The Stacks project

Proof. The implications (1) $\Rightarrow $ (2) $\Rightarrow $ (3) are formal. Assume (3) a assume given an arbitrary diagram (76.15.0.1). Choose a presentation $T' = U'/R'$, see Spaces, Definition 65.9.3. 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. $\square$

Proof. This is formal. Namely, by Lemma 76.15.1 it suffices to consider solid commutative diagrams (76.15.0.1) 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 étale. Hence the fact that $Z \subset Z'$ is a universal first order thickening over $Y$ produces the desired morphism $a' : T' \to Z'$. $\square$

Proof. Proof of (1). By Lemma 76.15.1 it suffices to consider solid commutative diagrams (76.15.0.1) 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 étale implies (see Lemma 76.13.5) that $T' \to Y'$ lifts to the desired morphism $a' : T' \to Z'$.

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 étale morphism such that $W = T \times _{T'} W'$, see Theorem 76.8.1. 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. $\square$

Proof. Choose any commutative diagram

where $V$ and $U$ are schemes and the vertical arrows are étale. Note that $V \to U$ is a formally unramified morphism of schemes, see Lemma 76.14.2. Combining Lemma 76.15.1 and More on Morphisms, Lemma 37.7.1 we see that a universal first order thickening $V \subset V'$ of $V$ over $U$ exists. By Lemma 76.15.2 part (1) $V'$ is a universal first order thickening of $V$ over $X$.

Fix a scheme $U$ and a surjective étale morphism $U \to X$. The argument above shows that for any $V \to Z$ étale 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 étale (see Spaces, Lemma 65.11.6). Then $R = V \times _ Z V$ a scheme étale 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 76.15.3 as $R'$ is the universal first order thickening of $R$ over $X$ these morphisms are étale. Then $(t', s') : R' \to V'$ is an étale equivalence relation and we can set $Z' = V'/R'$. Since $V' \to Z'$ is surjective étale and $v'$ is the universal first order thickening of $V$ over $X$ we conclude from Lemma 76.15.2 part (2) that $Z'$ is a universal first order thickening of $Z$ over $X$. $\square$

Proof. An immersion of algebraic spaces is by definition a representable morphism. Hence by Morphisms, Lemmas 29.35.7 and 29.35.8 an immersion is unramified (via the abstract principle of Spaces, Lemma 65.5.8). Hence it is formally unramified by Lemma 76.14.7. The other assertions follow by combining Lemmas 76.12.2 and 76.12.3 and the definitions. $\square$

Proof. Let $T \subset T'$ be a first order thickening of affine schemes over $X$. Let

be a commutative diagram. Set $T_0 = c^{-1}(Z) \subset T$ and $T'_ a = a^{-1}(Z)$ (scheme theoretically). Since $Z'$ is a first order thickening of $Z$, we see that $T'$ is a first order thickening of $T'_ a$. Moreover, since $c = a|_ T$ we see that $T_0 = T \cap T'_ a$ (scheme theoretically). As $T'$ is a first order thickening of $T$ it follows that $T'_ a$ is a first order thickening of $T_0$. Now $a|_{T'_ a}$ and $b|_{T'_ a}$ are morphisms of $T'_ a$ into $Z'$ over $X$ which agree on $T_0$ as morphisms into $Z$. Hence by the universal property of $Z'$ we conclude that $a|_{T'_ a} = b|_{T'_ a}$. Thus $a$ and $b$ are morphism from the first order thickening $T'$ of $T'_ a$ whose restrictions to $T'_ a$ agree as morphisms into $Z$. Thus using the universal property of $Z'$ once more we conclude that $a = b$. In other words, the defining property of a formally unramified morphism holds for $Z' \to X$ as desired. $\square$

Proof. The first assertion is clear from the universal property of $W'$. The induced map on conormal sheaves is the map of Lemma 76.5.3 applied to $(Z \subset Z') \to (W \subset W')$. $\square$

Proof. The morphism $h$ is formally unramified by Lemma 76.14.5. 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 76.5.5 for why this implies that the map of conormal sheaves is surjective. $\square$

Proof. Flatness is preserved under base change, see Morphisms of Spaces, Lemma 67.30.4. Hence the first statement follows from the description of $W'$ in Lemma 76.15.9. 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 76.5.5 for why this implies that the map of conormal sheaves is an isomorphism. $\square$

Proof. The first statement is Lemma 76.14.2. The compatibility of universal first order thickenings is a consequence of Lemmas 76.15.2 and 76.15.3. $\square$

Proof. The map $c_{h'}$ is the map defined in Lemma 76.7.6. 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 étale localization, see Lemma 76.15.11 and Lemma 76.7.3. In this case the result is More on Morphisms, Lemma 37.7.9. $\square$

Proof. We know that there is a canonical exact sequence

see Lemma 76.7.10. Hence the result follows on applying Lemma 76.15.12. $\square$

Proof. Since the maps have been defined, checking the sequence is exact reduces to the case of schemes by étale localization, see Lemma 76.15.11 and Lemma 76.7.3. In this case the result is More on Morphisms, Lemma 37.7.11. $\square$

Proof. The map $h : Z' \to Y'$ in (1) comes from Lemma 76.15.8. 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 76.5.6 we have an exact sequence

where $i' : Z \to Y \times _{Y'} Z'$ is the given morphism. By Lemma 76.5.5 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. $\square$