## 76.10 Morphisms of thickenings

If $(f, f') : (X \subset X') \to (Y \subset Y')$ is a morphism of thickenings of algebraic spaces, then often properties of the morphism $f$ are inherited by $f'$. There are several variants.

Lemma 76.10.1. Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \subset Y')$ be a morphism of thickenings of algebraic spaces over $S$. Then

1. $f$ is an affine morphism if and only if $f'$ is an affine morphism,

2. $f$ is a surjective morphism if and only if $f'$ is a surjective morphism,

3. $f$ is quasi-compact if and only if $f'$ quasi-compact,

4. $f$ is universally closed if and only if $f'$ is universally closed,

5. $f$ is integral if and only if $f'$ is integral,

6. $f$ is (quasi-)separated if and only if $f'$ is (quasi-)separated,

7. $f$ is universally injective if and only if $f'$ is universally injective,

8. $f$ is universally open if and only if $f'$ is universally open,

9. $f$ is representable if and only if $f'$ is representable, and

10. add more here.

Proof. Observe that $Y \to Y'$ and $X \to X'$ are integral and universal homeomorphisms. This immediately implies parts (2), (3), (4), (7), and (8). Part (1) follows from Limits of Spaces, Proposition 70.15.2 which tells us that there is a 1-to-1 correspondence between affine schemes étale over $X$ and $X'$ and between affine schemes étale over $Y$ and $Y'$. Part (5) follows from (1) and (4) by Morphisms of Spaces, Lemma 67.45.7. Finally, note that

$X \times _ Y X = X \times _{Y'} X \to X \times _{Y'} X' \to X' \times _{Y'} X'$

is a thickening (the two arrows are thickenings by Lemma 76.9.8). Hence applying (3) and (4) to the morphism $(X \subset X') \to (X \times _ Y X \to X' \times _{Y'} X')$ we obtain (6). Finally, part (9) follows from the fact that an algebraic space thickening of a scheme is again a scheme, see Lemma 76.9.5. $\square$

Lemma 76.10.2. Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \subset Y')$ be a morphism of thickenings of algebraic spaces over $S$ such that $X = Y \times _{Y'} X'$. If $X \subset X'$ is a finite order thickening, then

1. $f$ is a closed immersion if and only if $f'$ is a closed immersion,

2. $f$ is locally of finite type if and only if $f'$ is locally of finite type,

3. $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,

4. $f$ is locally of finite type of relative dimension $d$ if and only if $f'$ is locally of finite type of relative dimension $d$,

5. $\Omega _{X/Y} = 0$ if and only if $\Omega _{X'/Y'} = 0$,

6. $f$ is unramified if and only if $f'$ is unramified,

7. $f$ is proper if and only if $f'$ is proper,

8. $f$ is a finite morphism if and only if $f'$ is an finite morphism,

9. $f$ is a monomorphism if and only if $f'$ is a monomorphism,

10. $f$ is an immersion if and only if $f'$ is an immersion, and

11. add more here.

Proof. Choose a scheme $V'$ and a surjective étale morphism $V' \to Y'$. Choose a scheme $U'$ and a surjective étale morphism $U' \to X' \times _{Y'} V'$. Set $V = Y \times _{Y'} V'$ and $U = X \times _{X'} U'$. Then for étale local properties of morphisms we can reduce to the morphism of thickenings of schemes $(U \subset U') \to (V \subset V')$ and apply More on Morphisms, Lemma 37.3.3. This proves (2), (3), (4), (5), and (6).

The properties of morphisms in (1), (7), (8), (9), (10) are stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Spaces, Lemma 65.12.3, and Morphisms of Spaces, Lemmas 67.40.3, 67.45.5, and 67.10.5.

The interesting direction in (1), (7), (8), (9), (10) is to assume that $f$ has the property and deduce that $f'$ has it too. By induction on the order of the thickening we may assume that $Y \subset Y'$ is a first order thickening, see discussion on finite order thickenings above.

Proof of (1). Choose a scheme $V'$ and a surjective étale morphism $V' \to Y'$. Set $V = Y \times _{Y'} V'$, $U' = X' \times _{Y'} V'$ and $U = X \times _ Y V$. Then $U \to V$ is a closed immersion, which implies that $U$ is a scheme, which in turn implies that $U'$ is a scheme (Lemma 76.9.5). Thus we can apply the lemma in the case of schemes (More on Morphisms, Lemma 37.3.3) to $(U \subset U') \to (V \subset V')$ to conclude.

Proof of (7). Follows by combining (2) with results of Lemma 76.10.1 and the fact that proper equals quasi-compact $+$ separated $+$ locally of finite type $+$ universally closed.

Proof of (8). Follows by combining (2) with results of Lemma 76.10.1 and using the fact that finite equals integral $+$ locally of finite type (Morphisms, Lemma 29.44.4).

Proof of (9). As $f$ is a monomorphism we have $X = X \times _ Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times _ Y X \subset X' \times _{Y'} X')$. We conclude $X' \to X' \times _{Y'} X'$ is a closed immersion by (1). In fact, it is a first order thickening as the ideal defining the closed immersion $X' \to X' \times _{Y'} X'$ is contained in the pullback of the ideal $\mathcal{I} \subset \mathcal{O}_{Y'}$ cutting out $Y$ in $Y'$. Indeed, $X = X \times _ Y X = (X' \times _{Y'} X') \times _{Y'} Y$ is contained in $X'$. The conormal sheaf of the closed immersion $\Delta : X' \to X' \times _{Y'} X'$ is equal to $\Omega _{X'/Y'}$ (this is the analogue of Morphisms, Lemma 29.32.7 for algebraic spaces and follows either by étale localization or by combining Lemmas 76.7.11 and 76.7.13; some details omitted). Thus it suffices to show that $\Omega _{X'/Y'} = 0$ which follows from (5) and the corresponding statement for $X/Y$.

Proof of (10). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open subspace and $X \to V$ is a closed immersion, see Morphisms of Spaces, Remark 67.12.4. Let $V' \subset Y'$ be the open subspace whose underlying topological space $|V'|$ is the same as $|V| \subset |Y| = |Y'|$. Then $X' \to Y'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (1). This finishes the proof. $\square$

The following lemma is a variant on the preceding one. Rather than assume that the thickenings involved are finite order (which allows us to transfer the property of being locally of finite type from $f$ to $f'$), we instead take as given that each of $f$ and $f'$ is locally of finite type.

Lemma 76.10.3. Let $S$ be a scheme. Let $(f, f') : (X \subset X') \to (Y \to Y')$ be a morphism of thickenings of algebraic spaces over $S$. Assume $f$ and $f'$ are locally of finite type and $X = Y \times _{Y'} X'$. Then

1. $f$ is locally quasi-finite if and only if $f'$ is locally quasi-finite,

2. $f$ is finite if and only if $f'$ is finite,

3. $f$ is a closed immersion if and only if $f'$ is a closed immersion,

4. $\Omega _{X/Y} = 0$ if and only if $\Omega _{X'/Y'} = 0$,

5. $f$ is unramified if and only if $f'$ is unramified,

6. $f$ is a monomorphism if and only if $f'$ is a monomorphism,

7. $f$ is an immersion if and only if $f'$ is an immersion,

8. $f$ is proper if and only if $f'$ is proper, and

9. add more here.

Proof. Choose a scheme $V'$ and a surjective étale morphism $V' \to Y'$. Choose a scheme $U'$ and a surjective étale morphism $U' \to X' \times _{Y'} V'$. Set $V = Y \times _{Y'} V'$ and $U = X \times _{X'} U'$. Then for étale local properties of morphisms we can reduce to the morphism of thickenings of schemes $(U \subset U') \to (V \subset V')$ and apply More on Morphisms, Lemma 37.3.4. This proves (1), (4), and (5).

The properties in (2), (3), (6), (7), and (8) are stable under base change, hence if $f'$ has property $\mathcal{P}$, then so does $f$. See Spaces, Lemma 65.12.3, and Morphisms of Spaces, Lemmas 67.40.3, 67.45.5, and 67.10.5. Hence in each case we need only to prove that if $f$ has the desired property, so does $f'$.

Case (2) follows from case (5) of Lemma 76.10.1 and the fact that the finite morphisms are precisely the integral morphisms that are locally of finite type (Morphisms of Spaces, Lemma 67.45.6).

Case (3). This follows immediately from Limits of Spaces, Lemma 70.15.5.

Proof of (6). As $f$ is a monomorphism we have $X = X \times _ Y X$. We may apply the results proved so far to the morphism of thickenings $(X \subset X') \to (X \times _ Y X \subset X' \times _{Y'} X')$. We conclude $\Delta _{X'/Y'} : X' \to X' \times _{Y'} X'$ is a closed immersion by (3). In fact $\Delta _{X'/Y'}$ induces a bijection $|X'| \to |X' \times _{Y'} X'|$, hence $\Delta _{X'/Y'}$ is a thickening. On the other hand $\Delta _{X'/Y'}$ is locally of finite presentation by Morphisms of Spaces, Lemma 67.28.10. In other words, $\Delta _{X'/Y'}(X')$ is cut out by a quasi-coherent sheaf of ideals $\mathcal{J} \subset \mathcal{O}_{X' \times _{Y'} X'}$ of finite type. Since $\Omega _{X'/Y'} = 0$ by (5) we see that the conormal sheaf of $X' \to X' \times _{Y'} X'$ is zero. (The conormal sheaf of the closed immersion $\Delta _{X'/Y'}$ is equal to $\Omega _{X'/Y'}$; this is the analogue of Morphisms, Lemma 29.32.7 for algebraic spaces and follows either by étale localization or by combining Lemmas 76.7.11 and 76.7.13; some details omitted.) In other words, $\mathcal{J}/\mathcal{J}^2 = 0$. This implies $\Delta _{X'/Y'}$ is an isomorphism, for example by Algebra, Lemma 10.21.5.

Proof of (7). If $f : X \to Y$ is an immersion, then it factors as $X \to V \to Y$ where $V \to Y$ is an open subspace and $X \to V$ is a closed immersion, see Morphisms of Spaces, Remark 67.12.4. Let $V' \subset Y'$ be the open subspace whose underlying topological space $|V'|$ is the same as $|V| \subset |Y| = |Y'|$. Then $X' \to Y'$ factors through $V'$ and we conclude that $X' \to V'$ is a closed immersion by part (3).

Case (8) follows from Lemma 76.10.1 and the definition of proper morphisms as being the quasi-compact, universally closed, and separated morphisms that are locally of finite type. $\square$

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