mathcal{C}_{Z/Y \times_{Y'} Z'} \to 0 $$ where $i' : Z \to Y \times_{Y'} Z'$ is the given morphism. By Lemma \ref{lemma-conormal-functorial-flat} there exists a surjection $h^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{Y \times_{Y'} Z'/Z'}$. Combined with the equalities $\mathcal{C}_{Y/Y'} = \mathcal{C}_{Y/X}$, $\mathcal{C}_{Z/Z'} = \mathcal{C}_{Z/X}$, and $\mathcal{C}_{Z/Y \times_{Y'} Z'} = \mathcal{C}_{Z/Y}$ this proves the lemma.
Assume $f$ is formally \'etale. By Lemma \ref{lemma-representable-property-formally-property} the morphisms $U \to X$ and $V \to Y$ are formally \'etale. Thus by Lemma \ref{lemma-composition-formally-smooth-etale-unramified} the composition $U \to Y$ is formally \'etale. Then it follows from Lemma \ref{lemma-formally-permanence} that $U \to V$ is formally \'etale. Thus (1) implies (2). And (2) implies (3) trivially \medskip\noindent Assume given a diagram as in (3). By Lemma \ref{lemma-representable-property-formally-property} the morphism $V \to Y$ is formally \'etale. Thus by Lemma \ref{lemma-composition-formally-smooth-etale-unramified} the composition $U \to Y$ is formally \'etale. Then it follows from Lemma \ref{lemma-etale-on-top} that $X \to Y$ is formally \'etale, i.e., (1) holds.
Let $U' \to T'$ be a surjective \'etale morphism where $U' = \coprod U'_i$ is a disjoint union of affine schemes. Let $U_i = T \times_{T'} U'_i$. Then we get morphisms $a'_i : U'_i \to X$ such that $a'_i|_{U_i}$ equals the composition $U_i \to T \to X$. By uniqueness (see Lemma \ref{lemma-formally-unramified-not-affine}) we see that $a'_i$ and $a'_j$ agree on the fibre product $U'_i \times_{T'} U'_j$. Hence $\coprod a'_i : U' \to X$ descends to give a unique morphism $a' : T' \to X$.
Actually, the last assertion only make sense because $\Omega_{X/Y} = 0$ implies that $\mathcal{C}_{X/Y}$ is defined via Lemma \ref{lemma-characterize-formally-unramified} and Definition \ref{definition-universal-thickening}. This also makes it clear that (3) and (4) are equivalent. \medskip\noindent Either of the assumptions (1), (2), and (3) imply that $f$ is formally unramified. Hence we may assume $f$ is formally unramified. The equivalence of (1), (2), and (3) follow from the universal property of the universal first order thickening $X'$ of $X$ over $S$ and the fact that $X = X' \Leftrightarrow \mathcal{C}_{X/Y} = 0$ since after all by definition $\mathcal{C}_{X/Y} = \mathcal{C}_{X/X'}$ is the ideal sheaf of $X$ in $X'$.
Follows from the case of schemes, see More on Morphisms, Lemma \ref{more-morphisms-lemma-unramified-flat-formally-etale} and \'etale localization, see Lemmas \ref{lemma-formally-unramified} and \ref{lemma-formally-etale} and Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-local}.
Follows from the case of schemes, see More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-formally-etale} and \'etale localization, see Lemma \ref{lemma-formally-etale} and Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-finite-presentation-local} and \ref{spaces-morphisms-lemma-etale-local}.
Consider the map $$ \alpha = (a')^\sharp + D : a^{-1}\mathcal{O}_{Y'} \to \mathcal{O}_{X'} $$ where $D$ is as in Equation (\ref{equation-D}). As $D$ is an $\mathcal{O}_B$-derivation it follows that $\alpha$ is a map of sheaves of $\mathcal{O}_B$-algebras. By construction we have $i_X^\sharp \circ \alpha = a^\sharp \circ i_Y^\sharp$ where $i_X : X \to X'$ and $i_Y : Y \to Y'$ are the given closed immersions. By Lemma \ref{lemma-first-order-thickening-maps} we obtain a unique morphism $(a, b') : (X  \subset X') \to (Y \subset Y')$ of thickenings over $B$ such that $\alpha = (b')^\sharp$. Setting $b = a$ we win.
Denote $\mathcal{F}$ the rule of the lemma. The restriction mapping $\mathcal{F}(U') \to \mathcal{F}(V')$ for $V' \subset U' \subset X'$ of $\mathcal{F}$ is really the restriction map $a' \mapsto a'|_{V'}$. With this definition in place it is clear that $\mathcal{F}$ is a sheaf since morphisms of algebraic spaces satisfy \'etale descent, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-fpqc-universal-effective-epimorphisms}.
The equality sign comes from the identification $f^*\Omega_{X_1/S_1} = \Omega_{X_2/S_2}$ we get as the construction of the sheaf of differentials is compatible with \'etale localization (both on source and target), see Lemma \ref{lemma-localize-differentials}. Namely, using this we have $a_2^*\Omega_{X_2/S_2} = a_2^*f^*\Omega_{X_1/S_1} = h^*a_1^*\Omega_{X_1/S_1}$ because $f \circ a_2 = a_1 \circ h$. Having said this, the commutativity of the diagram may be checked on \'etale locally. Thus we may assume $T'_i$, $X'_i$, $B_2$, and $B_1$ are schemes and in this case the lemma follows from More on Morphisms, Lemma \ref{more-morphisms-lemma-action-by-derivations-etale-localization}. Alternative proof: using Lemma \ref{lemma-first-order-thickening-maps} it suffices to show a certain diagram of sheaves of rings on $X_1'$ is commutative; then argue exactly as in the proof of the aforementioned More on Morphisms, Lemma \ref{more-morphisms-lemma-action-by-derivations-etale-localization} to see that this is indeed the case.
Choose a scheme $V'$ and a surjective \'etale morphism $V' \to Y'$. Choose a scheme $U'$ and a surjective \'etale morphism $U' \to X' \times_{Y'} V'$. Set $U = X \times_{X'} U'$ and $V = Y \times_{Y'} V'$. According to our definition of a flat morphism of algebraic spaces we see that the induced map $g : U \to V$ is a flat morphism of schemes and that $f'$ is flat if and only if the corresponding morphism $g' : U' \to V'$ is flat. Also, $X = Y \times_{Y'} X'$ if and only if $U = V \times_{V'} V'$. Finally, the map $f^*\mathcal{C}_{Y/Y'} \to \mathcal{C}_{X/X'}$ is an isomorphism if and only if $g^*\mathcal{C}_{V/V'} \to \mathcal{C}_{U/U'}$ is an isomorphism. Hence the lemma follows from its analogue for morphisms of schemes, see More on Morphisms, Lemma \ref{more-morphisms-lemma-deform}.
Choose a scheme $U'$ and a surjective \'etale morphism $U' \to B'$. Choose a scheme $V'$ and a surjective \'etale morphism $V' \to U' \times_{B'} Y'$. Choose a scheme $W'$ and a surjective \'etale morphism $W' \to V' \times_{Y'} X'$. Let $U, V, W$ be the base change of $U', V', W'$ by $B \to B'$. Then flatness of $f'$ is equivalent to flatness of $W' \to V'$ and we are given that $W \to V$ is flat. Hence we may apply the lemma in the case of schemes to the diagram $$ \xymatrix{ (W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\ & (U \subset U') } $$ of thickenings of schemes. See More on Morphisms, Lemma \ref{more-morphisms-lemma-flatness-morphism-thickenings}. The statement about flatness of $Y'/B'$ at points in the image of $f'$ follows in the same manner.
Choose a scheme $U'$ and a surjective \'etale morphism $U' \to B'$. Choose a scheme $V'$ and a surjective \'etale morphism $V' \to U' \times_{B'} Y'$. Choose a scheme $W'$ and a surjective \'etale morphism $W' \to V' \times_{Y'} X'$. Let $U, V, W$ be the base change of $U', V', W'$ by $B \to B'$. Then flatness of $f'$ is equivalent to flatness of $W' \to V'$ and we are given that $W \to V$ is flat. Hence we may apply the lemma in the case of schemes to the diagram $$ \xymatrix{ (W \subset W') \ar[rr] \ar[rd] & & (V \subset V') \ar[ld] \\ & (U \subset U') } $$ of thickenings of schemes. See More on Morphisms, Lemma \ref{more-morphisms-lemma-flatness-morphism-thickenings-fp-over-ft}. The statement about flatness of $Y'/B'$ at points in the image of $f'$ follows in the same manner.
By Lemma \ref{lemma-representable-property-formally-property} the morphisms $U \to X$ and $V \to Y$ are formally \'etale. By Lemma \ref{lemma-composition-formally-smooth-etale-unramified} the composition $U \to Y$ is formally smooth. By Lemma \ref{lemma-formally-permanence} we see $\psi : U \to V$ is formally smooth.
Choose an \'etale covering $\{Y_i \to Y\}$ with each $Y_i$ affine. After replacing $T'$ by the induced \'etale covering we may assume $Y$ is affine. \medskip\noindent Assume $Y$ is affine. Choose an \'etale covering $\{X_i \to X\}$. This gives rise to an \'etale covering of $T$. This \'etale covering of $T$ comes from an \'etale covering of $T'$ (by Theorem \ref{theorem-topological-invariance}, see discussion in Section \ref{section-thickenings}). Hence we may assume $X$ is affine. \medskip\noindent Assume $X$ and $Y$ are affine. We can do one more \'etale covering of $T'$ and assume $T'$ is affine. In this case the lemma follows from Algebra, Lemma \ref{algebra-lemma-smooth-strong-lift}.
Choose a diagram $$ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^f & Y } $$ where $U$ and $V$ are affine(!) schemes and the vertical arrows are \'etale. By Lemma \ref{lemma-helper-formally-smooth} we see $\psi : U \to V$ is formally smooth. Hence $\Gamma(V, \mathcal{O}_V) \to \Gamma(U, \mathcal{O}_U)$ is a formally smooth ring map, see More on Morphisms, Lemma \ref{more-morphisms-lemma-affine-formally-smooth}. Hence by Algebra, Lemma \ref{algebra-lemma-characterize-formally-smooth-again} the $\Gamma(U, \mathcal{O}_U)$-module $\Omega_{\Gamma(U, \mathcal{O}_U)/\Gamma(V, \mathcal{O}_V)}$ is projective. Hence $\Omega_{U/V}$ is locally projective, see Properties, Section \ref{properties-section-locally-projective}. Since $\Omega_{X/Y}|_U = \Omega_{U/V}$ we see that $\Omega_{X/Y}$ is locally projective too. (Because we can find an \'etale covering of $X$ by the affine $U$'s fitting into diagrams as above -- details omitted.)
By the definition of a locally projective sheaf on an algebraic space (see Properties of Spaces, Definition \ref{spaces-properties-definition-locally-projective}) we see that $\mathcal{F}_{Zar} = \mathcal{F}|_{T_{Zar}}$ is a locally projective sheaf on the scheme $T$. Thus $\mathcal{F}_{Zar}$ is a direct summand of a free $\mathcal{O}_{T_{Zar}}$-module. Whereupon we conclude (as $\mathcal{F} = (\mathcal{F}_{Zar})^a$, see Descent, Proposition \ref{descent-proposition-equivalence-quasi-coherent}) that $\mathcal{F}$ is a direct summand of a free $\mathcal{O}_T$-module on $T_\etale$. Hence we may assume that $\mathcal{F} = \bigoplus_{i \in I} \mathcal{O}_T$ is a free module. In this case $\mathcal{H} = \prod_{i \in I} \mathcal{G}$ is a product of quasi-coherent modules. By Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-cohomology-products} we conclude that $H^1 = 0$ because the cohomology of a quasi-coherent sheaf on an affine scheme is zero, see Descent, Proposition \ref{descent-proposition-same-cohomology-quasi-coherent} and Cohomology of Schemes, Lemma \ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}.
We have seen that (1) implies (2) and (3) in Lemma \ref{lemma-helper-formally-smooth}. Assume (3). The proof that $f$ is formally smooth is entirely similar to the proof of (1) $\Rightarrow$ (2) of Lemma \ref{lemma-smooth-formally-smooth}. \medskip\noindent Consider a solid commutative diagram $$ \xymatrix{ X \ar[d]_f & T \ar[d]^i \ar[l]^a \\ Y & T' \ar[l] \ar@{-->}[lu] } $$ as in Definition \ref{definition-formally-smooth}. We will show the dotted arrow exists thereby proving that $f$ is formally smooth. Let $\mathcal{F}$ be the sheaf of sets on $(T')_{spaces, \etale}$ of Lemma \ref{lemma-sheaf} as in the special case discussed in Remark \ref{remark-special-case}. Let $$ \mathcal{H} = \SheafHom_{\mathcal{O}_T}(a^*\Omega_{X/Y}, \mathcal{C}_{T/T'}) $$ be the sheaf of $\mathcal{O}_T$-modules on $T_{spaces, \etale}$ with action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ as in Lemma \ref{lemma-action-sheaf}. The action $\mathcal{H} \times \mathcal{F} \to \mathcal{F}$ turns $\mathcal{F}$ into a pseudo $\mathcal{H}$-torsor, see Cohomology on Sites, Definition \ref{sites-cohomology-definition-torsor}. Our goal is to show that $\mathcal{F}$ is a trivial $\mathcal{H}$-torsor. There are two steps: (I) To show that $\mathcal{F}$ is a torsor we have to show that $\mathcal{F}$ has \'etale locally a section. (II) To show that $\mathcal{F}$ is the trivial torsor it suffices to show that $H^1(T_\etale, \mathcal{H}) = 0$, see Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-torsors-h1}. \medskip\noindent First we prove (I). To see this consider a diagram (which exists because we are assuming (3)) $$ \xymatrix{ U \ar[d] \ar[r]_\psi & V \ar[d] \\ X \ar[r]^f & Y } $$ where $U$ and $V$ are schemes, the vertical arrows are \'etale and surjective, and $\psi$ is formally smooth. By Lemma \ref{lemma-representable-property-formally-property} the morphism $V \to Y$ is formally \'etale. Thus by Lemma \ref{lemma-composition-formally-smooth-etale-unramified} the composition $U \to Y$ is formally smooth. Then (I) follows from Lemma \ref{lemma-etale-on-top} part (4). \medskip\noindent Finally we prove (II). By Lemma \ref{lemma-formally-smooth-sheaf-differentials} we see that $\Omega_{U/V}$ locally projective. Hence $\Omega_{X/Y}$ is locally projective, see Descent on Spaces, Lemma \ref{spaces-descent-lemma-locally-projective-descends}. Hence $a^*\Omega_{X/Y}$ is locally projective, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-locally-projective-pullback}. Hence $$ H^1(T_\etale, \mathcal{H}) = H^1(T_\etale, \SheafHom_{\mathcal{O}_T}(a^*\Omega_{X/Y}, \mathcal{C}_{T/T'}) = 0 $$ by Lemma \ref{lemma-h1-is-zero} as desired.
Let $f : X \to Y$ be a morphism of algebraic spaces over a scheme $S$. Choose an index set $I$ and diagrams $$ \xymatrix{ U_i \ar[d] \ar[r]_{\psi_i} & V_i \ar[d] \\ X \ar[r]^f & Y } $$ with \'etale vertical arrows and $U_i$, $V_i$ affine schemes. Moreover, assume that $\coprod U_i \to X$ and $\coprod V_i \to Y$ are surjective, see Properties of Spaces, Lemma \ref{spaces-properties-lemma-cover-by-union-affines}. By Lemma \ref{lemma-formally-smooth} we see that $f$ is formally smooth if and only if each of the morphisms $\psi_i$ are formally smooth. Hence we reduce to the case of a morphism of affine schemes. In this case the result follows from Algebra, Lemma \ref{algebra-lemma-descent-formally-smooth}. Some details omitted.
Let $Z \to Z'$ be the universal first order thickening of $Z$ over $X$. From the proof of Lemma \ref{lemma-universally-unramified-differentials-sequence} we see that our sequence is identified with the sequence $$ \mathcal{C}_{Z/Z'} \to \Omega_{Z'/B} \otimes \mathcal{O}_Z \to \Omega_{Z/B} \to 0. $$ Since $Z \to S$ is formally smooth we can \'etale locally on $Z'$ find a left inverse $Z' \to Z$ over $B$ to the inclusion map $Z \to Z'$. Thus the sequence is \'etale locally split, see Lemma \ref{lemma-differentials-relative-immersion-section}.
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 \ref{lemma-universally-unramified-differentials-sequence} here is a canonical morphism $Z' \to Z''$ so that we have a commutative diagram $$ \xymatrix{ Z \ar[r]_{i'} \ar[rd]_{j'} & Z' \ar[r]_a \ar[d]^k & X \ar[d]^f \\ & Z'' \ar[r]^b & Y } $$ The sequence above is identified with the sequence $$ \mathcal{C}_{Z/Z''} \to \mathcal{C}_{Z/Z'} \to (i')^*\Omega_{Z'/Z''} \to 0 $$ via our definitions concerning conormal sheaves of formally unramified morphisms. Let $U'' \to Z''$ be an \'etale morphism with $U''$ affine. Denote $U \to Z$ and $U' \to Z'$ the corresponding affine schemes \'etale over $Z$ and $Z'$. 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 $U$.